Abstract In the year 1994, Gupta (Approx Theory Appl (N.S.) 10(3):74–78, 1994)introduced the integral modification of well known Baskakov operators with weightsof Beta basis functions and obtained better approximation over the usual BaskakovDurrmeyer operators. The rate of convergence for Bézier variant of these operatorsfor functions of bounded variations were discussed in Gupta (Int J Math Math Sci32(8):471–479, 2002). The present paper is the extension of the previous work, herewe consider the Bézier variant of Baskakov-Beta-Stancu operators. We estimate therate of convergence of these operators for the bounded functions. In the end of thepaper we suggest an open problem.
To approximate Lebesgue integrable functions on the interval [0,∞), Gupta [3] intro-duced the integral modification of well known Baskakov operators by taking the weightfunctions of Beta basis functions. It was observed in [3] that by taking the weightsof Beta basis functions, we can have better approximation than the usual BaskakovDurrmeyer operators [6]. Recently, Buyukyazici [1] considered the Stancu type gen-eralization of Chlodowsky polynomials. Motivated by his work we now consider the
R. Yadav (B)School of Applied Sciences, Netaji Subhas Institute of Technology,Sector 3 Dwarka, New Delhi 110078, Indiae-mail: [email protected]
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218 Ann Univ Ferrara (2012) 58:217–227
Stancu variant of Baskakov-Beta operators. For α and β be two given real parameterssatisfying the conditions 0 ≤ α ≤ β and for f ∈ L1[0,∞), the Baskakov-Beta-Stancuoperators are defined as
Mn,α,β( f, x) =∞∑
k=0
pn,k(x)
∞∫
0
bn,k(t) f
(nt + α
n + β
)dt, (1.1)
where n ∈ N and the Baskakov and Beta basis functions are respectively defined as
pn,k (x) =(
n + k − 1k
)xk
(1 + x)n+k, bn,k (t) = 1
B(k + 1, n)
tk
(1 + t)n+k+1
and B(m, n) = (m − 1)!(n − 1)!(n + m − 1)! .
The rate of convergence for certain Durrmeyer type operators and their Béziervariants is one of the important areas of research in the recent years. Recently in [2]the authors have estimated the rate of convergence for function having derivatives ofbounded variation for Szasz-Mirakyan operators. Also it is obvious that Bézier curveplay an important role in computer aided geometric design, Zeng and collaboratorshave done commendable work in this direction and they estimated rate of convergencefor bounded/bounded variation functions (see [7–9]). Here we consider the Bézier vari-ant of the operators (1.1). For θ ≥ 1 and x ∈ [0,∞), the Baskakov-Beta-Stancu-Bézieroperators are defined as
Mθn,α,β( f, x) =
∞∑
k=0
Qθn,k(x)
∞∫
0
bn,k(t) f
(nt + α
n + β
)dt, (1.2)
where Qθn,k(x) = J θ
n,k(x) − J θn,k+1(x), Jn,k(x) = ∑∞
j=k pn, j (x).
For α = β = θ = 0, the operator Mθn,α,β( f, x) reduces to the operator discussed
in [3].In the present article, we extend the studies to estimate the rate of convergence for
the Bézier variant of the bounded functions.
2 Auxiliary results
In this section we present some basic results which will be used to prove the maintheorem.
Lemma 1 Let the m-th order moment μα,βn,m(x), m ∈ N ∪ {0} be defined as
μα,βn,m(x) = Mn((t − x)m, x) =
∞∑
k=0
pn,k(x)
∞∫
0
bn,k(t)
(nt + α
n + β− x
)m
dt,
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Ann Univ Ferrara (2012) 58:217–227 219
and there holds a recurrence relation
μα,βn,m+1(x)
[(n + β
n
)(n − m − 1)
]
= x(1 + x)[(μα,β
n,m)′(x) + mμα,βn,m−1(x)
]
+μα,βn,m(x)
[(m + 1 + nx) +
(−α + (n + β)x
n
)(2m − n + 1)
]
+μα,βn,m−1(x)
[(−α + (n + β)x
n
)2
m − m
(α
n + β− x
)]. (2.1)
Consequently for each x ∈ [0,∞)
μα,βn,m(x) = O(n−[(m+1)/2]). (2.2)
Proof Using the identities,
x(1 + x)p′n,k(x) = (k − nx)pn,k(x)
and
t (1 + t)b′n,k(t) = [k − (n + 1)t] bn,k(t)
we have
x(1 + x)(μα,βn,m)′(x) =
∞∑
k=0
x(1 + x)p′n,k(x)
∞∫
0
bn,k(t)
(nt + α
n + β− x
)m
dt
−mx(1 + x)
∞∑
k=0
pn,k(x)
∞∫
0
bn,k(t)
(nt + α
n + β− x
)m−1
dt
=∞∑
k=0
pn,k(x)
∞∫
0
(k − nx)bn,k(t)
(nt + α
n + β− x
)m
dt − mx(1 + x)μα,βn,m−1(x).
Thus
x(1 + x)[(μα,β
n,m)′(x) + mμα,βn,m−1(x)
]
=∞∑
k=0
pn,k(x)
∞∫
0
[{k − (n + 1)t} + (n + 1)t − nx] bn,k(t)
(nt + α
n + β− x
)m
dt
=∞∑
k=0
pn,k(x)
∞∫
0
t (1 + t)b′n,k(t)
(nt + α
n + β− x
)m
dt
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220 Ann Univ Ferrara (2012) 58:217–227
+(n + 1)
∞∑
k=0
pn,k(x)
∞∫
0
bn,k(t)t
(nt + α
n + β− x
)m
dt
−nx∞∑
k=0
pn,k(x)
∞∫
0
bn,k(t)
(nt + α
n + β− x
)m
dt.
Writing,
t =(
n + β
n
) (nt + α
n + β− x
)− α
n+ n + β
nx,
we have
x(1 + x)[(μα,β
n,m)′(x) + mμα,βn,m−1(x)
]
=∞∑
k=0
pn,k(x)
∞∫
0
n + β
nb′
n,k(t)
(nt + α
n + β− x
)m+1
dt
+∞∑
k=0
pn,k(x)
∞∫
0
(−α + (n + β)x
n
)b′
n,k(t)
(nt + α
n + β− x
)m
dt
+∞∑
k=0
pn,k(x)
∞∫
0
(n + β
n
)2
b′n,k(t)
(nt + α
n + β− x
)m+2
dt
+∞∑
k=0
pn,k(x)
∞∫
0
(−α + (n + β)x
n
)2
b′n,k(t)
(nt + α
n + β− x
)m
dt
+2∞∑
k=0
pn,k(x)
∞∫
0
(n + β
n
) (−α + (n + β)x
n
)b′
n,k(t)
(nt + α
n + β− x
)m+1
dt
+(n + 1)
∞∑
k=0
pn,k(x)
∞∫
0
(n + β
n
)bn,k(t)
(nt + α
n + β− x
)m+1
dt
+(n + 1)
∞∑
k=0
pn,k(x)
∞∫
0
(−α + (n + β)x
n
)bn,k(t)
(nt + α
n + β− x
)m
dt
−nxμα,βn,m(x).
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Ann Univ Ferrara (2012) 58:217–227 221
Integrating by parts, we have
x(1 + x)[(μα,β
n,m)′(x) + mμα,βn,m−1(x)
]
= −(m + 1)μα,βn,m(x) + m
(α
n + β− x
)μ
α,βn,m−1(x)
−(m + 2)n + β
nμ
α,βn,m+1(x) − m
(−α + (n + β)x)2
n(n + β)μ
α,βn,m−1(x)
−2(m + 1)(−α + (n + β)x)
nμα,β
n,m(x) − nxμα,βn,m(x)
which on solving gives (2.1) and (2.2) follows from (2.1).
Remark 1 By the recurrence relation (2.1), it can easily be verified that:
μα,βn,0 (x) = 1
μα,βn,1 (x) = (1 − α + αn) + x(1 + β − βn)
(n + β)(n − 1)(2.3)
μα,βn,2 (x) = n
(n + β)(n − 2)x(1 + x) + n2
(n + β)2(n − 1)(n − 2)
(1 + β
n− β
)
+ n2
(n + β)2(n − 1)(n − 2)
[(1 − α
n+ α
)+ x
(1 + β
n− β
)]
×[(α − βx)
(1 − 3
n
)+ 5
]+ n
(n + β)(n − 2).
×[
x − α
n + β+ α2 + (n + β)2x2 − 2α(n + β)x
n2
](2.4)
Remark 2 Given any n ∈ N , x ∈ (0,∞), λ > 2 then for n ≥ N (λ, x), by Lemma 1,we have
μα,βn,2 (x) < λ
x(1 + x)
n − 2. (2.5)
Lemma 2 For x ∈ (0,∞) and Kn,θ (x, t) =∞∑
k=0
Qθn,k(x)bn,k(t). Then for λ > 2 and
n ≥ N (λ, x), we have
γn,θ (x, y) =y∫
0
Kn,θ (x, t)dt ≤ λθx(1 + x)
(n − 2)(x − y)2 , 0 ≤ y < x,
1 − γn,θ (x, z) =∞∫
z
Kn,θ (x, t)dt ≤ λθx(1 + x)
(n − 2)(z − y)2 , x ≤ z < ∞, (2.6)
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222 Ann Univ Ferrara (2012) 58:217–227
where
1∫
0
Kn,θ (x, u)du = 1, and γn,θ (x, t) =t∫
0
Kn,θ (x, u)du.
Lemma 3 From Cauchy-Schwarz inequality and under the conditions of Remark 2,we have
Mθn,α,β(|t − x |, x) ≤
[Mθ
n,α,β((t − x)2, x)]1/2 ≤
√θλx(1 + x)
n − 2. (2.7)
Lemma 4 [4] For all x ∈ (0,∞), we have
Qθn,k(x) ≤ θpn,k(x) ≤ θ
√1 + x√2enx
,
where the constant 1/√
2e is the best possible.
Lemma 5 [4] For x ∈ (0,∞), we have
∞∫
x
bn,k(x) =k∑
j=0
pn, j (x).
3 Main theorem
We consider the class of functions �loc,γ and the quantity ωx ( f, λ), �loc,γ = { f :f is bounded in every finite sub interval of [0,∞), and f (t) = O(tγ ) for some γ >
0, t → ∞.}
ωx ( f, λ) = supt∈[x−λ,x+λ]
| f (t) − f (x)|.
It follows that
(i) If f is of bounded variation on [a, b] and∨b
a denotes the total variation of on[a, b], then
ωx ( f, λ) ≤x+λ∨
x−λ
( f ).
(ii) ωx ( f, λ) is monotone non-decreasing with respect to λ.
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Ann Univ Ferrara (2012) 58:217–227 223
Theorem 1 Let f ∈ �loc,γ and x ∈ [0,∞), and suppose θ ≥ 1, then for a sufficientlylarge n, we have
∣∣∣∣Mθn,α,β( f, x) −
[f (x+) + θ f (x−)
θ + 1
]∣∣∣∣ ≤[
4λθ(1 + x)
(n − 2)x+ 1
n
] n∑
k=1
ωx (gx , x/√
k)
+θ√
2(1 + x)√enx
| f (x+) − f (x−)|
+θ(2x)γ
x2mO(n−[m+1]/2),
where
fx (t) =⎧⎨
⎩
f (t) − f (x−), 0 ≤ t < x0, t = xf (t) − f (x+), x < t < ∞
, (3.1)
Proof It can be seen easily that
∣∣∣∣Mθn,α,β( f, x) −
[1
θ + 1f (x+) + θ
θ + 1f (x−)
]∣∣∣∣
≤∣∣∣Mθ
n,α,β(gx , x)
∣∣∣ + 1
2| f (x+) − f (x−)| .
∣∣∣∣Mθn,α,β(sign(t − x), x) + α − 1
α + 1
∣∣∣∣ .
(3.2)
First we estimate
Mθn,α,β(sign(t − x), x) =
∞∫
x
Kn,θ (x, t)dt +x∫
0
Kn,θ (x, t)dt
= 2
∞∫
x
Kn,θ (x, t)dt − 1, (3.3)
as∫ ∞
0Kn,θ (x, t)dt = 1.
Using Lemma 5, we get
Mθn,α,β(sign(t − x), x) = −1 + 2
∞∑
k=0
Qθn,k(x)
∞∫
x
bn,k(t)dt
= −1 + 2∞∑
k=0
Qθn,k(x)
k∑
j=0
pn, j (x)
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224 Ann Univ Ferrara (2012) 58:217–227
= −1 + 2k∑
j=0
pn, j (x)
∞∑
k= j
Qθn,k(x)
= −1 + 2k∑
j=0
pn, j (x)J θn, j (x).k
Thus,
∣∣∣∣Mθn,α,β(sign(t − x), x) + α − 1
α + 1
∣∣∣∣ = 2k∑
j=0
pn, j (x)J θn, j (x) − 2
α + 1
∞∑
j=0
Qθ+1n, j (x)
Since∞∑
k=0
Qθn,k(x) = 1 and using Lemma 4 ,we have
∣∣∣∣Mθn,α,β(sign(t − x), x) + α − 1
α + 1
∣∣∣∣ ≤ 2θ
√1 + x√2enx
∞∑
j=0
pn, j (x) = θ√
2(1 + x)√enx
.
(3.4)
Next, we have
Mθn,α,β(gx , x) =
∞∫
0
gx (t)Kn,θ (x, t)dt = 1,n + 2,n + 3,n + 4,n (3.5)
where
1,n =x−x/
√n∫
0
gx (t)dt Kn,θ (x, t), 2,n =x+x/
√n∫
x−x/√
n
gx (t)dt Kn,θ (x, t),
3,n =2x∫
x+x/√
n
gx (t)dt Kn,θ (x, t), 4,n =∞∫
2x
gx (t)dt Kn,θ (x, t),
We shall evaluate 1,n, 2,n, 3,n and 4,n with the quantity ωx (gx , λ).
First, for 2,n , note that gx (x) = 0, we have
|2,n| ≤x+x/
√n∫
x−x/√
n
|gx (t) − gx (x)| dt Kn,θ (x, t)
≤ ωx (gx , x/√
n) ≤ 1
n
n∑
k=1
ωx (gx , x/√
k). (3.6)
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Ann Univ Ferrara (2012) 58:217–227 225
For 1,n , note that ωx (gx , λ) is monotone non-decreasing with respect to λ,
|1,n| =
∣∣∣∣∣∣∣
x−x/√
n∫
0
gx (t)dt Kn,θ (x, t)
∣∣∣∣∣∣∣
≤x−x/
√n∫
0
ωx (gx , x − t)dt Kn,θ (x, t).
Using partial integration with y = x − x/√
n, we have
x−x/√
n∫
0
ωx (gx , x − t)dt Kn,θ (x, t)
≤ ωx (gx , x − y)K̂n,θ (x, y) +y∫
0
K̂n,θ (x, t)dt (−ωx (gx , x − t)),
where K̂n,θ (x, t) is normalized form of Kn,θ (x, t), and K̂n,θ (x, t) ≤ Kn,θ (x, t) on(0,∞).Therefore
|1,n| ≤ λθx(1+x)
(n−2)(x−y)2 ωx (gx , x−y)+λθx(1+x)
(n−2)
y∫
0
1
(x−t)2 dt (−ωx (gx , x−t)).
Since
y∫
0
1
(x − t)2 dt (−ωx (gx , x − t))
= − 1
(x − t)2 (−ωx (gx , x − t))|y0 +
y∫
0
(−ωx (gx , x − t))2
(x − t)3 dt
= −ωx (gx , x − y)
(x − y)2 + ωx (gx , x)
x2 +y∫
0
(−ωx (gx , x − t))2
(x − t)3 dt.
So we have
|1,n| ≤ λθx(1 + x)
(n − 2)x2 ωx (gx , x) + λθx(1 + x)
(n − 2)
x−x/√
n∫
0
ωx (gx , x − t)2
(x − t)3 dt.
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226 Ann Univ Ferrara (2012) 58:217–227
Putting t = x − x/√
u for the last integral, we get
x−x/√
n∫
0
ωx (gx , x − t)2
(x − t)3 dt = 1
x2
n∫
0
ωx (gx , x/√
u)du
≤ 1
x2
n∑
k=1
ωx (gx , x/√
k).
Thus
|1,n| ≤ λθ(1 + x)
(n − 2)x
(ωx (gx , x) +
n∑
k=1
ωx (gx , x/√
k)
)
≤ 2λθ(1 + x)
(n − 2)x
n∑
k=1
ωx (gx , x/√
k). (3.7)
Using similar estimates for 3,n , we get
|3,n| ≤ 2λθ(1 + x)
(n − 2)x
n∑
k=1
ωx (gx , x/√
k). (3.8)
Finally, by assumption gx (t) = O(tγ ) for γ > 0 as t → ∞, taking m ≥ γ /2 and
t ≥ 2x,tγ
(t − x)2mis monotone decreasing for variable t , and Qθ
n,k(x) ≤ θpn,k(x),
hence by Lemma 2, we get
|4,n| ≤∣∣∣∣∣∣
∞∫
2x
gx (t)dt Kn,θ (x, t)
∣∣∣∣∣∣
=∞∑
k=0
Qθn,k(x)
∞∫
2x
pn,k(t)(t)γ dt
= θ(2x)γ
x2m
∞∑
k=0
pn,k(x)
∞∫
2x
(t − x)2mbn,k(t)dt
= θ(2x)γ
x2mμα,β
n,m(x)
= θ(2x)γ
x2mO(n−[m+1]/2), (3.9)
Using (3.4), (3.6), (3.7), (3.8) and (3.9), in (3.5) the required result follows.
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Ann Univ Ferrara (2012) 58:217–227 227
4 Open problem
Very recently Gupta and Aral [5] introduced the q-analogue of original Baskakov-Betaoperators [3]. We can now propose the q-analogue of operator (1.1) as
Mqn,α,β( f, x) =
∞∑
k=0
pqn,k(x)
∞/A∫
0
bqn,k(t) f
( [n]q t + α
[n]q + β
),
where
pqn,k(x)=
[n + k − 1
k
]
qq
k22
(qx)k
(1 + qx)n + kq
, bqn,k(t)=
1
Bq(n, k + 1)q
k22
tk
(1 + t)n + k + 1q
and the q improper integral is defined as
∞/A∫
0
f (x)dq x = (1 − q)
∞∑
n=−∞f
(qn
A
)qn
A, A > 0.
For further details, we refer the readers to [5].It is observed here that the analogous result to our Theorem 1 is not possible for
the q-variant of Baskakov-Beta-Stancu operators as proposed above. There are sometechnical difficulties. At this moment it can be considered as an open problem.
Acknowledgments Author is thankful to Prof. Vijay Gupta for his valuable suggestions in preparing themanuscript. Thanks are also due to the reviewers for their valuable comments leading to overall improve-ments in the paper.
6. Sinha, R.P., Agrawal, P.N., Gupta, V.: On simultaneous approximation by modified Baskakov operator.Bull. Soc. Math Belg. Ser. B43(2), 217–231 (1991)
7. Walczak, Z.: Baskakov type operator. Rocky Mountain J. Math. 39(3), 981–993 (2009)8. Zeng, X.M.: On the rate of convergence of the generalized Szász type operators for functions of bounded
variation. J. Math. Anal. Appl. 226, 309–325 (1998)9. Zeng, X.M., Tao, W.: Rate of convergence of integral type Lupas Bézier operators. Kyungpook Math.
