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Bivariate Bernstein type operators
Applied Mathematics and Computation 273 (2016) 543–552
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Bivariate Bernstein type operators a b ˙ ˙ Gülen Bascanbaz-Tunca ¸ , Hatice Gül InceIlarslan , Aysegül ¸ Erençin c,∗ a b c
Ankara University, Faculty of Science, Department of Mathematics, Tando˘gan, Ankara 06100, Turkey Gazi University, Faculty of Arts and Science, Department of Mathematics, Teknikokullar 06500, Ankara, Turkey ˙ Abant Izzet Baysal University, Faculty of Arts and Science, Department of Mathematics, Bolu 14280, Turkey
a r t i c l e
i n f o
a b s t r a c t
MSC: 41A25 41A36
In this paper, we introduce bivariate extension of Bernstein type operators defined in [11]. We show that these operators preserve some properties of the original function f, such as Lipschitz constant and monotonicity. Furthermore, we present the monotonicity of the sequence of bivariate Bernstein type operators for n when f is τ -convex.
Keywords: Lipschitz continuous function Monotony Bernstein type operator
© 2015 Elsevier Inc. All rights reserved.
1. Introduction In [11], Cárdenas-Morales et al. presented the following Bernstein type operators for f ∈ C[0, 1] and x ∈ [0, 1],
Bn ( f ; τ (x))
n n = k
(τ (x)) (1 − τ (x)) k
n−k
k (f ◦ τ ) −1
n
k=0
= B∗n ( f ◦ τ −1 ) ◦ τ
n n = k
(τ (x)) (1 − τ (x)) k
k=0
n−k
k f τ −1 , n
(1.1)
where B∗n is the classical Bernstein operators, n ∈ N and τ is a function defined on [0, 1] and having the properties: (τ 1 ) τ is ∞-times continuously differentiable on [0, 1] (τ 2 ) τ (0) = 0, τ (1) = 1 and τ (x) > 0 on [0, 1]. These conditions ensure that τ is strictly increasing and the inverse τ −1 of τ exists on [0, 1]. The authors discussed shape preserving and convergence properties and also introduced comparative results. Note that if τ (x) = x, then the operators given by (1.1) reduce to classical Bernstein operators. Before the construction of bivariate Bernstein type operators, we recall some usual notations and definitions which are essential for our work. For x = (x1 , x2 ) ∈ R2 , k = (k1 , k2 ) ∈ N20 and n ∈ N, we will write
|x| := x1 + x2 , xk := xk11 xk22 , |k| := k1 + k2 , k! := k1 !k2 ! ∗
Corresponding author. fax.: +903742541000. ˙ ˙ E-mail addresses: [email protected] (G. Bascanbaz-Tunca), ¸ [email protected] (H.G. InceIlarslan), [email protected] (A. Erençin).
http://dx.doi.org/10.1016/j.amc.2015.10.037 0096-3003/© 2015 Elsevier Inc. All rights reserved.
544
and
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
n k
=
n! n , := k1 , k2 k!(n − |k|)!
k := i
k1 i1
k2 , i2
k
:=
k1 k2
.
i1 =0 i2 =0
i=0
Now let S := {x = (x1 , x2 ) ∈ R2 : 0 ≤ x1 , x2 ≤ 1, |τ(x)| = τ (x1 ) + τ (x2 ) ≤ 1}. In this paper, for a function f defined on S, we consider the bivariate extension of the operators defined by (1.1) as follows:
Bn ( f ; τ(x)) =
n n−k 1 k1 =0 k2 =0
n k k n−|k| −1 (f ◦ τ ) , (τ(x)) (1 − |τ(x)|) n
k
where τ (x) := (τ (x1 ), τ (x2 )) such that τ (x1 ) and τ (x2 ) have the properties given by (τ 1 ) and (τ 2 ). Moreover, τ −1 (x) := (τ −1 (x1 ), τ −1 (x2 )) and ( f ◦ τ −1 )( kn ) = ( f ◦ τ −1 )( kn1 , kn2 ) := f (τ −1 ( kn1 ), τ −1 ( kn2 )). We observe that these operators are not the tensor product of the operators given by (1.1) obtained by a natural way. If τ(x) = x, then one obtains the bivariate operators defined on triangle (see [1], p. 109). Definition 1. Let f be a real valued continuous function defined on D ⊂ R2 and also let τ be a function satisfying the conditions (τ 1 ) and (τ 2 ). We say that f is a τ -Lipschitz continuous function of order μ on D, if
| f (x) − f (y)| ≤ A
2
|τ (xi ) − τ (yi )|μ
i=1
for x, y ∈ D with A > 0 and 0 < μ ≤ 1. We denote the set of τ -Lipschitz continuous functions by LipτA (μ, D). When τ (x) = x, LipτA (μ, D) reduces to the Lipschitz class defined in [10]. Definition 2. A real valued continuous function f(x) is said to be convex on the convex set D ⊂ R2 , if
f
r
αi xi ≤
i=1
r
αi f (xi )
i=1
for any x1 , x2 , . . . , xr in D and for any non-negative numbers α1 , α2 , . . . , αr such that α1 + α2 + · · · + αr = 1 (see [13]). Definition 3. We call f is a convex with respect to τ or τ -convex in D ⊂ R2 , if f ◦ τ −1 is convex in the sense of Definition 2. Note that this definition is an analogue of the τ -convexity given in [11]. 2. Main results In [10], Cao et al. introduced multivariate Baskakov operators and showed that these operators preserve some properties of the original function f such as monotonicity, semi-additivity and Lipschitz constant. Furthermore, they studied monotonicity of the sequence of multivariate Baskakov operators for n when the attached function f is convex. Finally, the authors computed the rate of approximation with the help of the K-functional and modulus of smoothness. In this part, inspired by the work [10], we firstly show that the Lipschitz constant and monotonicity properties of function f can be retained by the operators Bn ( f ; τ(x)) := Bτn f . After that, we study monotonicity of the sequence of bivariate operators Bτn f for n when f ◦ τ −1 is convex. For some preservation and approximation properties of univariate or multivariate Bernstein and some other classical positive linear operators we refer the papers [2–10,12–21] and references therein. Theorem 4. If f ∈ LipτA (μ, S), then Bτn f ∈ LipτA (μ, S) for all n ∈ N. Proof. Let x, y ∈ S and x ≤ y which means that x1 ≤ y1 and x2 ≤ y2 . Using the definition of the operators Bτn f and Binomial formula, we get
Bn ( f ; τ(y)) =
n n−k 1 k1 =0 k2 =0
n k (τ(y))k (1 − |τ(y)|)n−|k| ( f ◦ τ −1 ) k
n
n n−k 1 n k = ((τ(y) − τ(x)) + τ(x))k (1 − |τ(y)|)n−|k| ( f ◦ τ −1 ) k1 =0 k2 =0
n
k
k n n−k 1 n k k i k−i n−|k| −1 = (f ◦ τ ) . (τ(x)) (τ(y) − τ(x)) (1 − |τ(y)|) k1 =0 k2 =0 i=0
k
n
i
If we change the order of the above summations, then we can write
Bn ( f ; τ(y)) =
n n−k n 1 n−k 1 i1 =0 k1 =i1 i2 =0 k2 =i2
n k k (τ(x))i (τ(y) − τ(x))k−i (1 − |τ(y)|)n−|k| ( f ◦ τ −1 ) . k
i
n
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
Now letting k − i = m, one gets
Bn ( f ; τ(y)) =
|i|−m1 n−i 1 −m1 n−
n n−i 1 i1 =0 m1 =0
×(τ(y) − τ(x))
i1 =0 m1 =0
(1 − |τ(y)|)
i+m (f ◦ τ ) −1
n
n! (τ(x))i i!m!(n − |i| − |m|)!
m2 =0
i2 =0
(τ(x))i
i+m i
n−|i|−|m|
|i|−m1 n−i 1 −m1 n−
n n−i 1
n i+m
m2 =0
i2 =0
m
=
×(τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| ( f ◦ τ −1 ) Similarly,
Bn ( f ; τ(x)) =
n n−i 1
n i
i1 =0 i2 =0
545
i+m . n
(2.1)
i (τ(x))i (1 − |τ(x)|)n−|i| ( f ◦ τ −1 ) n
n−i1 n−|i| n−|i|−m1
=
n
i1 =0 i2 =0 m1 =0
m2 =0
n − |i| n − |i| − m1
n i
m1
m2
×(τ(x))i (τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| ( f ◦ τ −1 ) Change of the order of the second and third summations gives
Bn ( f ; τ(x)) =
n n−i 1
|i|−m1 n−i 1 −m1 n−
i1 =0 m1 =0
i2 =0
m2 =0
×(τ(x)) (τ(y) − τ(x)) i
=
n n−i 1 i1 =0 m1 =0
m
|i|−m1 n−i 1 −m1 n− i2 =0
m2 =0
n − |i| m1
n i
n − |i| − m1 m2
n−|i|−|m|
(1 − |τ(y)|)
i . n
i (f ◦ τ ) −1
n
n! (τ(x))i i!m!(n − |i| − |m|)!
×(τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| ( f ◦ τ −1 )
i . n
(2.2)
Thus, from (2.1) and (2.2), it follows that
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤
n n−i 1
|i|−m1 n−i 1 −m1 n−
i1 =0 m1 =0
m2 =0
i2 =0
× (τ(y) − τ(x))
m
n! (τ(x))i i!m!(n − |i| − |m|)! n−|i|−|m|
(1 − |τ(y)|)
( f ◦ τ −1 ) i + m − ( f ◦ τ −1 ) i . n
n
Since f ∈ LipτA (μ, S), we have
−1 i1 −1 i2 ( f ◦ τ −1 ) i + m − ( f ◦ τ −1 ) i = f τ −1 i1 + m1 , τ −1 i2 + m2 −f τ ,τ n
n
≤A
m μ 1
n
n
+
m μ
n
n
2
n
and so
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤ A
n n−i 1
|i|−m1 n−i 1 −m1 n−
i1 =0 m1 =0
m2 =0
i2 =0
×(τ(y) − τ(x))
m
Now, changing order of the summations, one has
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))|
n! (τ(x))i i!m!(n − |i| − |m|)!
(1 − |τ(y)|)n−|i|−|m|
m μ 1
n
+
m μ 2
n
.
n
546
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
≤A
n n−m 1 −|m| 1 −m1 n−i 1 n−i m1 =0 i1 =0
m2 =0
i2 =0
n! (τ(x))i i!m!(n − |i| − |m|)!
×(τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| =A
|m| n−i n n−m 1 −|m| 1 n− m1 =0 m2 =0 i1 =0
i2 =0
m μ 1
n
n n−m 1
2
n
m μ 1
n
m μ m μ n 1 2 + (τ(y) − τ(x))m n
m
m1 =0 m2 =0
n−|m| n−i1 −|m|
×
m μ
n! (n − |m|)! m!(n − |m|)! i!(n − |i| − |m|)!
×(τ(x))i (τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| =A
+
i1 =0
i2 =0
n − |m| i
+
m μ 2
n
n
(τ(x))i (1 − |τ(y)|)n−|i|−|m| .
By direct computations, we obtain n n−m 1
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤ A
m μ m μ n 1 2 + (τ(y) − τ(x))m (1 − |τ(y) − τ(x)|)n−|m|
m1 =0 m2 =0
m
n
n
= ABn ((τ (t1 ))μ + (τ (t2 ))μ ; τ(y) − τ(x))
= A{Bn ((τ (t1 ))μ ; τ (y1 ) − τ (x1 )) + Bn ((τ (t2 ))μ ; τ (y2 ) − τ (x2 ))}
=A
n m1 =0
+
n m2 =0
n m1
m μ (τ (y1 ) − τ (x1 ))m1 [1 − (τ (y1 ) − τ (x1 ))]n−m1 1 n
m μ n . (τ (y2 ) − τ (x2 ))m2 [1 − (τ (y2 ) − τ (x2 ))]n−m2 2
m2
1 Application of the Hölder inequality with p = μ and q =
n
1 1−μ
yields
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤ A{[Bn (τ (t1 ); τ (y1 ) − τ (x1 ))]μ [Bn (1; τ (y1 ) − τ (x1 ))]1−μ + [Bn (τ (t2 ); τ (y2 ) − τ (x2 ))]μ [Bn (1; τ (y2 ) − τ (x2 ))]1−μ }. From [11], we have Bn (τ (t ); τ (x)) = τ (x) and Bn (1; τ (x)) = 1. Thus, one can write
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤ A{[τ (y1 ) − τ (x1 )]μ + [τ (y2 ) − τ (x2 )]μ }, which gives Bτn ∈ LipτA (μ, S). The proof of the case x1 ≥ y1 , x2 ≥ y2 or x1 ≥ y1 , x2 ≤ y2 or x1 ≤ y1 x2 ≥ y2 is similar. Hence, we have proved the required result. Theorem 5. Suppose that f is defined on S and f(x) ≥ 0. If τf ((xx)) is non-increasing with respect to xi on (0, 1], then i non-increasing with respect to xi on (0, 1] for each n ∈ N and i = 1, 2. Proof. We prove for i = 1. Using the definition of Bn (f; τ (x)), one gets n n−k 1 Bn ( f ; τ(x)) = τ (x1 )
k1 =0 k2 =0
n k
k (τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| ( f ◦ τ −1 ) n
n n−k 1 n k k (τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| f τ −1 1 , τ −1 2 = k1 =1 k2 =0
k
n
n
n n k + . (τ (x1 ))−1 (τ (x2 ))k2 (1 − |τ(x)|)n−k2 f τ −1 (0), τ −1 2 k2 =0
n
k2
From the hypotheses on τ we have τ −1 (0) = 0. So we can write n n−k 1 Bn ( f ; τ(x)) = τ (x1 )
k1 =1 k2 =0
n k k (τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| f τ −1 1 , τ −1 2 k
n
n
Bn ( f ;τ(x)) τ (xi )
is also
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
+
n
n k2
k2 =0
and
547
k (τ (x1 ))−1 (τ (x2 ))k2 (1 − |τ(x)|)n−k2 f 0, τ −1 2 n
n n−k 1 n ∂ Bn ( f ; τ(x)) k1 −2 k2 n−|k| −1 k1 −1 k2 = τ (x1 ) ×f τ ,τ (k1 − 1)(τ (x1 )) (τ (x2 )) (1 − |τ(x)|) ∂ x1 n n k τ (x1 ) k1 =2 k2 =0 n−1 n−k 1 −1 n − τ (x1 ) (n − |k|)(τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k|−1
τ −1
×f
k1 n
n
− τ (x1 )
k2 =0 n
− τ (x1 ) = τ (x1 )
, τ −1 n k2
k (τ (x1 ))−2 (τ (x2 ))k2 (1 − |τ(x)|)n−k2 f 0, τ −1 2
−1
− τ (x1 )
k1 + 1 , τ −1 n
k1 =1 k2 =0
τ
k2 n
n−1 nk1 (τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k|−1 k1 + 1 k
×f
(n − k2 )(τ (x1 ))−1 (τ (x2 ))k2 (1 − |τ(x)|)n−k2 −1 f 0, τ −1
k2 n
n−1 n−k 1 −1 n−1 n(τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k|−1 k
− τ (x1 )
n
n k2
n−1 n−k 1 −1
τ −1
k2 n
k1 =1 k2 =0
k2 =0
×f
k
k1 =1 k2 =0
k1 n
n k2 =0
× [(τ (x1 ))
−1
,τ
−1
n k2
k2 n
(τ (x1 ))−1 (τ (x2 ))k2 (1 − |τ(x)|)n−k2
+ (n − k2 )(1 − |τ(x)|)
−1
] f 0, τ
−1
k2 n
k (τ(x))k (1 − |τ(x)|)n−|k|−1 1 τ ( x1 ) k1 =1 k2 =0 n n k1 + 1 k2 k1 k2 f τ −1 × , τ −1 − f τ −1 , τ −1 k1 + 1 n n k1 n n n τ (x1 ) n − (τ (x2 ))k2 (1 − |τ(x)|)n−k2 τ (x1 ) k =0 k2 2 k2 × [(τ (x1 ))−1 + (n − k2 )(1 − |τ(x)|)−1 ] f 0, τ −1 . = τ (x1 )
n−1 n−k 1 −1 n−1 k
n
Since τ is increasing, f(x) ≥ 0 and τf((xx)) is non-increasing with respect to x1 on (0, 1], we can write ∂∂x ( Bn (τf(;xτ()x)) ) ≤ 0. This leads 1 1 1 to the desired result. We can prove the similar result for i = 2. Therefore, the proof is completed. Theorem 6. Let f be a τ -convex function defined on S. Then Bn (f; τ (x)) is monotonically non-increasing in n. Proof. From the definition of Bn (f; τ (x)), we have
Bn ( f ; τ(x)) =
n n−k 1 k1 =0 k2 =0
=
n n−k 1 k1 =0 k2 =0
n k k n−|k| −1 (f ◦ τ ) (τ(x)) (1 − |τ(x)|) k
n
n k (|τ(x)| + 1 − |τ(x)|)(τ(x))k (1 − |τ(x)|)n−|k| ( f ◦ τ −1 ) k
n
548
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552 n n−k 1
=
k1 =0 k2 =0
n k (τ (x1 ))k1 +1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| ( f ◦ τ −1 ) k
n
n n−k 1 n k k1 k2 +1 n−|k| −1 (τ (x1 )) (τ (x2 )) (1 − |τ(x)|) + (f ◦ τ ) k
k1 =0 k2 =0
+
n k . (τ(x))k (1 − |τ(x)|)n−|k|+1 ( f ◦ τ −1 )
n n−k 1
n n−k 1
S1 :=
n
k
k1 =0 k2 =0
Now let,
n
n k (τ (x1 ))k1 +1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| ( f ◦ τ −1 ) n
k
k1 =0 k2 =0
n n−k 1 n k k1 k2 +1 n−|k| −1 S2 := (f ◦ τ ) (τ (x1 )) (τ (x2 )) (1 − |τ(x)|) k
k1 =0 k2 =0
and
S3 :=
n n−k 1
n k . (τ(x))k (1 − |τ(x)|)n−|k|+1 ( f ◦ τ −1 )
S1 =
n−1 n−k 1
n
k
k1 =0 k2 =0
For S1 , we can write
n
n k
k1 =0 k2 =0
k k + (τ (x1 ))n+1 f (τ −1 (1), τ −1 (0)) (τ (x1 ))k1 +1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| f τ −1 1 , τ −1 2 n
n
n−1 n−k 1 n k1 +1 k2 n−|k| −1 k1 −1 k2 = f τ ,τ (τ (x1 )) (τ (x2 )) (1 − |τ(x)|) k
k1 =0 k2 =1
n
n−1
+
n k1
k1 =0
=
n−2 n−k 1 −1
n k
n−1
k1 =0
+
n−1
n k1
k1 =0
=
n−1 n−k 1
k1 =1 k2 =1
+
n k1 =1
+
n
k1 =1
k1 n
n
−1 , τ (0) + (τ (x1 ))n+1 f (τ −1 (1), τ −1 (0))
k k (τ (x1 ))k1 +1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| f τ −1 1 , τ −1 2 n
(τ (x1 ))k1 +1 (τ (x2 ))n−k1 f τ −1
n k1
(τ (x1 ))k1 +1 (1 − |τ(x)|)n−k1 f τ −1
k1 =0 k2 =1
+
k1 n
(τ (x1 ))
k1 +1
n k1 − 1, k2
(1 − |τ(x)|)
n−k1
f
, τ −1
τ
−1
k1 n
n − k1 n
n
, τ (0) + (τ (x1 ))n+1 f (τ −1 (1), τ −1 (0)) −1
k −1 k2 , τ −1 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 1 n
n
k1 n−k1 +1 −1 k1 − 1 −1 n − k1 + 1 f τ ,τ (τ (x1 )) (τ (x2 ))
k1 n−k1 +1 −1 k1 − 1 −1 f τ , τ (0) + (τ (x1 ))n+1 f (τ −1 (1), τ −1 (0)). (τ (x1 )) (1 − |τ(x)|)
n k1 − 1 n k1 − 1
n
n
n
Since τ −1 (0) = 0 and τ −1 (1) = 1,
S1 =
n−1 n−k 1
k1 =1 k2 =1
+
n k1 =1
+
n k1 =1
n k1 − 1, k2
k −1 k2 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 1 , τ −1 n
n
k −1 n − k1 + 1 , τ −1 (τ (x1 ))k1 (τ (x2 ))n−k1 +1 f τ −1 1
k −1 (τ (x1 ))k1 (1 − |τ(x)|)n−k1 +1 f τ −1 1 , 0 + (τ (x1 ))n+1 f (1, 0).
n k1 − 1 n k1 − 1
n
n
n
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
549
Applying the similar steps for S2 and S3 , we reach to
S2 =
n−1 n−k 1
n k1 , k2 − 1
k1 =1 k2 =1
k k −1 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 1 , τ −1 2 n
n
n n k1 n−k1 +1 −1 k1 −1 n − k1 + f τ ,τ (τ (x1 )) (τ (x2 )) k1 =1
+
n
k2 =1
k1
n
n
n k2 n−k2 +1 −1 k2 − 1 (τ (x2 )) (1 − |τ(x)|) f 0, τ + (τ (x2 ))n+1 f (0, 1)
k2 − 1
n
and
S3 =
n−1 n−k 1
n k1 , k2
k1 =1 k2 =1
k n−|k|+1 −1 k1 −1 k2 f τ ,τ (τ(x)) (1 − |τ(x)|) n
n
n n k1 k1 n−k1 +1 + f τ −1 ,0 (τ (x1 )) (1 − |τ(x)|) k1 =1
k1
n
n n k + + (1 − |τ(x)|)n+1 f (0, 0). (τ (x2 ))k2 (1 − |τ(x)|)n−k2 +1 f 0, τ −1 2 k2 =1
n
k2
Therefore, we have
Bn ( f ; τ(x)) =
n−1 n−k 1
k1 =1 k2 =1
+
n k1 =1
+
n k1 =1
n k1 − 1, k2
n
n
n k −1 n − k1 + 1 , τ −1 (τ (x1 ))k1 (τ (x2 ))n−k1 +1 f τ −1 1
k1 − 1
n
n
n k −1 ,0 (τ (x1 ))k1 (1 − |τ(x)|)n−k1 +1 f τ −1 1
k1 − 1
n
n−1 n−k1
+
k −1 k2 , τ −1 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 1
n k1 , k2 − 1
k1 =1 k2 =1
k n−|k|+1 −1 k1 −1 k2 − 1 f τ ,τ (τ(x)) (1 − |τ(x)|) n
n
n n k1 n−k1 +1 −1 k1 −1 n − k1 (τ (x1 )) (τ (x2 )) + f τ ,τ k1 =1
+
n k2 =1
k1
n
k2 − 1
n−1 n−k1
+
+
k1 =1
n
n k1 , k2
k1 =1 k2 =1 n
n
n k2 n−k2 +1 −1 k2 − 1 f 0, τ (τ (x2 )) (1 − |τ(x)|)
n−|k|+1
(τ(x)) (1 − |τ(x)|) k
f
τ
−1
k1 n
,τ
−1
k2 n
n k1 k1 n−k1 +1 f τ −1 ,0 (τ (x1 )) (1 − |τ(x)|) k1
n
n n k2 n−k2 +1 −1 k2 + f 0, τ (τ (x2 )) (1 − |τ(x)|) k2 =1
n
k2
+ (τ (x1 ))n+1 f (1, 0) + (τ (x2 ))n+1 f (0, 1) + (1 − |τ(x)|)n+1 f (0, 0). Similarly,
Bn+1 ( f ; τ(x)) =
k1 =0 k2 =0
=
n+1 n+1−k 1
n−1 n−k 1 k1 =1 k2 =1
n+1 k
n+1 k1 , k2
k k n−|k|+1 −1 (f ◦ τ ) (τ(x)) (1 − |τ(x)|) n+1
k1 k2 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 , τ −1 n+1
n+1
550
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
+
n k1 =1
+
n k1 =1
+
n k2 =1
n+1 k1
k1 n − k1 + 1 , τ −1 (τ (x1 ))k1 (τ (x2 ))n−k1 +1 f τ −1 n+1
n
n+1 k1 ,0 (τ (x1 ))k1 (1 − |τ(x)|)n−k1 +1 f τ −1
k1
n+1 k2
n+1
(τ (x2 )) (1 − |τ(x)|)
n−k2 +1
k2
f 0, τ
−1
k2 n+1
+(τ (x1 ))n+1 f (1, 0) + (τ (x2 ))n+1 f (0, 1) + (1 − |τ(x)|)n+1 f (0, 0). Thus, we can write
Bn ( f ; τ(x)) − Bn+1 ( f ; τ(x)) =
n−1 n−k 1
n k1 − 1, k2
k1 =1 k2 =1
f
n + f k1 , k2 − 1
k1 − 1 , τ −1 n
k1 , τ −1 n
τ
−1
τ −1
k1 − 1
k1 =1
k
k2 n
k2 − 1 n
n k1 , k2
+
−1 k1 −1 k2 f τ ,τ n
n+1 k1 k2 −1 −1 − f τ ,τ (τ(x))k (1 − |τ(x)|)n−|k|+1 n+1 n+1 k1 , k2 n n k1 − 1 n − k1 + 1 + f τ −1 , τ −1 n k1
n
n−k
n
1 τ −1 1 , τ −1 n n n+1 k1 n − k1 + 1 − f τ −1 , τ −1 τ (x1 )k1 τ (x2 )n−k1 +1 n+1 n+1 k1 n n k1 − 1 n k1 + f τ −1 ,0 + f τ −1 ,0
+
f
k1 − 1
k1 =1
k2 =1
n
n
k1
n+1 k1 − f τ −1 , 0 τ (x1 )k1 (1 − |τ(x)|)n−k1 +1 n+1 k1 n n k2 − 1 n k2 + f 0, τ −1 + f 0, τ −1 − =
k2 − 1
n+1 f 0, τ −1 k2
n
n+1
τ (x1 )τ (x2 )(1 − |τ(x)|)
k2
k2 τ (x2 )k2 (1 − |τ(x)|)n−k2 +1 n−2 n−2−k 1
k1 =0 k2 =0
n k1 , k2 + 1
n
k1 k2 + 1 −1 , (f ◦ τ ) n
n
n k + 1 k2 , + ( f ◦ τ −1 ) 1 k1 + 1, k2 n n n k1 + 1 k2 + 1 −1 , + (f ◦ τ ) n n k1 + 1, k2 + 1 n+1 k1 + 1 k2 + 1 −1 − , (f ◦ τ ) (τ(x))k (1 − |τ(x)|)n−|k|−2
k1 + 1, k2 + 1
+ τ (x1 )τ (x2 )
n−1 k1 =0
n+1
n+1
n k n − k1 ( f ◦ τ −1 ) 1 , k1
n
n
n k1 + 1 n − k1 − 1 −1 , + (f ◦ τ ) k1 + 1 n n n+1 k + 1 n − k1 , ( f ◦ τ −1 ) 1 τ (x1 )k1 τ (x2 )n−k1 −1 −
k1 + 1
n+1
n+1
n
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552 n−1
+ τ (x1 )(1 − |τ(x)|)
n k n k +1 ,0 ( f ◦ τ −1 ) 1 , 0 + ( f ◦ τ −1 ) 1 k1 + 1
n
k1
k1 =0
551
n+1 k +1 , 0 τ (x1 )k1 (1 − |τ(x)|)n−k1 −1 ( f ◦ τ −1 ) 1 − k1 + 1
n+1
n−1
+ τ (x2 )(1 − |τ(x)|)
k2 k2 + 1 n n −1 −1 + ( f ◦ τ ) 0, ( f ◦ τ ) 0, n
k2
k2 =0
k2 + 1
k +1 n+1 − ( f ◦ τ −1 ) 0, 2 τ (x2 )k2 (1 − |τ(x)|)n−k2 −1 .
k2 + 1
Now set
n
n+1
n
(2.3)
k k +1 n k + 1 k2 , + ( f ◦ τ −1 ) 1 , 2 ( f ◦ τ −1 ) 1 k1 + 1, k2 n n n n n k + 1 k2 + 1 n+1 k + 1 k2 + 1 , , ( f ◦ τ −1 ) 1 ( f ◦ τ −1 ) 1 + − , k1 + 1, k2 + 1 k1 + 1, k2 + 1 n n n+1 n+1 n k1 n − k1 n k1 + 1 n − k1 − 1 −1 −1 I2 : = , , + (f ◦ τ ) (f ◦ τ ) k1 k1 + 1 n n n n n+1 k + 1 n − k1 , − , ( f ◦ τ −1 ) 1 n+1 n+1 k1 + 1 n k1 n k1 + 1 n+1 k1 + 1 −1 −1 −1 I3 : = ,0 + ,0 − ,0 (f ◦ τ ) (f ◦ τ ) (f ◦ τ )
n k1 , k2 + 1
I1 : =
and
k1 + 1
n
k1
k1 + 1
n
n+1
k2 k2 + 1 k2 + 1 n n n+1 −1 −1 −1 I4 : = + − . ( f ◦ τ ) 0, ( f ◦ τ ) 0, ( f ◦ τ ) 0, k2 + 1
n
k2
n
k2 + 1
n+1
We firstly consider I1 . Let
(k1 ,kn2 +1)
α1 =
=
k1 + 1
≥ 0,
n+1 n+1 (k1 +1,k ) 2 +1 n ( +1,k2 ) k2 + 1 ≥ 0, α2 = k1n+1 = n+1 (k1 +1,k2 +1) ( n 2 +1) n − |k| − 1 ≥0 α3 = k1 +1,k = n+1 n+1 (k1 +1,k ) 2 +1
and
x1 =
k1 k2 + 1 , , n n
x2 =
k1 + 1 k2 , n n
,
x3 =
Then it is easily seen that α1 + α2 + α3 = 1 and α1 x1 + α2 x2 + α3 x3 = ( readily follows that I1 ≥ 0. For I2 , we set
α1 =
(kn1 )
( ( α2 = ( and
x1 =
=
k1 + 1 k2 + 1 , . n n k1 +1 k2 +1 n+1 , n+1 ).
Thus, from the definition of τ -convexity it
k1 + 1 ≥ 0, n+1
) ) n − k1 = ≥0 n+1 )
n+1 k1 +1 n k1 +1 n+1 k1 +1
k1 n − k1 , n n
,
x2 =
k1 + 1 n − k1 − 1 , . n n
Thus we have α1 + α2 = 1 and α1 x1 + α2 x2 = (
k1 +1 n−k1 n+1 , n+1 ).
Since f is a τ -convex function, one gets I2 ≥ 0.
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G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
For I3 , if we take
α1 =
(kn1 )
( ( α2 = ( and
x1 =
=
k1 + 1 ≥ 0, n+1
) ) n − k1 ≥0 = n+1 )
n+1 k1 +1 n k1 +1 n+1 k1 +1
k1 ,0 , n
x2 =
k1 + 1 ,0 , n k +1
1 then we have α1 + α2 = 1 and α1 x1 + α2 x2 = ( n+1 , 0). These lead to I3 ≥ 0. Similarly, we have I4 ≥ 0. Therefore, from (2.3), we reach to the desired result Bn ( f ; τ(x)) ≥ Bn+1 ( f ; τ(x)) for all n ∈ N.
We note that for τ(x) = x from Theorem 6 we get: if f is a convex function, then for every x ∈ S := {x = (x1 , x2 ) ∈ R2 : 0 ≤ x1 , x2 ≤ 1, x1 + x2 ≤ 1} one has Bn ( f ; x) ≥ Bn+1 ( f ; x). This result was proved in [13]. Remark 1. The all definitions and results presented in this paper can be extended to the case of n-variate with n ≥ 3. Acknowledgment The authors thank the referee for valuable comments and suggestions which improved the quality of the paper. References [1] F. Altomare, M.C. Montano, V. Leonessa, I. Rasa, ¸ Markov Operators, Positive Semigroups and Approximation Processes, 61, De Gruyter Studies in Mathematics, De Gruyter, Berlin, 2014. [2] G. Bascanbaz-Tunca, ¸ Y. Tuncer, Some properties of multivariate beta operator, Fasc. Math. 41 (2009) 31–43. [3] G. Bascanbaz-Tunca, ¸ Y. Tuncer, On a Chlodovsky variant of a multivariate beta operator, J. Comput. Appl. Math. 235 (16) (2011) 4816–4824. [4] G. Bascanbaz-Tunca, ¸ F. Tasdelen, ¸ On Chlodovsky form of the Meyer–König and Zeller operators, Ann. Univ. Vest Timis. ¸ Ser. Mat.-Inform., XLIX 49 (2) (2011) 137–144. [5] M. Birou, Some blending surfaces on rhombus, Automat. Comput. Appl. Math. 19 (1) (2010) 47–55. [6] M. Birou, Blending surfaces on ellipse generated using the Bernstein operators, Stud. Univ. Babes-Bolyai ¸ Math. 56 (2) (2011a) 237–246. [7] M. Birou, Bezier blending surfaces on astroid, Stud. Univ. Babes-Bolyai ¸ Math. 56 (3) (2011b) 127–134. [8] M. Birou, Blending surfaces generated using the Bernstein operator, Creat. Math. Inform. 21 (1) (2012) 35–40. [9] B.M. Brown, D. Elliott, D.F. Paget, Lipschitz constants for the Bernstein polynomials of a Lipschitz continuous function, J. Approx. Theory 49 (2) (1987) 196–199. [10] F. Cao, C. Ding, Z. Xu, On multivariate Baskakov operators, J. Math. Anal. Appl. 307 (1) (2005) 274–291. [11] D. Cárdenas-Morales, P. Garrancho, I. Rasa, ¸ Bernstein-type operators which preserve polynomials, Comput. Math. Appl. 62 (1) (2011a) 158–163. [12] D. Cárdenas-Morales, A. Fraguela, P. Garrancho, Saturation in multivariate simultaneous approximation, Math. Comput. Simul. 81 (10) (2011b) 2098–2102. [13] G.Z. Chang, P.J. Davis, The convexity of Bernstein polynomials over triangles, J. Approx. Theory 40 (1) (1984) 11–28. [14] A. Erençin, G. Bascanbaz-Tunca, ¸ F. Tasdelen, ¸ Some preservation properties of MKZ-Stancu type operators, Sarejevo J. Math. 22 (1) (2011a) 93–102. [15] A. Erençin, G. Bascanbaz-Tunca, ¸ F. Tasdelen, ¸ Some properties of the operators defined by Lupas, ¸ Rev. Anal. Numér. Théor. Approx. 43 (2) (2011b) 168–174. [16] M.D. Farcas, ¸ About Bernstein polynomials, Ann. Univ. Craiova Ser. Math. Inform. 35 (2008) 117–121. [17] M.K. Khan, Approximation Properties of Beta Operators, Progress in Approximation Theory, Academic Press, Boston, MA, 1991, pp. 483–495. [18] Z. Li, Bernstein polynomials and modulus of continuity, J. Approx. Theory 102 (1) (2000) 171–174. [19] T. Lindvail, Bernstein polynomials and the law of large numbers, Math. Sci. 7 (2) (1982) 127–139. [20] T. Trif, An Elementary proof of the preservation of Lipschitz constants by the Meyer–König and Zeller operators, J. Inequal. Pure Appl. Math. 4 (5) (2003) 3.Article90 [21] Y. Zhang, F. Cao, Z. Xu, K-functionals and multivariate Bernstein polynomials, J. Approx. Theory 155 (2) (2008) 125–135.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Bivariate Bernstein type operators a b ˙ ˙ Gülen Bascanbaz-Tunca ¸ , Hatice Gül InceIlarslan , Aysegül ¸ Erençin c,∗ a b c
Ankara University, Faculty of Science, Department of Mathematics, Tando˘gan, Ankara 06100, Turkey Gazi University, Faculty of Arts and Science, Department of Mathematics, Teknikokullar 06500, Ankara, Turkey ˙ Abant Izzet Baysal University, Faculty of Arts and Science, Department of Mathematics, Bolu 14280, Turkey
a r t i c l e
i n f o
a b s t r a c t
MSC: 41A25 41A36
In this paper, we introduce bivariate extension of Bernstein type operators defined in [11]. We show that these operators preserve some properties of the original function f, such as Lipschitz constant and monotonicity. Furthermore, we present the monotonicity of the sequence of bivariate Bernstein type operators for n when f is τ -convex.
Keywords: Lipschitz continuous function Monotony Bernstein type operator
© 2015 Elsevier Inc. All rights reserved.
1. Introduction In [11], Cárdenas-Morales et al. presented the following Bernstein type operators for f ∈ C[0, 1] and x ∈ [0, 1],
Bn ( f ; τ (x))
n n = k
(τ (x)) (1 − τ (x)) k
n−k
k (f ◦ τ ) −1
n
k=0
= B∗n ( f ◦ τ −1 ) ◦ τ
n n = k
(τ (x)) (1 − τ (x)) k
k=0
n−k
k f τ −1 , n
(1.1)
where B∗n is the classical Bernstein operators, n ∈ N and τ is a function defined on [0, 1] and having the properties: (τ 1 ) τ is ∞-times continuously differentiable on [0, 1] (τ 2 ) τ (0) = 0, τ (1) = 1 and τ (x) > 0 on [0, 1]. These conditions ensure that τ is strictly increasing and the inverse τ −1 of τ exists on [0, 1]. The authors discussed shape preserving and convergence properties and also introduced comparative results. Note that if τ (x) = x, then the operators given by (1.1) reduce to classical Bernstein operators. Before the construction of bivariate Bernstein type operators, we recall some usual notations and definitions which are essential for our work. For x = (x1 , x2 ) ∈ R2 , k = (k1 , k2 ) ∈ N20 and n ∈ N, we will write
|x| := x1 + x2 , xk := xk11 xk22 , |k| := k1 + k2 , k! := k1 !k2 ! ∗
Corresponding author. fax.: +903742541000. ˙ ˙ E-mail addresses: [email protected] (G. Bascanbaz-Tunca), ¸ [email protected] (H.G. InceIlarslan), [email protected] (A. Erençin).
http://dx.doi.org/10.1016/j.amc.2015.10.037 0096-3003/© 2015 Elsevier Inc. All rights reserved.
544
and
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
n k
=
n! n , := k1 , k2 k!(n − |k|)!
k := i
k1 i1
k2 , i2
k
:=
k1 k2
.
i1 =0 i2 =0
i=0
Now let S := {x = (x1 , x2 ) ∈ R2 : 0 ≤ x1 , x2 ≤ 1, |τ(x)| = τ (x1 ) + τ (x2 ) ≤ 1}. In this paper, for a function f defined on S, we consider the bivariate extension of the operators defined by (1.1) as follows:
Bn ( f ; τ(x)) =
n n−k 1 k1 =0 k2 =0
n k k n−|k| −1 (f ◦ τ ) , (τ(x)) (1 − |τ(x)|) n
k
where τ (x) := (τ (x1 ), τ (x2 )) such that τ (x1 ) and τ (x2 ) have the properties given by (τ 1 ) and (τ 2 ). Moreover, τ −1 (x) := (τ −1 (x1 ), τ −1 (x2 )) and ( f ◦ τ −1 )( kn ) = ( f ◦ τ −1 )( kn1 , kn2 ) := f (τ −1 ( kn1 ), τ −1 ( kn2 )). We observe that these operators are not the tensor product of the operators given by (1.1) obtained by a natural way. If τ(x) = x, then one obtains the bivariate operators defined on triangle (see [1], p. 109). Definition 1. Let f be a real valued continuous function defined on D ⊂ R2 and also let τ be a function satisfying the conditions (τ 1 ) and (τ 2 ). We say that f is a τ -Lipschitz continuous function of order μ on D, if
| f (x) − f (y)| ≤ A
2
|τ (xi ) − τ (yi )|μ
i=1
for x, y ∈ D with A > 0 and 0 < μ ≤ 1. We denote the set of τ -Lipschitz continuous functions by LipτA (μ, D). When τ (x) = x, LipτA (μ, D) reduces to the Lipschitz class defined in [10]. Definition 2. A real valued continuous function f(x) is said to be convex on the convex set D ⊂ R2 , if
f
r
αi xi ≤
i=1
r
αi f (xi )
i=1
for any x1 , x2 , . . . , xr in D and for any non-negative numbers α1 , α2 , . . . , αr such that α1 + α2 + · · · + αr = 1 (see [13]). Definition 3. We call f is a convex with respect to τ or τ -convex in D ⊂ R2 , if f ◦ τ −1 is convex in the sense of Definition 2. Note that this definition is an analogue of the τ -convexity given in [11]. 2. Main results In [10], Cao et al. introduced multivariate Baskakov operators and showed that these operators preserve some properties of the original function f such as monotonicity, semi-additivity and Lipschitz constant. Furthermore, they studied monotonicity of the sequence of multivariate Baskakov operators for n when the attached function f is convex. Finally, the authors computed the rate of approximation with the help of the K-functional and modulus of smoothness. In this part, inspired by the work [10], we firstly show that the Lipschitz constant and monotonicity properties of function f can be retained by the operators Bn ( f ; τ(x)) := Bτn f . After that, we study monotonicity of the sequence of bivariate operators Bτn f for n when f ◦ τ −1 is convex. For some preservation and approximation properties of univariate or multivariate Bernstein and some other classical positive linear operators we refer the papers [2–10,12–21] and references therein. Theorem 4. If f ∈ LipτA (μ, S), then Bτn f ∈ LipτA (μ, S) for all n ∈ N. Proof. Let x, y ∈ S and x ≤ y which means that x1 ≤ y1 and x2 ≤ y2 . Using the definition of the operators Bτn f and Binomial formula, we get
Bn ( f ; τ(y)) =
n n−k 1 k1 =0 k2 =0
n k (τ(y))k (1 − |τ(y)|)n−|k| ( f ◦ τ −1 ) k
n
n n−k 1 n k = ((τ(y) − τ(x)) + τ(x))k (1 − |τ(y)|)n−|k| ( f ◦ τ −1 ) k1 =0 k2 =0
n
k
k n n−k 1 n k k i k−i n−|k| −1 = (f ◦ τ ) . (τ(x)) (τ(y) − τ(x)) (1 − |τ(y)|) k1 =0 k2 =0 i=0
k
n
i
If we change the order of the above summations, then we can write
Bn ( f ; τ(y)) =
n n−k n 1 n−k 1 i1 =0 k1 =i1 i2 =0 k2 =i2
n k k (τ(x))i (τ(y) − τ(x))k−i (1 − |τ(y)|)n−|k| ( f ◦ τ −1 ) . k
i
n
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
Now letting k − i = m, one gets
Bn ( f ; τ(y)) =
|i|−m1 n−i 1 −m1 n−
n n−i 1 i1 =0 m1 =0
×(τ(y) − τ(x))
i1 =0 m1 =0
(1 − |τ(y)|)
i+m (f ◦ τ ) −1
n
n! (τ(x))i i!m!(n − |i| − |m|)!
m2 =0
i2 =0
(τ(x))i
i+m i
n−|i|−|m|
|i|−m1 n−i 1 −m1 n−
n n−i 1
n i+m
m2 =0
i2 =0
m
=
×(τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| ( f ◦ τ −1 ) Similarly,
Bn ( f ; τ(x)) =
n n−i 1
n i
i1 =0 i2 =0
545
i+m . n
(2.1)
i (τ(x))i (1 − |τ(x)|)n−|i| ( f ◦ τ −1 ) n
n−i1 n−|i| n−|i|−m1
=
n
i1 =0 i2 =0 m1 =0
m2 =0
n − |i| n − |i| − m1
n i
m1
m2
×(τ(x))i (τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| ( f ◦ τ −1 ) Change of the order of the second and third summations gives
Bn ( f ; τ(x)) =
n n−i 1
|i|−m1 n−i 1 −m1 n−
i1 =0 m1 =0
i2 =0
m2 =0
×(τ(x)) (τ(y) − τ(x)) i
=
n n−i 1 i1 =0 m1 =0
m
|i|−m1 n−i 1 −m1 n− i2 =0
m2 =0
n − |i| m1
n i
n − |i| − m1 m2
n−|i|−|m|
(1 − |τ(y)|)
i . n
i (f ◦ τ ) −1
n
n! (τ(x))i i!m!(n − |i| − |m|)!
×(τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| ( f ◦ τ −1 )
i . n
(2.2)
Thus, from (2.1) and (2.2), it follows that
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤
n n−i 1
|i|−m1 n−i 1 −m1 n−
i1 =0 m1 =0
m2 =0
i2 =0
× (τ(y) − τ(x))
m
n! (τ(x))i i!m!(n − |i| − |m|)! n−|i|−|m|
(1 − |τ(y)|)
( f ◦ τ −1 ) i + m − ( f ◦ τ −1 ) i . n
n
Since f ∈ LipτA (μ, S), we have
−1 i1 −1 i2 ( f ◦ τ −1 ) i + m − ( f ◦ τ −1 ) i = f τ −1 i1 + m1 , τ −1 i2 + m2 −f τ ,τ n
n
≤A
m μ 1
n
n
+
m μ
n
n
2
n
and so
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤ A
n n−i 1
|i|−m1 n−i 1 −m1 n−
i1 =0 m1 =0
m2 =0
i2 =0
×(τ(y) − τ(x))
m
Now, changing order of the summations, one has
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))|
n! (τ(x))i i!m!(n − |i| − |m|)!
(1 − |τ(y)|)n−|i|−|m|
m μ 1
n
+
m μ 2
n
.
n
546
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
≤A
n n−m 1 −|m| 1 −m1 n−i 1 n−i m1 =0 i1 =0
m2 =0
i2 =0
n! (τ(x))i i!m!(n − |i| − |m|)!
×(τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| =A
|m| n−i n n−m 1 −|m| 1 n− m1 =0 m2 =0 i1 =0
i2 =0
m μ 1
n
n n−m 1
2
n
m μ 1
n
m μ m μ n 1 2 + (τ(y) − τ(x))m n
m
m1 =0 m2 =0
n−|m| n−i1 −|m|
×
m μ
n! (n − |m|)! m!(n − |m|)! i!(n − |i| − |m|)!
×(τ(x))i (τ(y) − τ(x))m (1 − |τ(y)|)n−|i|−|m| =A
+
i1 =0
i2 =0
n − |m| i
+
m μ 2
n
n
(τ(x))i (1 − |τ(y)|)n−|i|−|m| .
By direct computations, we obtain n n−m 1
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤ A
m μ m μ n 1 2 + (τ(y) − τ(x))m (1 − |τ(y) − τ(x)|)n−|m|
m1 =0 m2 =0
m
n
n
= ABn ((τ (t1 ))μ + (τ (t2 ))μ ; τ(y) − τ(x))
= A{Bn ((τ (t1 ))μ ; τ (y1 ) − τ (x1 )) + Bn ((τ (t2 ))μ ; τ (y2 ) − τ (x2 ))}
=A
n m1 =0
+
n m2 =0
n m1
m μ (τ (y1 ) − τ (x1 ))m1 [1 − (τ (y1 ) − τ (x1 ))]n−m1 1 n
m μ n . (τ (y2 ) − τ (x2 ))m2 [1 − (τ (y2 ) − τ (x2 ))]n−m2 2
m2
1 Application of the Hölder inequality with p = μ and q =
n
1 1−μ
yields
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤ A{[Bn (τ (t1 ); τ (y1 ) − τ (x1 ))]μ [Bn (1; τ (y1 ) − τ (x1 ))]1−μ + [Bn (τ (t2 ); τ (y2 ) − τ (x2 ))]μ [Bn (1; τ (y2 ) − τ (x2 ))]1−μ }. From [11], we have Bn (τ (t ); τ (x)) = τ (x) and Bn (1; τ (x)) = 1. Thus, one can write
|Bn ( f ; τ(y)) − Bn ( f ; τ(x))| ≤ A{[τ (y1 ) − τ (x1 )]μ + [τ (y2 ) − τ (x2 )]μ }, which gives Bτn ∈ LipτA (μ, S). The proof of the case x1 ≥ y1 , x2 ≥ y2 or x1 ≥ y1 , x2 ≤ y2 or x1 ≤ y1 x2 ≥ y2 is similar. Hence, we have proved the required result. Theorem 5. Suppose that f is defined on S and f(x) ≥ 0. If τf ((xx)) is non-increasing with respect to xi on (0, 1], then i non-increasing with respect to xi on (0, 1] for each n ∈ N and i = 1, 2. Proof. We prove for i = 1. Using the definition of Bn (f; τ (x)), one gets n n−k 1 Bn ( f ; τ(x)) = τ (x1 )
k1 =0 k2 =0
n k
k (τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| ( f ◦ τ −1 ) n
n n−k 1 n k k (τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| f τ −1 1 , τ −1 2 = k1 =1 k2 =0
k
n
n
n n k + . (τ (x1 ))−1 (τ (x2 ))k2 (1 − |τ(x)|)n−k2 f τ −1 (0), τ −1 2 k2 =0
n
k2
From the hypotheses on τ we have τ −1 (0) = 0. So we can write n n−k 1 Bn ( f ; τ(x)) = τ (x1 )
k1 =1 k2 =0
n k k (τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| f τ −1 1 , τ −1 2 k
n
n
Bn ( f ;τ(x)) τ (xi )
is also
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
+
n
n k2
k2 =0
and
547
k (τ (x1 ))−1 (τ (x2 ))k2 (1 − |τ(x)|)n−k2 f 0, τ −1 2 n
n n−k 1 n ∂ Bn ( f ; τ(x)) k1 −2 k2 n−|k| −1 k1 −1 k2 = τ (x1 ) ×f τ ,τ (k1 − 1)(τ (x1 )) (τ (x2 )) (1 − |τ(x)|) ∂ x1 n n k τ (x1 ) k1 =2 k2 =0 n−1 n−k 1 −1 n − τ (x1 ) (n − |k|)(τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k|−1
τ −1
×f
k1 n
n
− τ (x1 )
k2 =0 n
− τ (x1 ) = τ (x1 )
, τ −1 n k2
k (τ (x1 ))−2 (τ (x2 ))k2 (1 − |τ(x)|)n−k2 f 0, τ −1 2
−1
− τ (x1 )
k1 + 1 , τ −1 n
k1 =1 k2 =0
τ
k2 n
n−1 nk1 (τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k|−1 k1 + 1 k
×f
(n − k2 )(τ (x1 ))−1 (τ (x2 ))k2 (1 − |τ(x)|)n−k2 −1 f 0, τ −1
k2 n
n−1 n−k 1 −1 n−1 n(τ (x1 ))k1 −1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k|−1 k
− τ (x1 )
n
n k2
n−1 n−k 1 −1
τ −1
k2 n
k1 =1 k2 =0
k2 =0
×f
k
k1 =1 k2 =0
k1 n
n k2 =0
× [(τ (x1 ))
−1
,τ
−1
n k2
k2 n
(τ (x1 ))−1 (τ (x2 ))k2 (1 − |τ(x)|)n−k2
+ (n − k2 )(1 − |τ(x)|)
−1
] f 0, τ
−1
k2 n
k (τ(x))k (1 − |τ(x)|)n−|k|−1 1 τ ( x1 ) k1 =1 k2 =0 n n k1 + 1 k2 k1 k2 f τ −1 × , τ −1 − f τ −1 , τ −1 k1 + 1 n n k1 n n n τ (x1 ) n − (τ (x2 ))k2 (1 − |τ(x)|)n−k2 τ (x1 ) k =0 k2 2 k2 × [(τ (x1 ))−1 + (n − k2 )(1 − |τ(x)|)−1 ] f 0, τ −1 . = τ (x1 )
n−1 n−k 1 −1 n−1 k
n
Since τ is increasing, f(x) ≥ 0 and τf((xx)) is non-increasing with respect to x1 on (0, 1], we can write ∂∂x ( Bn (τf(;xτ()x)) ) ≤ 0. This leads 1 1 1 to the desired result. We can prove the similar result for i = 2. Therefore, the proof is completed. Theorem 6. Let f be a τ -convex function defined on S. Then Bn (f; τ (x)) is monotonically non-increasing in n. Proof. From the definition of Bn (f; τ (x)), we have
Bn ( f ; τ(x)) =
n n−k 1 k1 =0 k2 =0
=
n n−k 1 k1 =0 k2 =0
n k k n−|k| −1 (f ◦ τ ) (τ(x)) (1 − |τ(x)|) k
n
n k (|τ(x)| + 1 − |τ(x)|)(τ(x))k (1 − |τ(x)|)n−|k| ( f ◦ τ −1 ) k
n
548
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552 n n−k 1
=
k1 =0 k2 =0
n k (τ (x1 ))k1 +1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| ( f ◦ τ −1 ) k
n
n n−k 1 n k k1 k2 +1 n−|k| −1 (τ (x1 )) (τ (x2 )) (1 − |τ(x)|) + (f ◦ τ ) k
k1 =0 k2 =0
+
n k . (τ(x))k (1 − |τ(x)|)n−|k|+1 ( f ◦ τ −1 )
n n−k 1
n n−k 1
S1 :=
n
k
k1 =0 k2 =0
Now let,
n
n k (τ (x1 ))k1 +1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| ( f ◦ τ −1 ) n
k
k1 =0 k2 =0
n n−k 1 n k k1 k2 +1 n−|k| −1 S2 := (f ◦ τ ) (τ (x1 )) (τ (x2 )) (1 − |τ(x)|) k
k1 =0 k2 =0
and
S3 :=
n n−k 1
n k . (τ(x))k (1 − |τ(x)|)n−|k|+1 ( f ◦ τ −1 )
S1 =
n−1 n−k 1
n
k
k1 =0 k2 =0
For S1 , we can write
n
n k
k1 =0 k2 =0
k k + (τ (x1 ))n+1 f (τ −1 (1), τ −1 (0)) (τ (x1 ))k1 +1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| f τ −1 1 , τ −1 2 n
n
n−1 n−k 1 n k1 +1 k2 n−|k| −1 k1 −1 k2 = f τ ,τ (τ (x1 )) (τ (x2 )) (1 − |τ(x)|) k
k1 =0 k2 =1
n
n−1
+
n k1
k1 =0
=
n−2 n−k 1 −1
n k
n−1
k1 =0
+
n−1
n k1
k1 =0
=
n−1 n−k 1
k1 =1 k2 =1
+
n k1 =1
+
n
k1 =1
k1 n
n
−1 , τ (0) + (τ (x1 ))n+1 f (τ −1 (1), τ −1 (0))
k k (τ (x1 ))k1 +1 (τ (x2 ))k2 (1 − |τ(x)|)n−|k| f τ −1 1 , τ −1 2 n
(τ (x1 ))k1 +1 (τ (x2 ))n−k1 f τ −1
n k1
(τ (x1 ))k1 +1 (1 − |τ(x)|)n−k1 f τ −1
k1 =0 k2 =1
+
k1 n
(τ (x1 ))
k1 +1
n k1 − 1, k2
(1 − |τ(x)|)
n−k1
f
, τ −1
τ
−1
k1 n
n − k1 n
n
, τ (0) + (τ (x1 ))n+1 f (τ −1 (1), τ −1 (0)) −1
k −1 k2 , τ −1 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 1 n
n
k1 n−k1 +1 −1 k1 − 1 −1 n − k1 + 1 f τ ,τ (τ (x1 )) (τ (x2 ))
k1 n−k1 +1 −1 k1 − 1 −1 f τ , τ (0) + (τ (x1 ))n+1 f (τ −1 (1), τ −1 (0)). (τ (x1 )) (1 − |τ(x)|)
n k1 − 1 n k1 − 1
n
n
n
Since τ −1 (0) = 0 and τ −1 (1) = 1,
S1 =
n−1 n−k 1
k1 =1 k2 =1
+
n k1 =1
+
n k1 =1
n k1 − 1, k2
k −1 k2 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 1 , τ −1 n
n
k −1 n − k1 + 1 , τ −1 (τ (x1 ))k1 (τ (x2 ))n−k1 +1 f τ −1 1
k −1 (τ (x1 ))k1 (1 − |τ(x)|)n−k1 +1 f τ −1 1 , 0 + (τ (x1 ))n+1 f (1, 0).
n k1 − 1 n k1 − 1
n
n
n
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
549
Applying the similar steps for S2 and S3 , we reach to
S2 =
n−1 n−k 1
n k1 , k2 − 1
k1 =1 k2 =1
k k −1 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 1 , τ −1 2 n
n
n n k1 n−k1 +1 −1 k1 −1 n − k1 + f τ ,τ (τ (x1 )) (τ (x2 )) k1 =1
+
n
k2 =1
k1
n
n
n k2 n−k2 +1 −1 k2 − 1 (τ (x2 )) (1 − |τ(x)|) f 0, τ + (τ (x2 ))n+1 f (0, 1)
k2 − 1
n
and
S3 =
n−1 n−k 1
n k1 , k2
k1 =1 k2 =1
k n−|k|+1 −1 k1 −1 k2 f τ ,τ (τ(x)) (1 − |τ(x)|) n
n
n n k1 k1 n−k1 +1 + f τ −1 ,0 (τ (x1 )) (1 − |τ(x)|) k1 =1
k1
n
n n k + + (1 − |τ(x)|)n+1 f (0, 0). (τ (x2 ))k2 (1 − |τ(x)|)n−k2 +1 f 0, τ −1 2 k2 =1
n
k2
Therefore, we have
Bn ( f ; τ(x)) =
n−1 n−k 1
k1 =1 k2 =1
+
n k1 =1
+
n k1 =1
n k1 − 1, k2
n
n
n k −1 n − k1 + 1 , τ −1 (τ (x1 ))k1 (τ (x2 ))n−k1 +1 f τ −1 1
k1 − 1
n
n
n k −1 ,0 (τ (x1 ))k1 (1 − |τ(x)|)n−k1 +1 f τ −1 1
k1 − 1
n
n−1 n−k1
+
k −1 k2 , τ −1 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 1
n k1 , k2 − 1
k1 =1 k2 =1
k n−|k|+1 −1 k1 −1 k2 − 1 f τ ,τ (τ(x)) (1 − |τ(x)|) n
n
n n k1 n−k1 +1 −1 k1 −1 n − k1 (τ (x1 )) (τ (x2 )) + f τ ,τ k1 =1
+
n k2 =1
k1
n
k2 − 1
n−1 n−k1
+
+
k1 =1
n
n k1 , k2
k1 =1 k2 =1 n
n
n k2 n−k2 +1 −1 k2 − 1 f 0, τ (τ (x2 )) (1 − |τ(x)|)
n−|k|+1
(τ(x)) (1 − |τ(x)|) k
f
τ
−1
k1 n
,τ
−1
k2 n
n k1 k1 n−k1 +1 f τ −1 ,0 (τ (x1 )) (1 − |τ(x)|) k1
n
n n k2 n−k2 +1 −1 k2 + f 0, τ (τ (x2 )) (1 − |τ(x)|) k2 =1
n
k2
+ (τ (x1 ))n+1 f (1, 0) + (τ (x2 ))n+1 f (0, 1) + (1 − |τ(x)|)n+1 f (0, 0). Similarly,
Bn+1 ( f ; τ(x)) =
k1 =0 k2 =0
=
n+1 n+1−k 1
n−1 n−k 1 k1 =1 k2 =1
n+1 k
n+1 k1 , k2
k k n−|k|+1 −1 (f ◦ τ ) (τ(x)) (1 − |τ(x)|) n+1
k1 k2 (τ(x))k (1 − |τ(x)|)n−|k|+1 f τ −1 , τ −1 n+1
n+1
550
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
+
n k1 =1
+
n k1 =1
+
n k2 =1
n+1 k1
k1 n − k1 + 1 , τ −1 (τ (x1 ))k1 (τ (x2 ))n−k1 +1 f τ −1 n+1
n
n+1 k1 ,0 (τ (x1 ))k1 (1 − |τ(x)|)n−k1 +1 f τ −1
k1
n+1 k2
n+1
(τ (x2 )) (1 − |τ(x)|)
n−k2 +1
k2
f 0, τ
−1
k2 n+1
+(τ (x1 ))n+1 f (1, 0) + (τ (x2 ))n+1 f (0, 1) + (1 − |τ(x)|)n+1 f (0, 0). Thus, we can write
Bn ( f ; τ(x)) − Bn+1 ( f ; τ(x)) =
n−1 n−k 1
n k1 − 1, k2
k1 =1 k2 =1
f
n + f k1 , k2 − 1
k1 − 1 , τ −1 n
k1 , τ −1 n
τ
−1
τ −1
k1 − 1
k1 =1
k
k2 n
k2 − 1 n
n k1 , k2
+
−1 k1 −1 k2 f τ ,τ n
n+1 k1 k2 −1 −1 − f τ ,τ (τ(x))k (1 − |τ(x)|)n−|k|+1 n+1 n+1 k1 , k2 n n k1 − 1 n − k1 + 1 + f τ −1 , τ −1 n k1
n
n−k
n
1 τ −1 1 , τ −1 n n n+1 k1 n − k1 + 1 − f τ −1 , τ −1 τ (x1 )k1 τ (x2 )n−k1 +1 n+1 n+1 k1 n n k1 − 1 n k1 + f τ −1 ,0 + f τ −1 ,0
+
f
k1 − 1
k1 =1
k2 =1
n
n
k1
n+1 k1 − f τ −1 , 0 τ (x1 )k1 (1 − |τ(x)|)n−k1 +1 n+1 k1 n n k2 − 1 n k2 + f 0, τ −1 + f 0, τ −1 − =
k2 − 1
n+1 f 0, τ −1 k2
n
n+1
τ (x1 )τ (x2 )(1 − |τ(x)|)
k2
k2 τ (x2 )k2 (1 − |τ(x)|)n−k2 +1 n−2 n−2−k 1
k1 =0 k2 =0
n k1 , k2 + 1
n
k1 k2 + 1 −1 , (f ◦ τ ) n
n
n k + 1 k2 , + ( f ◦ τ −1 ) 1 k1 + 1, k2 n n n k1 + 1 k2 + 1 −1 , + (f ◦ τ ) n n k1 + 1, k2 + 1 n+1 k1 + 1 k2 + 1 −1 − , (f ◦ τ ) (τ(x))k (1 − |τ(x)|)n−|k|−2
k1 + 1, k2 + 1
+ τ (x1 )τ (x2 )
n−1 k1 =0
n+1
n+1
n k n − k1 ( f ◦ τ −1 ) 1 , k1
n
n
n k1 + 1 n − k1 − 1 −1 , + (f ◦ τ ) k1 + 1 n n n+1 k + 1 n − k1 , ( f ◦ τ −1 ) 1 τ (x1 )k1 τ (x2 )n−k1 −1 −
k1 + 1
n+1
n+1
n
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552 n−1
+ τ (x1 )(1 − |τ(x)|)
n k n k +1 ,0 ( f ◦ τ −1 ) 1 , 0 + ( f ◦ τ −1 ) 1 k1 + 1
n
k1
k1 =0
551
n+1 k +1 , 0 τ (x1 )k1 (1 − |τ(x)|)n−k1 −1 ( f ◦ τ −1 ) 1 − k1 + 1
n+1
n−1
+ τ (x2 )(1 − |τ(x)|)
k2 k2 + 1 n n −1 −1 + ( f ◦ τ ) 0, ( f ◦ τ ) 0, n
k2
k2 =0
k2 + 1
k +1 n+1 − ( f ◦ τ −1 ) 0, 2 τ (x2 )k2 (1 − |τ(x)|)n−k2 −1 .
k2 + 1
Now set
n
n+1
n
(2.3)
k k +1 n k + 1 k2 , + ( f ◦ τ −1 ) 1 , 2 ( f ◦ τ −1 ) 1 k1 + 1, k2 n n n n n k + 1 k2 + 1 n+1 k + 1 k2 + 1 , , ( f ◦ τ −1 ) 1 ( f ◦ τ −1 ) 1 + − , k1 + 1, k2 + 1 k1 + 1, k2 + 1 n n n+1 n+1 n k1 n − k1 n k1 + 1 n − k1 − 1 −1 −1 I2 : = , , + (f ◦ τ ) (f ◦ τ ) k1 k1 + 1 n n n n n+1 k + 1 n − k1 , − , ( f ◦ τ −1 ) 1 n+1 n+1 k1 + 1 n k1 n k1 + 1 n+1 k1 + 1 −1 −1 −1 I3 : = ,0 + ,0 − ,0 (f ◦ τ ) (f ◦ τ ) (f ◦ τ )
n k1 , k2 + 1
I1 : =
and
k1 + 1
n
k1
k1 + 1
n
n+1
k2 k2 + 1 k2 + 1 n n n+1 −1 −1 −1 I4 : = + − . ( f ◦ τ ) 0, ( f ◦ τ ) 0, ( f ◦ τ ) 0, k2 + 1
n
k2
n
k2 + 1
n+1
We firstly consider I1 . Let
(k1 ,kn2 +1)
α1 =
=
k1 + 1
≥ 0,
n+1 n+1 (k1 +1,k ) 2 +1 n ( +1,k2 ) k2 + 1 ≥ 0, α2 = k1n+1 = n+1 (k1 +1,k2 +1) ( n 2 +1) n − |k| − 1 ≥0 α3 = k1 +1,k = n+1 n+1 (k1 +1,k ) 2 +1
and
x1 =
k1 k2 + 1 , , n n
x2 =
k1 + 1 k2 , n n
,
x3 =
Then it is easily seen that α1 + α2 + α3 = 1 and α1 x1 + α2 x2 + α3 x3 = ( readily follows that I1 ≥ 0. For I2 , we set
α1 =
(kn1 )
( ( α2 = ( and
x1 =
=
k1 + 1 k2 + 1 , . n n k1 +1 k2 +1 n+1 , n+1 ).
Thus, from the definition of τ -convexity it
k1 + 1 ≥ 0, n+1
) ) n − k1 = ≥0 n+1 )
n+1 k1 +1 n k1 +1 n+1 k1 +1
k1 n − k1 , n n
,
x2 =
k1 + 1 n − k1 − 1 , . n n
Thus we have α1 + α2 = 1 and α1 x1 + α2 x2 = (
k1 +1 n−k1 n+1 , n+1 ).
Since f is a τ -convex function, one gets I2 ≥ 0.
552
G. Bascanbaz-Tunca ¸ et al. / Applied Mathematics and Computation 273 (2016) 543–552
For I3 , if we take
α1 =
(kn1 )
( ( α2 = ( and
x1 =
=
k1 + 1 ≥ 0, n+1
) ) n − k1 ≥0 = n+1 )
n+1 k1 +1 n k1 +1 n+1 k1 +1
k1 ,0 , n
x2 =
k1 + 1 ,0 , n k +1
1 then we have α1 + α2 = 1 and α1 x1 + α2 x2 = ( n+1 , 0). These lead to I3 ≥ 0. Similarly, we have I4 ≥ 0. Therefore, from (2.3), we reach to the desired result Bn ( f ; τ(x)) ≥ Bn+1 ( f ; τ(x)) for all n ∈ N.
We note that for τ(x) = x from Theorem 6 we get: if f is a convex function, then for every x ∈ S := {x = (x1 , x2 ) ∈ R2 : 0 ≤ x1 , x2 ≤ 1, x1 + x2 ≤ 1} one has Bn ( f ; x) ≥ Bn+1 ( f ; x). This result was proved in [13]. Remark 1. The all definitions and results presented in this paper can be extended to the case of n-variate with n ≥ 3. Acknowledgment The authors thank the referee for valuable comments and suggestions which improved the quality of the paper. References [1] F. Altomare, M.C. Montano, V. Leonessa, I. Rasa, ¸ Markov Operators, Positive Semigroups and Approximation Processes, 61, De Gruyter Studies in Mathematics, De Gruyter, Berlin, 2014. [2] G. Bascanbaz-Tunca, ¸ Y. Tuncer, Some properties of multivariate beta operator, Fasc. Math. 41 (2009) 31–43. [3] G. Bascanbaz-Tunca, ¸ Y. Tuncer, On a Chlodovsky variant of a multivariate beta operator, J. Comput. Appl. Math. 235 (16) (2011) 4816–4824. [4] G. Bascanbaz-Tunca, ¸ F. Tasdelen, ¸ On Chlodovsky form of the Meyer–König and Zeller operators, Ann. Univ. Vest Timis. ¸ Ser. Mat.-Inform., XLIX 49 (2) (2011) 137–144. [5] M. Birou, Some blending surfaces on rhombus, Automat. Comput. Appl. Math. 19 (1) (2010) 47–55. [6] M. Birou, Blending surfaces on ellipse generated using the Bernstein operators, Stud. Univ. Babes-Bolyai ¸ Math. 56 (2) (2011a) 237–246. [7] M. Birou, Bezier blending surfaces on astroid, Stud. Univ. Babes-Bolyai ¸ Math. 56 (3) (2011b) 127–134. [8] M. Birou, Blending surfaces generated using the Bernstein operator, Creat. Math. Inform. 21 (1) (2012) 35–40. [9] B.M. Brown, D. Elliott, D.F. Paget, Lipschitz constants for the Bernstein polynomials of a Lipschitz continuous function, J. Approx. Theory 49 (2) (1987) 196–199. [10] F. Cao, C. Ding, Z. Xu, On multivariate Baskakov operators, J. Math. Anal. Appl. 307 (1) (2005) 274–291. [11] D. Cárdenas-Morales, P. Garrancho, I. Rasa, ¸ Bernstein-type operators which preserve polynomials, Comput. Math. Appl. 62 (1) (2011a) 158–163. [12] D. Cárdenas-Morales, A. Fraguela, P. Garrancho, Saturation in multivariate simultaneous approximation, Math. Comput. Simul. 81 (10) (2011b) 2098–2102. [13] G.Z. Chang, P.J. Davis, The convexity of Bernstein polynomials over triangles, J. Approx. Theory 40 (1) (1984) 11–28. [14] A. Erençin, G. Bascanbaz-Tunca, ¸ F. Tasdelen, ¸ Some preservation properties of MKZ-Stancu type operators, Sarejevo J. Math. 22 (1) (2011a) 93–102. [15] A. Erençin, G. Bascanbaz-Tunca, ¸ F. Tasdelen, ¸ Some properties of the operators defined by Lupas, ¸ Rev. Anal. Numér. Théor. Approx. 43 (2) (2011b) 168–174. [16] M.D. Farcas, ¸ About Bernstein polynomials, Ann. Univ. Craiova Ser. Math. Inform. 35 (2008) 117–121. [17] M.K. Khan, Approximation Properties of Beta Operators, Progress in Approximation Theory, Academic Press, Boston, MA, 1991, pp. 483–495. [18] Z. Li, Bernstein polynomials and modulus of continuity, J. Approx. Theory 102 (1) (2000) 171–174. [19] T. Lindvail, Bernstein polynomials and the law of large numbers, Math. Sci. 7 (2) (1982) 127–139. [20] T. Trif, An Elementary proof of the preservation of Lipschitz constants by the Meyer–König and Zeller operators, J. Inequal. Pure Appl. Math. 4 (5) (2003) 3.Article90 [21] Y. Zhang, F. Cao, Z. Xu, K-functionals and multivariate Bernstein polynomials, J. Approx. Theory 155 (2) (2008) 125–135.