Return to Article Details Exact orders in simultaneous approximation by complex Bernstein-Stancu polynomials

EXACT ORDERS IN SIMULTANEOUS APPROXIMATION BY COMPLEX BERNSTEIN-STANCU POLYNOMIALS † † ^(†){ }^{\dagger}†

SORIN G. GAL*

Abstract

In this paper the exact orders in approximation by the complex Bernstein-Stancu polynomials (depending on two parameters) and their derivatives on compact disks are obtained.

MSC 2000. Primary: 30E10; Secondary: 41A25, 41A28.
Keywords. Complex Bernstein-Stancu polynomials, exact orders in simultaneous approximation.

1. INTRODUCTION

In the recent paper [1] the following upper estimates and Voronovskaja's theorem in approximation by complex Bernstein-Stancu polynomials depending on two parameters were proved.
Theorem 1.1. Let D R = { z ∈ C ; | z | < R } D R = { z ∈ C ; | z | < R } D_(R)={z inC;|z| < R}\mathbb{D}_{R}=\{z \in \mathbb{C} ;|z|<R\}DR={z∈C;|z|<R} be with R > 1 R > 1 R > 1R>1R>1 and let us suppose that f : D R → C f : D R → C f:D_(R)rarrCf: \mathbb{D}_{R} \rightarrow \mathbb{C}f:DR→C is analytic in D R D R D_(R)\mathbb{D}_{R}DR, i.e. f ( z ) = ∑ k = 0 ∞ c k z k f ( z ) = ∑ k = 0 ∞   c k z k f(z)=sum_(k=0)^(oo)c_(k)z^(k)f(z)=\sum_{k=0}^{\infty} c_{k} z^{k}f(z)=∑k=0∞ckzk, for all z ∈ D R z ∈ D R z inD_(R)z \in \mathbb{D}_{R}z∈DR. Also, for 0 ≤ α ≤ β 0 ≤ α ≤ β 0 <= alpha <= beta0 \leq \alpha \leq \beta0≤α≤β (independent of n n nnn ) let us define the complex Bernstein-Stancu polynomials by
S n ( α , β ) ( f ) ( z ) = ∑ k = 0 n ( n k ) z k ( 1 − z ) n − k f [ ( k + α ) / ( n + β ) ] , z ∈ C S n ( α , β ) ( f ) ( z ) = ∑ k = 0 n   ( n k ) z k ( 1 − z ) n − k f [ ( k + α ) / ( n + β ) ] , z ∈ C S_(n)^((alpha,beta))(f)(z)=sum_(k=0)^(n)((n)/(k))z^(k)(1-z)^(n-k)f[(k+alpha)//(n+beta)],quad z inCS_{n}^{(\alpha, \beta)}(f)(z)=\sum_{k=0}^{n}\binom{n}{k} z^{k}(1-z)^{n-k} f[(k+\alpha) /(n+\beta)], \quad z \in \mathbb{C}Sn(α,β)(f)(z)=∑k=0n(nk)zk(1−z)n−kf[(k+α)/(n+β)],z∈C
(i) For 1 ≤ r < R 1 ≤ r < R 1 <= r < R1 \leq r<R1≤r<R and n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N, we have
‖ S n ( α , β ) ( f ) − f ‖ r ≤ M 2 , r ( β ) ( f ) n + β , where 0 < M 2 , r ( β ) ( f ) = 2 r 2 ∑ j = 2 ∞ j ( j − 1 ) | c j | r j − 2 + 2 β r ∑ j = 1 ∞ j | c j | r j − 1 < ∞ Here ‖ f ‖ r = sup { | f ( z ) | ; | z | ≤ r } S n ( α , β ) ( f ) − f r ≤ M 2 , r ( β ) ( f ) n + β ,   where  0 < M 2 , r ( β ) ( f ) = 2 r 2 ∑ j = 2 ∞   j ( j − 1 ) c j r j − 2 + 2 β r ∑ j = 1 ∞   j c j r j − 1 < ∞  Here  ‖ f ‖ r = sup { | f ( z ) | ; | z | ≤ r } {:[qquad||S_(n)^((alpha,beta))(f)-f||_(r) <= (M_(2,r)^((beta))(f))/(n+beta)", "],[" where "0 < M_(2,r)^((beta))(f)=2r^(2)sum_(j=2)^(oo)j(j-1)|c_(j)|r^(j-2)+2beta rsum_(j=1)^(oo)j|c_(j)|r^(j-1) < oo],[" Here "||f||_(r)=s u p{|f(z)|;|z| <= r}]:}\begin{aligned} & \qquad\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r} \leq \frac{M_{2, r}^{(\beta)}(f)}{n+\beta} \text {, } \\ & \text { where } 0<M_{2, r}^{(\beta)}(f)=2 r^{2} \sum_{j=2}^{\infty} j(j-1)\left|c_{j}\right| r^{j-2}+2 \beta r \sum_{j=1}^{\infty} j\left|c_{j}\right| r^{j-1}<\infty \\ & \text { Here }\|f\|_{r}=\sup \{|f(z)| ;|z| \leq r\} \end{aligned}‖Sn(α,β)(f)−f‖r≤M2,r(β)(f)n+β,  where 0<M2,r(β)(f)=2r2∑j=2∞j(j−1)|cj|rj−2+2βr∑j=1∞j|cj|rj−1<∞ Here ‖f‖r=sup{|f(z)|;|z|≤r}
(ii) If 1 ≤ r < r 1 < R 1 ≤ r < r 1 < R 1 <= r < r_(1) < R1 \leq r<r_{1}<R1≤r<r1<R, then for all n , p ∈ N n , p ∈ N n,p inNn, p \in \mathbb{N}n,p∈N, we have
‖ [ S n ( α , β ) ( f ) ] ( p ) − f ( p ) ‖ r ≤ M 2 , r 1 ( β ) ( f ) p ! r 1 ( n + β ) ( r 1 − r ) p + 1 S n ( α , β ) ( f ) ( p ) − f ( p ) r ≤ M 2 , r 1 ( β ) ( f ) p ! r 1 ( n + β ) r 1 − r p + 1 ||[S_(n)^((alpha,beta))(f)]^((p))-f^((p))||_(r) <= (M_(2,r_(1))^((beta))(f)p!r_(1))/((n+beta)(r_(1)-r)^(p+1))\left\|\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}-f^{(p)}\right\|_{r} \leq \frac{M_{2, r_{1}}^{(\beta)}(f) p!r_{1}}{(n+\beta)\left(r_{1}-r\right)^{p+1}}‖[Sn(α,β)(f)](p)−f(p)‖r≤M2,r1(β)(f)p!r1(n+β)(r1−r)p+1
(iii) For all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N, we have
‖ S n ( α , β ) ( f ) − f + β e 1 − α n + β f ′ − n e 1 ( 1 − e 1 ) 2 ( n + β ) 2 f ′ ′ ‖ 1 ≤ M 1 ( α , β ) ( f ) ( n + β ) 2 S n ( α , β ) ( f ) − f + β e 1 − α n + β f ′ − n e 1 1 − e 1 2 ( n + β ) 2 f ′ ′ 1 ≤ M 1 ( α , β ) ( f ) ( n + β ) 2 ||S_(n)^((alpha,beta))(f)-f+(betae_(1)-alpha)/(n+beta)f^(')-(ne_(1)(1-e_(1)))/(2(n+beta)^(2))f^('')||_(1) <= (M_(1)^((alpha,beta))(f))/((n+beta)^(2))\left\|S_{n}^{(\alpha, \beta)}(f)-f+\frac{\beta e_{1}-\alpha}{n+\beta} f^{\prime}-\frac{n e_{1}\left(1-e_{1}\right)}{2(n+\beta)^{2}} f^{\prime \prime}\right\|_{1} \leq \frac{M_{1}^{(\alpha, \beta)}(f)}{(n+\beta)^{2}}‖Sn(α,β)(f)−f+βe1−αn+βf′−ne1(1−e1)2(n+β)2f′′‖1≤M1(α,β)(f)(n+β)2
where e 1 ( z ) = z e 1 ( z ) = z e_(1)(z)=ze_{1}(z)=ze1(z)=z and 0 < M 1 ( α , β ) ( f ) < ∞ 0 < M 1 ( α , β ) ( f ) < ∞ 0 < M_(1)^((alpha,beta))(f) < oo0<M_{1}^{(\alpha, \beta)}(f)<\infty0<M1(α,β)(f)<∞ depends only on α , β α , β alpha,beta\alpha, \betaα,β and f f fff.
Remark 1.2. Following exactly the lines in the proof of Theorem 1.1, (iii) in [1], it is immediate that in fact for any 1 ≤ r < R 1 ≤ r < R 1 <= r < R1 \leq r<R1≤r<R we have an upper estimate of the form
‖ S n ( α , β ) ( f ) − f + β e 1 − α n + β f ′ − n e 1 ( 1 − e 1 ) 2 ( n + β ) 2 f ′ ′ ‖ r ≤ M r ( α , β ) ( f ) ( n + β ) 2 S n ( α , β ) ( f ) − f + β e 1 − α n + β f ′ − n e 1 1 − e 1 2 ( n + β ) 2 f ′ ′ r ≤ M r ( α , β ) ( f ) ( n + β ) 2 ||S_(n)^((alpha,beta))(f)-f+(betae_(1)-alpha)/(n+beta)f^(')-(ne_(1)(1-e_(1)))/(2(n+beta)^(2))f^('')||_(r) <= (M_(r)^((alpha,beta))(f))/((n+beta)^(2))\left\|S_{n}^{(\alpha, \beta)}(f)-f+\frac{\beta e_{1}-\alpha}{n+\beta} f^{\prime}-\frac{n e_{1}\left(1-e_{1}\right)}{2(n+\beta)^{2}} f^{\prime \prime}\right\|_{r} \leq \frac{M_{r}^{(\alpha, \beta)}(f)}{(n+\beta)^{2}}‖Sn(α,β)(f)−f+βe1−αn+βf′−ne1(1−e1)2(n+β)2f′′‖r≤Mr(α,β)(f)(n+β)2
where the constant M r ( α , β ) ( f ) > 0 M r ( α , β ) ( f ) > 0 M_(r)^((alpha,beta))(f) > 0M_{r}^{(\alpha, \beta)}(f)>0Mr(α,β)(f)>0 is independent of n n nnn and depends on f , r , α f , r , α f,r,alphaf, r, \alphaf,r,α and β β beta\betaβ. This estimate will be useful in Section 3.
The goal of this paper is to show that in Theorem 1.1, (i) and (ii), also lower estimates hold. Thus, in Section 2 we prove that if the analytic function f f fff is not a polynomial of degree ≤ 0 ≤ 0 <= 0\leq 0≤0 and 1 ≤ r < R 1 ≤ r < R 1 <= r < R1 \leq r<R1≤r<R, then we have ‖ S n ( α , β ) ( f ) − f ‖ r ≥ C r ( α , β ) ( f ) n , n ∈ N S n ( α , β ) ( f ) − f r ≥ C r ( α , β ) ( f ) n , n ∈ N ||S_(n)^((alpha,beta))(f)-f||_(r) >= (C_(r)^((alpha,beta))(f))/(n),n inN\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r} \geq \frac{C_{r}^{(\alpha, \beta)}(f)}{n}, n \in \mathbb{N}‖Sn(α,β)(f)−f‖r≥Cr(α,β)(f)n,n∈N, that is in Theorem 1.1, (i), in fact the equivalence ‖ S n ( α , β ) ( f ) − f ‖ r ∼ 1 n ‖ S n ( α , β ) ( f ) − f ‖ r ∼ 1 n ||S_(n)^((alpha,beta))(f)-f||_(r)∼(1)/(n)\| S_{n}^{(\alpha, \beta)}(f)- f \|_{r} \sim \frac{1}{n}‖Sn(α,β)(f)−f‖r∼1n holds. In Section 3 we prove that for any p ∈ N p ∈ N p inNp \in \mathbb{N}p∈N and 1 ≤ r < R 1 ≤ r < R 1 <= r < R1 \leq r<R1≤r<R, if f f fff is not a polynomial of degree ≤ p − 1 ≤ p − 1 <= p-1\leq p-1≤p−1 then we have ‖ [ S n ( α , β ) ( f ) ] ( p ) − f ( p ) ‖ r ∼ 1 n S n ( α , β ) ( f ) ( p ) − f ( p ) r ∼ 1 n ||[S_(n)^((alpha,beta))(f)]^((p))-f^((p))||_(r)∼(1)/(n)\left\|\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}-f^{(p)}\right\|_{r} \sim \frac{1}{n}‖[Sn(α,β)(f)](p)−f(p)‖r∼1n, where the constants in the equivalence depend only on f , α , β , r f , α , β , r f,alpha,beta,rf, \alpha, \beta, rf,α,β,r and p p ppp.
Since the case α = β = 0 α = β = 0 alpha=beta=0\alpha=\beta=0α=β=0 (i.e. the case of classical Bernstein polynomials) was already considered in [2], in the rest of the paper we will exclude it.

2. EXACT ORDER OF APPROXIMATION FOR COMPLEX BERNSTEIN-STANCU POLYNOMIALS

The main result of this section is the following.
Theorem 2.1. Let R > 1 , 0 ≤ α ≤ β R > 1 , 0 ≤ α ≤ β R > 1,0 <= alpha <= betaR>1,0 \leq \alpha \leq \betaR>1,0≤α≤β with α + β > 0 , D R = { z ∈ C ; | z | < R } α + β > 0 , D R = { z ∈ C ; | z | < R } alpha+beta > 0,D_(R)={z inC;|z| < R}\alpha+\beta>0, \mathbb{D}_{R}=\{z \in \mathbb{C} ;|z|<R\}α+β>0,DR={z∈C;|z|<R} and let us suppose that f : D R → C f : D R → C f:D_(R)rarrCf: \mathbb{D}_{R} \rightarrow \mathbb{C}f:DR→C is analytic in D R D R D_(R)\mathbb{D}_{R}DR, that is we can write f ( z ) = ∑ k = 0 ∞ c k z k f ( z ) = ∑ k = 0 ∞   c k z k f(z)=sum_(k=0)^(oo)c_(k)z^(k)f(z)=\sum_{k=0}^{\infty} c_{k} z^{k}f(z)=∑k=0∞ckzk, for all z ∈ D R z ∈ D R z inD_(R)z \in \mathbb{D}_{R}z∈DR. If f f fff is not a polynomial of degree 0 and 1 ≤ r < R 1 ≤ r < R 1 <= r < R1 \leq r<R1≤r<R, then we have
‖ S n ( α , β ) ( f ) − f ‖ r ≥ C r ( α , β ) ( f ) n + β , n ∈ N S n ( α , β ) ( f ) − f r ≥ C r ( α , β ) ( f ) n + β , n ∈ N ||S_(n)^((alpha,beta))(f)-f||_(r) >= (C_(r)^((alpha,beta))(f))/(n+beta),n inN\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r} \geq \frac{C_{r}^{(\alpha, \beta)}(f)}{n+\beta}, n \in \mathbb{N}‖Sn(α,β)(f)−f‖r≥Cr(α,β)(f)n+β,n∈N
where the constant C r ( α , β ) ( f ) C r ( α , β ) ( f ) C_(r)^((alpha,beta))(f)C_{r}^{(\alpha, \beta)}(f)Cr(α,β)(f) depends only on f , r , α f , r , α f,r,alphaf, r, \alphaf,r,α and β β beta\betaβ.
Proof. For all z ∈ D R z ∈ D R z inD_(R)z \in \mathbb{D}_{R}z∈DR and n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N we have
S n ( α , β ) ( f ) ( z ) − f ( z ) = 1 n + β { − ( β z − α ) f ′ ( z ) + z ( 1 − z ) 2 f ′ ′ ( z ) + 1 n + β ⋅ ⋅ [ ( n + β ) 2 ( S n ( α , β ) ( f ) ( z ) − f ( z ) + β z − α n + β f ′ ( z ) − n z ( 1 − z ) 2 ( n + β ) 2 f ′ ′ ( z ) ) − β z ( 1 − z ) 2 f ′ ′ ( z ) ] } . S n ( α , β ) ( f ) ( z ) − f ( z ) = 1 n + β − ( β z − α ) f ′ ( z ) + z ( 1 − z ) 2 f ′ ′ ( z ) + 1 n + β ⋅ ⋅ ( n + β ) 2 S n ( α , β ) ( f ) ( z ) − f ( z ) + β z − α n + β f ′ ( z ) − n z ( 1 − z ) 2 ( n + β ) 2 f ′ ′ ( z ) − β z ( 1 − z ) 2 f ′ ′ ( z ) . {:[S_(n)^((alpha,beta))(f)(z)-f(z)=(1)/(n+beta){-(beta z-alpha)f^(')(z)+(z(1-z))/(2)f^('')(z)+(1)/(n+beta)*:}],[{: quad*[(n+beta)^(2)(S_(n)^((alpha,beta))(f)(z)-f(z)+(beta z-alpha)/(n+beta)f^(')(z)-(nz(1-z))/(2(n+beta)^(2))f^('')(z))-(beta z(1-z))/(2)f^('')(z)]}.]:}\begin{aligned} & S_{n}^{(\alpha, \beta)}(f)(z)-f(z)=\frac{1}{n+\beta}\left\{-(\beta z-\alpha) f^{\prime}(z)+\frac{z(1-z)}{2} f^{\prime \prime}(z)+\frac{1}{n+\beta} \cdot\right. \\ & \left.\quad \cdot\left[(n+\beta)^{2}\left(S_{n}^{(\alpha, \beta)}(f)(z)-f(z)+\frac{\beta z-\alpha}{n+\beta} f^{\prime}(z)-\frac{n z(1-z)}{2(n+\beta)^{2}} f^{\prime \prime}(z)\right)-\frac{\beta z(1-z)}{2} f^{\prime \prime}(z)\right]\right\} . \end{aligned}Sn(α,β)(f)(z)−f(z)=1n+β{−(βz−α)f′(z)+z(1−z)2f′′(z)+1n+β⋅⋅[(n+β)2(Sn(α,β)(f)(z)−f(z)+βz−αn+βf′(z)−nz(1−z)2(n+β)2f′′(z))−βz(1−z)2f′′(z)]}.
Note that in the case α = β = 0 α = β = 0 alpha=beta=0\alpha=\beta=0α=β=0 in [2], necessarily f f fff was supposed to be not a polynomial of degree ≤ 1 ≤ 1 <= 1\leq 1≤1.
In what follows we will apply to the above identity the following obvious property:
‖ F + G ‖ r ≥ | ‖ F ‖ r − ‖ G ‖ r | ≥ ‖ F ‖ r − ‖ G ‖ r ‖ F + G ‖ r ≥ ‖ F ‖ r − ‖ G ‖ r ≥ ‖ F ‖ r − ‖ G ‖ r ||F+G||_(r) >= |||F||_(r)-||G||_(r)| >= ||F||_(r)-||G||_(r)\|F+G\|_{r} \geq\left|\|F\|_{r}-\|G\|_{r}\right| \geq\|F\|_{r}-\|G\|_{r}‖F+G‖r≥|‖F‖r−‖G‖r|≥‖F‖r−‖G‖r
It follows
‖ S n ( α , β ) ( f ) − f ‖ r ≥ 1 n + β { ‖ − ( β e 1 − α ) f ′ + e 1 ( 1 − e 1 ) 2 f ′ ′ ‖ r − 1 n + β ⋅ ⋅ [ ‖ ( n + β ) 2 ( S n ( α , β ) ( f ) − f + β e 1 − α n + β f ′ − n e 1 ( 1 − e 1 ) 2 ( n + β ) 2 f ′ ′ ) − β e 1 ( 1 − e 1 ) 2 f ′ ′ ‖ r ] } . S n ( α , β ) ( f ) − f r ≥ 1 n + β − β e 1 − α f ′ + e 1 1 − e 1 2 f ′ ′ r − 1 n + β ⋅ ⋅ ( n + β ) 2 S n ( α , β ) ( f ) − f + β e 1 − α n + β f ′ − n e 1 1 − e 1 2 ( n + β ) 2 f ′ ′ − β e 1 1 − e 1 2 f ′ ′ r . {:[||S_(n)^((alpha,beta))(f)-f||_(r) >= (1)/(n+beta){||-(betae_(1)-alpha)f^(')+(e_(1)(1-e_(1)))/(2)f^('')||_(r)-(1)/(n+beta)*:}],[{: quad*[||(n+beta)^(2)(S_(n)^((alpha,beta))(f)-f+(betae_(1)-alpha)/(n+beta)f^(')-(ne_(1)(1-e_(1)))/(2(n+beta)^(2))f^(''))-(betae_(1)(1-e_(1)))/(2)f^('')||_(r)]}.]:}\begin{aligned} & \left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r} \geq \frac{1}{n+\beta}\left\{\left\|-\left(\beta e_{1}-\alpha\right) f^{\prime}+\frac{e_{1}\left(1-e_{1}\right)}{2} f^{\prime \prime}\right\|_{r}-\frac{1}{n+\beta} \cdot\right. \\ & \left.\quad \cdot\left[\left\|(n+\beta)^{2}\left(S_{n}^{(\alpha, \beta)}(f)-f+\frac{\beta e_{1}-\alpha}{n+\beta} f^{\prime}-\frac{n e_{1}\left(1-e_{1}\right)}{2(n+\beta)^{2}} f^{\prime \prime}\right)-\frac{\beta e_{1}\left(1-e_{1}\right)}{2} f^{\prime \prime}\right\|_{r}\right]\right\} . \end{aligned}‖Sn(α,β)(f)−f‖r≥1n+β{‖−(βe1−α)f′+e1(1−e1)2f′′‖r−1n+β⋅⋅[‖(n+β)2(Sn(α,β)(f)−f+βe1−αn+βf′−ne1(1−e1)2(n+β)2f′′)−βe1(1−e1)2f′′‖r]}.
Since by Remark 1.2 we have
‖ ( n + β ) 2 ( S n ( α , β ) ( f ) − f + β e 1 − α n + β f ′ − n e 1 ( 1 − e 1 ) 2 ( n + β ) 2 f ′ ′ ) − β e 1 ( 1 − e 1 ) 2 f ′ ′ ‖ r ≤ ≤ M r ( α , β ) ( f ) + β ‖ f ′ ′ ‖ r ( n + β ) 2 S n ( α , β ) ( f ) − f + β e 1 − α n + β f ′ − n e 1 1 − e 1 2 ( n + β ) 2 f ′ ′ − β e 1 1 − e 1 2 f ′ ′ r ≤ ≤ M r ( α , β ) ( f ) + β f ′ ′ r {:[||(n+beta)^(2)(S_(n)^((alpha,beta))(f)-f+(betae_(1)-alpha)/(n+beta)f^(')-(ne_(1)(1-e_(1)))/(2(n+beta)^(2))f^(''))-(betae_(1)(1-e_(1)))/(2)f^('')||_(r) <= ],[ <= M_(r)^((alpha,beta))(f)+beta||f^('')||_(r)]:}\begin{gathered} \left\|(n+\beta)^{2}\left(S_{n}^{(\alpha, \beta)}(f)-f+\frac{\beta e_{1}-\alpha}{n+\beta} f^{\prime}-\frac{n e_{1}\left(1-e_{1}\right)}{2(n+\beta)^{2}} f^{\prime \prime}\right)-\frac{\beta e_{1}\left(1-e_{1}\right)}{2} f^{\prime \prime}\right\|_{r} \leq \\ \leq M_{r}^{(\alpha, \beta)}(f)+\beta\left\|f^{\prime \prime}\right\|_{r} \end{gathered}‖(n+β)2(Sn(α,β)(f)−f+βe1−αn+βf′−ne1(1−e1)2(n+β)2f′′)−βe1(1−e1)2f′′‖r≤≤Mr(α,β)(f)+β‖f′′‖r
and denoting H ( z ) = − ( β z − α ) f ′ ( z ) + z ( 1 − z ) 2 f ′ ′ ( z ) H ( z ) = − ( β z − α ) f ′ ( z ) + z ( 1 − z ) 2 f ′ ′ ( z ) H(z)=-(beta z-alpha)f^(')(z)+(z(1-z))/(2)f^('')(z)H(z)=-(\beta z-\alpha) f^{\prime}(z)+\frac{z(1-z)}{2} f^{\prime \prime}(z)H(z)=−(βz−α)f′(z)+z(1−z)2f′′(z), if we prove that ‖ H ‖ r > 0 ‖ H ‖ r > 0 ||H||_(r) > 0\|H\|_{r}>0‖H‖r>0, then it is clear that there exists an index n 0 n 0 n_(0)n_{0}n0 depending only on f , α f , α f,alphaf, \alphaf,α and β β beta\betaβ, such that
‖ S n ( α , β ) ( f ) − f ‖ r ≥ 1 n + β ⋅ ‖ H ‖ r 2 , ∀ n ≥ n 0 S n ( α , β ) ( f ) − f r ≥ 1 n + β ⋅ ‖ H ‖ r 2 , ∀ n ≥ n 0 ||S_(n)^((alpha,beta))(f)-f||_(r) >= (1)/(n+beta)*(||H||_(r))/(2),AA n >= n_(0)\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r} \geq \frac{1}{n+\beta} \cdot \frac{\|H\|_{r}}{2}, \forall n \geq n_{0}‖Sn(α,β)(f)−f‖r≥1n+β⋅‖H‖r2,∀n≥n0
For n ∈ { 1 , 2 , … , n 0 − 1 } n ∈ 1 , 2 , … , n 0 − 1 n in{1,2,dots,n_(0)-1}n \in\left\{1,2, \ldots, n_{0}-1\right\}n∈{1,2,…,n0−1} we have ‖ S n ( α , β ) ( f ) − f ‖ r ≥ A n , r ( α , β ) ( f ) n + β S n ( α , β ) ( f ) − f r ≥ A n , r ( α , β ) ( f ) n + β ||S_(n)^((alpha,beta))(f)-f||_(r) >= (A_(n,r)^((alpha,beta))(f))/(n+beta)\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r} \geq \frac{A_{n, r}^{(\alpha, \beta)}(f)}{n+\beta}‖Sn(α,β)(f)−f‖r≥An,r(α,β)(f)n+β with A n , r ( α , β ) ( f ) = ( n + β ) ⋅ ‖ S n ( α , β ) ( f ) − f ‖ r > 0 A n , r ( α , β ) ( f ) = ( n + β ) ⋅ S n ( α , β ) ( f ) − f r > 0 A_(n,r)^((alpha,beta))(f)=(n+beta)*||S_(n)^((alpha,beta))(f)-f||_(r) > 0A_{n, r}^{(\alpha, \beta)}(f)= (n+\beta) \cdot\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r}>0An,r(α,β)(f)=(n+β)⋅‖Sn(α,β)(f)−f‖r>0, which finally implies ‖ S n ( α , β ) ( f ) − f ‖ r ≥ C r ( α , β ) ( f ) n + β S n ( α , β ) ( f ) − f r ≥ C r ( α , β ) ( f ) n + β ||S_(n)^((alpha,beta))(f)-f||_(r) >= (C_(r)^((alpha,beta))(f))/(n+beta)\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r} \geq \frac{C_{r}^{(\alpha, \beta)}(f)}{n+\beta}‖Sn(α,β)(f)−f‖r≥Cr(α,β)(f)n+β for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N, with C r ( α , β ) ( f ) = min { A 1 , r ( α , β ) , A 2 , r ( α , β ) ( f ) , … , A n 0 − 1 , r ( α , β ) ( f ) , ‖ H ‖ r 2 } C r ( α , β ) ( f ) = min A 1 , r ( α , β ) , A 2 , r ( α , β ) ( f ) , … , A n 0 − 1 , r ( α , β ) ( f ) , ‖ H ‖ r 2 C_(r)^((alpha,beta))(f)=min{A_(1,r)^((alpha,beta)),A_(2,r)^((alpha,beta))(f),dots,A_(n_(0)-1,r)^((alpha,beta))(f),(||H||_(r))/(2)}C_{r}^{(\alpha, \beta)}(f)=\min \left\{A_{1, r}^{(\alpha, \beta)}, A_{2, r}^{(\alpha, \beta)}(f), \ldots, A_{n_{0}-1, r}^{(\alpha, \beta)}(f), \frac{\|H\|_{r}}{2}\right\}Cr(α,β)(f)=min{A1,r(α,β),A2,r(α,β)(f),…,An0−1,r(α,β)(f),‖H‖r2}.
Therefore it remains to show that ‖ H ‖ r > 0 ‖ H ‖ r > 0 ||H||_(r) > 0\|H\|_{r}>0‖H‖r>0. Indeed, suppose that ‖ H ‖ r = ‖ H ‖ r = ||H||_(r)=\|H\|_{r}=‖H‖r= 0 . We have two possibilities: 1) 0 = α < β 0 = α < β 0=alpha < beta0=\alpha<\beta0=α<β or 2) 0 < α ≤ β 0 < α ≤ β 0 < alpha <= beta0<\alpha \leq \beta0<α≤β.
Case 1). We obtain H ( z ) = − β z f ′ ( z ) + z ( 1 − z ) 2 f ′ ′ ( z ) = 0 H ( z ) = − β z f ′ ( z ) + z ( 1 − z ) 2 f ′ ′ ( z ) = 0 H(z)=-beta zf^(')(z)+(z(1-z))/(2)f^('')(z)=0H(z)=-\beta z f^{\prime}(z)+\frac{z(1-z)}{2} f^{\prime \prime}(z)=0H(z)=−βzf′(z)+z(1−z)2f′′(z)=0, for all | z | ≤ r | z | ≤ r |z| <= r|z| \leq r|z|≤r and denoting y ( z ) = f ′ ( z ) y ( z ) = f ′ ( z ) y(z)=f^(')(z)y(z)=f^{\prime}(z)y(z)=f′(z), it follows that y ( z ) y ( z ) y(z)y(z)y(z) is an analytic function in D R D R D_(R)\mathbb{D}_{R}DR, solution of the differential equation − β z y ( z ) + z ( 1 − z ) 2 y ′ ( z ) = 0 , | z | ≤ r − β z y ( z ) + z ( 1 − z ) 2 y ′ ( z ) = 0 , | z | ≤ r -beta zy(z)+(z(1-z))/(2)y^(')(z)=0,|z| <= r-\beta z y(z)+\frac{z(1-z)}{2} y^{\prime}(z)=0,|z| \leq r−βzy(z)+z(1−z)2y′(z)=0,|z|≤r, which after simplification with z ≠ 0 z ≠ 0 z!=0z \neq 0z≠0 becomes − β y ( z ) + ( 1 − z ) 2 y ′ ( z ) = 0 , | z | ≤ r − β y ( z ) + ( 1 − z ) 2 y ′ ( z ) = 0 , | z | ≤ r -beta y(z)+((1-z))/(2)y^(')(z)=0,|z| <= r-\beta y(z)+\frac{(1-z)}{2} y^{\prime}(z)=0,|z| \leq r−βy(z)+(1−z)2y′(z)=0,|z|≤r. Now, seeking y ( z ) y ( z ) y(z)y(z)y(z) in the form y ( z ) = ∑ k = 0 ∞ b k z k y ( z ) = ∑ k = 0 ∞   b k z k y(z)=sum_(k=0)^(oo)b_(k)z^(k)y(z)=\sum_{k=0}^{\infty} b_{k} z^{k}y(z)=∑k=0∞bkzk and replacing it in the differential equation, by the identification of the coefficients we easily obtain b k = 0 b k = 0 b_(k)=0b_{k}=0bk=0 for all k = 0 , 1 , … k = 0 , 1 , … k=0,1,dotsk=0,1, \ldotsk=0,1,…. Therefore y ( z ) = 0 y ( z ) = 0 y(z)=0y(z)=0y(z)=0 for all | z | ≤ r | z | ≤ r |z| <= r|z| \leq r|z|≤r, which by the identity theorem on analytic (holomorphic) functions implies y ( z ) = 0 y ( z ) = 0 y(z)=0y(z)=0y(z)=0 for all z ∈ D R z ∈ D R z inD_(R)z \in \mathbb{D}_{R}z∈DR and the contradiction that f f fff is a polynomial of degree ≤ 0 ≤ 0 <= 0\leq 0≤0.
Case 2). Denoting y ( z ) = f ′ ( z ) y ( z ) = f ′ ( z ) y(z)=f^(')(z)y(z)=f^{\prime}(z)y(z)=f′(z) by hypothesis it follows that y ( z ) y ( z ) y(z)y(z)y(z) is an analytic function in D R D R D_(R)\mathbb{D}_{R}DR solution of the differential equation ( − β z + α ) y ( z ) + z ( 1 − z ) 2 y ′ ( z ) = 0 , | z | ≤ r ( − β z + α ) y ( z ) + z ( 1 − z ) 2 y ′ ( z ) = 0 , | z | ≤ r (-beta z+alpha)y(z)+(z(1-z))/(2)y^(')(z)=0,|z| <= r(-\beta z+\alpha) y(z)+ \frac{z(1-z)}{2} y^{\prime}(z)=0,|z| \leq r(−βz+α)y(z)+z(1−z)2y′(z)=0,|z|≤r.
Taking z = 0 z = 0 z=0z=0z=0 it follows α y ( 0 ) = 0 α y ( 0 ) = 0 alpha y(0)=0\alpha y(0)=0αy(0)=0, which means y ( 0 ) = 0 y ( 0 ) = 0 y(0)=0y(0)=0y(0)=0. Seeking y ( z ) y ( z ) y(z)y(z)y(z) in the form y ( z ) = ∑ k = 1 ∞ b k z k y ( z ) = ∑ k = 1 ∞   b k z k y(z)=sum_(k=1)^(oo)b_(k)z^(k)y(z)=\sum_{k=1}^{\infty} b_{k} z^{k}y(z)=∑k=1∞bkzk and replacing it in the differential equation, by the
identification of the coefficients we easily obtain b k = 0 b k = 0 b_(k)=0b_{k}=0bk=0 for all k = 1 , 2 , … k = 1 , 2 , … k=1,2,dotsk=1,2, \ldotsk=1,2,…, which finally leads to the contradiction that f f fff is a constant.
Combining now Theorem 2.1 with Theorem 1.1, (i), we immediately get the following.
Corollary 2.2. Let R > 1 , 0 ≤ α ≤ β R > 1 , 0 ≤ α ≤ β R > 1,0 <= alpha <= betaR>1,0 \leq \alpha \leq \betaR>1,0≤α≤β with α + β > 0 , D R = { z ∈ C ; | z | < R } α + β > 0 , D R = { z ∈ C ; | z | < R } alpha+beta > 0,D_(R)={z inC;|z| < R}\alpha+\beta>0, \mathbb{D}_{R}=\{z \in \mathbb{C} ;|z|<R\}α+β>0,DR={z∈C;|z|<R} and let us suppose that f : D R → C f : D R → C f:D_(R)rarrCf: \mathbb{D}_{R} \rightarrow \mathbb{C}f:DR→C is analytic in D R D R D_(R)\mathbb{D}_{R}DR. If f f fff is not a polynomial of degree 0 and 1 ≤ r < R 1 ≤ r < R 1 <= r < R1 \leq r<R1≤r<R, then we have
‖ S n ( α , β ) ( f ) − f ‖ r ∼ 1 n + β , n ∈ N S n ( α , β ) ( f ) − f r ∼ 1 n + β , n ∈ N ||S_(n)^((alpha,beta))(f)-f||_(r)∼(1)/(n+beta),n inN\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r} \sim \frac{1}{n+\beta}, n \in \mathbb{N}‖Sn(α,β)(f)−f‖r∼1n+β,n∈N
where the constants in the equivalence depend on f , r , α f , r , α f,r,alphaf, r, \alphaf,r,α and β β beta\betaβ.

3. EXACT ORDERS OF APPROXIMATION FOR DERIVATIVES OF COMPLEX BERNSTEIN-STANCU POLYNOMIALS

The main result of this section is the following.
Theorem 3.1. Let D R = { z ∈ C ; | z | < R } D R = { z ∈ C ; | z | < R } D_(R)={z inC;|z| < R}\mathbb{D}_{R}=\{z \in \mathbb{C} ;|z|<R\}DR={z∈C;|z|<R} be with R > 1 , 0 ≤ α ≤ β R > 1 , 0 ≤ α ≤ β R > 1,0 <= alpha <= betaR>1,0 \leq \alpha \leq \betaR>1,0≤α≤β with α + β > 0 α + β > 0 alpha+beta > 0\alpha+\beta>0α+β>0 and let us suppose that f : D R → C f : D R → C f:D_(R)rarrCf: \mathbb{D}_{R} \rightarrow \mathbb{C}f:DR→C is analytic in D R D R D_(R)\mathbb{D}_{R}DR, i.e. f ( z ) = ∑ k = 0 ∞ c k z k f ( z ) = ∑ k = 0 ∞   c k z k f(z)=sum_(k=0)^(oo)c_(k)z^(k)f(z)=\sum_{k=0}^{\infty} c_{k} z^{k}f(z)=∑k=0∞ckzk, for all z ∈ D R z ∈ D R z inD_(R)z \in \mathbb{D}_{R}z∈DR. Also, let 1 ≤ r < r 1 < R 1 ≤ r < r 1 < R 1 <= r < r_(1) < R1 \leq r<r_{1}<R1≤r<r1<R and p ∈ N p ∈ N p inNp \in \mathbb{N}p∈N be fixed. If f f fff is not a polynomial of degree ≤ p − 1 ≤ p − 1 <= p-1\leq p-1≤p−1, then we have
‖ [ S n ( α , β ) ( f ) ] ( p ) − f ( p ) ‖ r ∼ 1 n + β S n ( α , β ) ( f ) ( p ) − f ( p ) r ∼ 1 n + β ||[S_(n)^((alpha,beta))(f)]^((p))-f^((p))||_(r)∼(1)/(n+beta)\left\|\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}-f^{(p)}\right\|_{r} \sim \frac{1}{n+\beta}‖[Sn(α,β)(f)](p)−f(p)‖r∼1n+β
where the constants in the equivalence depend on f , α , β , r , r 1 f , α , β , r , r 1 f,alpha,beta,r,r_(1)f, \alpha, \beta, r, r_{1}f,α,β,r,r1 and p p ppp.
Proof. Taking into account Theorem 1.1, (ii), it remains only to prove the lower estimate for ‖ [ S n ( α , β ) ( f ) ] ( p ) − f ( p ) ‖ r S n ( α , β ) ( f ) ( p ) − f ( p ) r ||[S_(n)^((alpha,beta))(f)]^((p))-f^((p))||_(r)\left\|\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}-f^{(p)}\right\|_{r}‖[Sn(α,β)(f)](p)−f(p)‖r.
Denoting by Γ Γ Gamma\GammaΓ the circle of radius r 1 > r r 1 > r r_(1) > rr_{1}>rr1>r (with r ≥ 1 r ≥ 1 r >= 1r \geq 1r≥1 ) and center 0 , by the Cauchy's formulas it follows that for all | z | ≤ r | z | ≤ r |z| <= r|z| \leq r|z|≤r and n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N we have
[ S n ( α , β ) ( f ) ] ( p ) ( z ) − f ( p ) ( z ) = p ! 2 π i ∫ Γ S n ( α , β ) ( f ) ( v ) − f ( v ) ( v − z ) p + 1 d v S n ( α , β ) ( f ) ( p ) ( z ) − f ( p ) ( z ) = p ! 2 π i ∫ Γ   S n ( α , β ) ( f ) ( v ) − f ( v ) ( v − z ) p + 1 d v [S_(n)^((alpha,beta))(f)]^((p))(z)-f^((p))(z)=(p!)/(2pii)int_(Gamma)(S_(n)^((alpha,beta))(f)(v)-f(v))/((v-z)^(p+1))dv\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}(z)-f^{(p)}(z)=\frac{p!}{2 \pi \mathrm{i}} \int_{\Gamma} \frac{S_{n}^{(\alpha, \beta)}(f)(v)-f(v)}{(v-z)^{p+1}} \mathrm{~d} v[Sn(α,β)(f)](p)(z)−f(p)(z)=p!2πi∫ΓSn(α,β)(f)(v)−f(v)(v−z)p+1 dv
where we have the inequality | v − z | ≥ r 1 − r | v − z | ≥ r 1 − r |v-z| >= r_(1)-r|v-z| \geq r_{1}-r|v−z|≥r1−r valid for all | z | ≤ r | z | ≤ r |z| <= r|z| \leq r|z|≤r and v ∈ Γ v ∈ Γ v in Gammav \in \Gammav∈Γ.
As in the proof of Theorem 2.1 (keeping the notation for H H HHH ), for all v ∈ Γ v ∈ Γ v in Gammav \in \Gammav∈Γ and n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N we have
S n ( α , β ) ( f ) ( v ) − f ( v ) = = 1 n + β { H ( v ) + 1 n + β [ ( n + β ) 2 ( S n ( α , β ) ( f ) ( v ) − f ( v ) + β v − α n + β f ′ ( v ) − n v ( 1 − v ) 2 ( n + β ) 2 f ′ ′ ( v ) ) − β v ( 1 − v ) 2 f ′ ′ ( v ) ] } S n ( α , β ) ( f ) ( v ) − f ( v ) = = 1 n + β H ( v ) + 1 n + β ( n + β ) 2 S n ( α , β ) ( f ) ( v ) − f ( v ) + β v − α n + β f ′ ( v ) − n v ( 1 − v ) 2 ( n + β ) 2 f ′ ′ ( v ) − β v ( 1 − v ) 2 f ′ ′ ( v ) {:[S_(n)^((alpha,beta))(f)(v)-f(v)=],[=(1)/(n+beta){H(v)+(1)/(n+beta)[(n+beta)^(2)(S_(n)^((alpha,beta))(f)(v)-f(v)+:}],[{: quad(beta v-alpha)/(n+beta)f^(')(v)-(nv(1-v))/(2(n+beta)^(2))f^('')(v))-(beta v(1-v))/(2)f^('')(v)]}]:}\begin{aligned} & S_{n}^{(\alpha, \beta)}(f)(v)-f(v)= \\ & =\frac{1}{n+\beta}\left\{H(v)+\frac{1}{n+\beta}\left[( n + \beta ) ^ { 2 } \left(S_{n}^{(\alpha, \beta)}(f)(v)-f(v)+\right.\right.\right. \\ & \left.\left.\left.\quad \frac{\beta v-\alpha}{n+\beta} f^{\prime}(v)-\frac{n v(1-v)}{2(n+\beta)^{2}} f^{\prime \prime}(v)\right)-\frac{\beta v(1-v)}{2} f^{\prime \prime}(v)\right]\right\} \end{aligned}Sn(α,β)(f)(v)−f(v)==1n+β{H(v)+1n+β[(n+β)2(Sn(α,β)(f)(v)−f(v)+βv−αn+βf′(v)−nv(1−v)2(n+β)2f′′(v))−βv(1−v)2f′′(v)]}
which replaced in the above Cauchy's formula implies
[ S n ( α , β ) ( f ) ] ( p ) ( z ) − f ( p ) ( z ) = 1 n + β { H ( p ) ( z ) + 1 n + β ⋅ [ p ! 2 π i ∫ Γ ( n + β ) 2 ( S n ( α , β ) ( f ) ( v ) − f ( v ) + β v − α n + β f ′ ( v ) − n v ( 1 − v ) 2 ( n + β ) 2 f ′ ′ ( v ) ) ( v − z ) p + 1 d v − − p ! 2 π i ∫ Γ β v ( 1 − v ) 2 ( v − z ) p + 1 f ′ ′ ( v ) d v ] } S n ( α , β ) ( f ) ( p ) ( z ) − f ( p ) ( z ) = 1 n + β H ( p ) ( z ) + 1 n + β ⋅ p ! 2 π i ∫ Γ   ( n + β ) 2 S n ( α , β ) ( f ) ( v ) − f ( v ) + β v − α n + β f ′ ( v ) − n v ( 1 − v ) 2 ( n + β ) 2 f ′ ′ ( v ) ( v − z ) p + 1 d v − − p ! 2 π i ∫ Γ   β v ( 1 − v ) 2 ( v − z ) p + 1 f ′ ′ ( v ) d v {:[[S_(n)^((alpha,beta))(f)]^((p))(z)-f^((p))(z)=(1)/(n+beta){H^((p))(z)+(1)/(n+beta):}],[quad*[(p!)/(2pii)int_(Gamma)((n+beta)^(2)(S_(n)^((alpha,beta))(f)(v)-f(v)+(beta v-alpha)/(n+beta)f^(')(v)-(nv(1-v))/(2(n+beta)^(2))f^('')(v)))/((v-z)^(p+1))(d)v-:}],[{: quad-(p!)/(2pii)int_(Gamma)(beta v(1-v))/(2(v-z)^(p+1))f^('')(v)dv]}]:}\begin{aligned} & {\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}(z)-f^{(p)}(z)=\frac{1}{n+\beta}\left\{H^{(p)}(z)+\frac{1}{n+\beta}\right.} \\ & \quad \cdot\left[\frac{p!}{2 \pi \mathrm{i}} \int_{\Gamma} \frac{(n+\beta)^{2}\left(S_{n}^{(\alpha, \beta)}(f)(v)-f(v)+\frac{\beta v-\alpha}{n+\beta} f^{\prime}(v)-\frac{n v(1-v)}{2(n+\beta)^{2}} f^{\prime \prime}(v)\right)}{(v-z)^{p+1}} \mathrm{~d} v-\right. \\ & \left.\left.\quad-\frac{p!}{2 \pi \mathrm{i}} \int_{\Gamma} \frac{\beta v(1-v)}{2(v-z)^{p+1}} f^{\prime \prime}(v) \mathrm{d} v\right]\right\} \end{aligned}[Sn(α,β)(f)](p)(z)−f(p)(z)=1n+β{H(p)(z)+1n+β⋅[p!2πi∫Γ(n+β)2(Sn(α,β)(f)(v)−f(v)+βv−αn+βf′(v)−nv(1−v)2(n+β)2f′′(v))(v−z)p+1 dv−−p!2πi∫Γβv(1−v)2(v−z)p+1f′′(v)dv]}
Passing now to absolute value, for all | z | ≤ r | z | ≤ r |z| <= r|z| \leq r|z|≤r and n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N it follows
| [ S n ( α , β ) ( f ) ] ( p ) ( z ) − f ( p ) ( z ) | ≥ 1 n + β { | H ( p ) ( z ) | − 1 n + β ⋅ [ | p ! 2 π i ∫ Γ ( n + β ) 2 ( S n ( α , β ) ( f ) ( v ) − f ( v ) + β v − α n + β f ′ ( v ) − n v ( 1 − v ) 2 ( n + β ) 2 f ′ ′ ( v ) ) ( v − z ) p + 1 d v − p ! 2 π i ∫ Γ β v ( 1 − v ) 2 ( v − z ) p + 1 f ′ ′ ( v ) d v | ] } S n ( α , β ) ( f ) ( p ) ( z ) − f ( p ) ( z ) ≥ 1 n + β H ( p ) ( z ) − 1 n + β ⋅ p ! 2 π i ∫ Γ   ( n + β ) 2 S n ( α , β ) ( f ) ( v ) − f ( v ) + β v − α n + β f ′ ( v ) − n v ( 1 − v ) 2 ( n + β ) 2 f ′ ′ ( v ) ( v − z ) p + 1 d v − p ! 2 π i ∫ Γ   β v ( 1 − v ) 2 ( v − z ) p + 1 f ′ ′ ( v ) d v {:[|[S_(n)^((alpha,beta))(f)]^((p))(z)-f^((p))(z)| >= (1)/(n+beta){|H^((p))(z)|-(1)/(n+beta)*:}],[[|(p!)/(2pii)int_(Gamma)((n+beta)^(2)(S_(n)^((alpha,beta))(f)(v)-f(v)+(beta v-alpha)/(n+beta)f^(')(v)-(nv(1-v))/(2(n+beta)^(2))f^('')(v)))/((v-z)^(p+1))(d)v-:}],[{: quad(p!)/(2pii)int_(Gamma)(beta v(1-v))/(2(v-z)^(p+1))f^('')(v)dv|]}]:}\begin{aligned} & \left|\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}(z)-f^{(p)}(z)\right| \geq \frac{1}{n+\beta}\left\{\left|H^{(p)}(z)\right|-\frac{1}{n+\beta} \cdot\right. \\ & {\left[\left\lvert\, \frac{p!}{2 \pi \mathrm{i}} \int_{\Gamma} \frac{(n+\beta)^{2}\left(S_{n}^{(\alpha, \beta)}(f)(v)-f(v)+\frac{\beta v-\alpha}{n+\beta} f^{\prime}(v)-\frac{n v(1-v)}{2(n+\beta)^{2}} f^{\prime \prime}(v)\right)}{(v-z)^{p+1}} \mathrm{~d} v-\right.\right.} \\ & \left.\left.\left.\quad \frac{p!}{2 \pi \mathrm{i}} \int_{\Gamma} \frac{\beta v(1-v)}{2(v-z)^{p+1}} f^{\prime \prime}(v) \mathrm{d} v \right\rvert\,\right]\right\} \end{aligned}|[Sn(α,β)(f)](p)(z)−f(p)(z)|≥1n+β{|H(p)(z)|−1n+β⋅[|p!2πi∫Γ(n+β)2(Sn(α,β)(f)(v)−f(v)+βv−αn+βf′(v)−nv(1−v)2(n+β)2f′′(v))(v−z)p+1 dv−p!2πi∫Γβv(1−v)2(v−z)p+1f′′(v)dv|]}
where by using the Remark 1.2, for all | z | ≤ r | z | ≤ r |z| <= r|z| \leq r|z|≤r and n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N we get
| p ! 2 π i ∫ Γ ( n + β ) 2 ( S n ( α , β ) ( f ) ( v ) − f ( v ) + β v − α n + β f ′ ( v ) − n v ( 1 − v ) 2 ( n + β ) 2 f ′ ′ ( v ) ) ( v − z ) p + 1 d v − − p ! 2 π i ∫ Γ β v ( 1 − v ) 2 ( v − z ) p + 1 f ′ ′ ( v ) d v | ≤ p ! 2 π ⋅ 2 π r 1 M r ( α , β ) ( r 1 − r ) p + 1 + p ! 2 π ⋅ 2 π r 1 β r 1 ( 1 + r 1 ) ‖ f ′ ′ ‖ r 1 2 ( r 1 − r ) p + 1 p ! 2 π i ∫ Γ   ( n + β ) 2 S n ( α , β ) ( f ) ( v ) − f ( v ) + β v − α n + β f ′ ( v ) − n v ( 1 − v ) 2 ( n + β ) 2 f ′ ′ ( v ) ( v − z ) p + 1 d v − − p ! 2 π i ∫ Γ   β v ( 1 − v ) 2 ( v − z ) p + 1 f ′ ′ ( v ) d v ≤ p ! 2 π ⋅ 2 π r 1 M r ( α , β ) r 1 − r p + 1 + p ! 2 π ⋅ 2 π r 1 β r 1 1 + r 1 f ′ ′ r 1 2 r 1 − r p + 1 {:[|(p!)/(2pii)int_(Gamma)((n+beta)^(2)(S_(n)^((alpha,beta))(f)(v)-f(v)+(beta v-alpha)/(n+beta)f^(')(v)-(nv(1-v))/(2(n+beta)^(2))f^('')(v)))/((v-z)^(p+1))(d)v-:}],[quad-(p!)/(2pii)int_(Gamma)(beta v(1-v))/(2(v-z)^(p+1))f^('')(v)dv| <= (p!)/(2pi)*(2pir_(1)M_(r)^((alpha,beta)))/((r_(1)-r)^(p+1))+(p!)/(2pi)*(2pir_(1)betar_(1)(1+r_(1))||f^('')||_(r_(1)))/(2(r_(1)-r)^(p+1)):}]:}\begin{aligned} & \left\lvert\, \frac{p!}{2 \pi \mathrm{i}} \int_{\Gamma} \frac{(n+\beta)^{2}\left(S_{n}^{(\alpha, \beta)}(f)(v)-f(v)+\frac{\beta v-\alpha}{n+\beta} f^{\prime}(v)-\frac{n v(1-v)}{2(n+\beta)^{2}} f^{\prime \prime}(v)\right)}{(v-z)^{p+1}} \mathrm{~d} v-\right. \\ & \quad-\frac{p!}{2 \pi \mathrm{i}} \int_{\Gamma} \frac{\beta v(1-v)}{2(v-z)^{p+1}} f^{\prime \prime}(v) \mathrm{d} v \left\lvert\, \leq \frac{p!}{2 \pi} \cdot \frac{2 \pi r_{1} M_{r}^{(\alpha, \beta)}}{\left(r_{1}-r\right)^{p+1}}+\frac{p!}{2 \pi} \cdot \frac{2 \pi r_{1} \beta r_{1}\left(1+r_{1}\right)\left\|f^{\prime \prime}\right\|_{r_{1}}}{2\left(r_{1}-r\right)^{p+1}}\right. \end{aligned}|p!2πi∫Γ(n+β)2(Sn(α,β)(f)(v)−f(v)+βv−αn+βf′(v)−nv(1−v)2(n+β)2f′′(v))(v−z)p+1 dv−−p!2πi∫Γβv(1−v)2(v−z)p+1f′′(v)dv|≤p!2π⋅2πr1Mr(α,β)(r1−r)p+1+p!2π⋅2πr1βr1(1+r1)‖f′′‖r12(r1−r)p+1
Denoting now F p ( z ) = H ( p ) ( z ) F p ( z ) = H ( p ) ( z ) F_(p)(z)=H^((p))(z)F_{p}(z)=H^{(p)}(z)Fp(z)=H(p)(z), we prove that ‖ F p ‖ r > 0 F p r > 0 ||F_(p)||_(r) > 0\left\|F_{p}\right\|_{r}>0‖Fp‖r>0. Indeed, if we suppose that ‖ F p ‖ r = 0 F p r = 0 ||F_(p)||_(r)=0\left\|F_{p}\right\|_{r}=0‖Fp‖r=0 then it follows that f f fff satisfies the differential equation
− β z f ′ ( z ) + z ( 1 − z ) 2 f ′ ′ ( z ) = Q p − 1 ( z ) , ∀ | z | ≤ r − β z f ′ ( z ) + z ( 1 − z ) 2 f ′ ′ ( z ) = Q p − 1 ( z ) , ∀ | z | ≤ r -beta zf^(')(z)+(z(1-z))/(2)f^('')(z)=Q_(p-1)(z),AA|z| <= r-\beta z f^{\prime}(z)+\frac{z(1-z)}{2} f^{\prime \prime}(z)=Q_{p-1}(z), \forall|z| \leq r−βzf′(z)+z(1−z)2f′′(z)=Qp−1(z),∀|z|≤r
where Q p − 1 ( z ) Q p − 1 ( z ) Q_(p-1)(z)Q_{p-1}(z)Qp−1(z) is a polynomial of degree ≤ p − 1 ≤ p − 1 <= p-1\leq p-1≤p−1. Simplifying with z z zzz, making the substitution y ( z ) = f ′ ( z ) y ( z ) = f ′ ( z ) y(z)=f^(')(z)y(z)=f^{\prime}(z)y(z)=f′(z), searching y ( z ) y ( z ) y(z)y(z)y(z) in the form y ( z ) = ∑ k = 0 ∞ b k z k y ( z ) = ∑ k = 0 ∞   b k z k y(z)=sum_(k=0)^(oo)b_(k)z^(k)y(z)=\sum_{k=0}^{\infty} b_{k} z^{k}y(z)=∑k=0∞bkzk and then replacing in the differential equation, by simple calculations we easily obtain that b k = 0 b k = 0 b_(k)=0b_{k}=0bk=0 for all k ≥ p − 1 k ≥ p − 1 k >= p-1k \geq p-1k≥p−1, that is y ( z ) y ( z ) y(z)y(z)y(z) is a polynomial of degree ≤ p − 2 ≤ p − 2 <= p-2\leq p-2≤p−2. This implies the contradiction that f f fff is a polynomial of degree ≤ p − 1 ≤ p − 1 <= p-1\leq p-1≤p−1.
Continuing exactly as in the proof of Theorem 2.1 (with ‖ S n ( α , β ) ( f ) − f ‖ r S n ( α , β ) ( f ) − f r ||S_(n)^((alpha,beta))(f)-f||_(r)\left\|S_{n}^{(\alpha, \beta)}(f)-f\right\|_{r}‖Sn(α,β)(f)−f‖r replaced by ‖ [ S n ( α , β ) ( f ) ] ( p ) − f ( p ) ‖ r S n ( α , β ) ( f ) ( p ) − f ( p ) r ||[S_(n)^((alpha,beta))(f)]^((p))-f^((p))||_(r)\left\|\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}-f^{(p)}\right\|_{r}‖[Sn(α,β)(f)](p)−f(p)‖r ), finally there exists an index n 0 ∈ N n 0 ∈ N n_(0)inNn_{0} \in \mathbb{N}n0∈N depending on f , r , r 1 f , r , r 1 f,r,r_(1)f, r, r_{1}f,r,r1 and p p ppp, such that for all n ≥ n 0 n ≥ n 0 n >= n_(0)n \geq n_{0}n≥n0 we have
‖ [ S n ( α , β ) ( f ) ] ( p ) − f ( p ) ‖ r ≥ 1 n ⋅ C 0 2 S n ( α , β ) ( f ) ( p ) − f ( p ) r ≥ 1 n ⋅ C 0 2 ||[S_(n)^((alpha,beta))(f)]^((p))-f^((p))||_(r) >= (1)/(n)*(C_(0))/(2)\left\|\left[S_{n}^{(\alpha, \beta)}(f)\right]^{(p)}-f^{(p)}\right\|_{r} \geq \frac{1}{n} \cdot \frac{C_{0}}{2}‖[Sn(α,β)(f)](p)−f(p)‖r≥1n⋅C02
Also, the cases when n ∈ { 1 , 2 , … , n 0 − 1 } n ∈ 1 , 2 , … , n 0 − 1 n in{1,2,dots,n_(0)-1}n \in\left\{1,2, \ldots, n_{0}-1\right\}n∈{1,2,…,n0−1} are similar with those in the proof of Theorem 2.1.

REFERENCES

[1] Gal, S.G., Approximation by complex Bernstein-Stancu polynomials in compact disks, Results in Mathematics, 2008, accepted for publication.
[2] Gal, S.G., Exact orders in simultaneous approximation by complex Bernstein polynomials, J. Concr. Applic. Math., 2009, accepted for publication.
Received by the editors: March 2, 2008.
  1. † † ^(†){ }^{\dagger}† This work has been supported by the Romanian Ministry of Education and Research, under CEEX grant: 2-CEx 06-11-96.
    *Department of Mathematics and Computer Science, University of Oradea, Universităţii str., no. 1, 410087 Oradea, Romania, e-mail: galso@uoradea.ro.

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