Rate of approximation for certain Durrmeyer operators

Maria Crăciun
(original), Approximation Theory, paper

Abstract

In the present note, we study a certain Durrmeyer type integral modification of Bernstein polynomials. We investigate simultaneous approximation and estimate the rate of convergence in simultaneous approximation.

Authors

V. GUPTA
School of Applied Sciences, Netaji Subhas Institute of Technology, Sector 3 Dwarka, New Delhi 110075, India

T. SHERVASHIDZE
A. Razmadze Mathematical Institute, Georgian Academy of Science 1, M. Aleksidze St., Tbilisi 0193 Georgia

M. Craciun
Tiberiu Popoviciu Institute of Numerical Analysis (Romanian Academy)

Keywords

Lebesgue integrable functions; Bernstein polynomials; functions of bounded variation.

PDF

http://www.heldermann-verlag.de/gmj/gmj13/gmj13024.pdf

Cite this paper as:

V. Gupta, T. Shervashidze, M. Crăciun, Rate of approximation for certain Durrmeyer operators, Georgian Mathematical Journal, vol. 13 (2006), no.2, 277-284.

About this paper

Journal

Georgian Mathematical Journal

Publisher Name

De Gruyter

DOI

10.1515/GMJ.2006.277

Print ISSN

1072-947X

Online ISSN

Not available yet.

References

[1] O. Agratini, On the rate of convergence of some integral operators for functions of bounded variation. Studia Sci. Math. Hungar. 42(2005), No. 2, 235–252

[2] R. N. Bhattacharya and R. Ranga Rao, Normal approximation and asymptotic expansions. Wiley Series in Probability and Mathematical Statistics. John Wiley & Sons, New York–London–Sydney, 1976.

[3] J. L. Durrmeyer, Une formule d’inversion de la transformee de Laplace: Application a la Theorie des Moments. These de 3e cycles, Faculte des Sciences de l’ Universite de Paris, 1967.

[4] S. S. Guo, On the rate of convergence of the Durrmeyer operator for functions of bounded variation. J. Approx. Theory 51(1987), No. 2, 183–192.

[5] V. Gupta, A note on the rate of convergence of Durrmeyer type operators for function of bounded variation. Soochow J. Math. 23(1997), No. 1, 115–118.

[6] V. Gupta and P. Maheshwari, Bezier variant of a new Durrmeyer type operators. Riv. Mat. Univ. Parma (7) 2(2003), 9–21.

[7] V. Gupta and G. S. Srivastava, Approximation by Durrmeyer-type operators. Ann. Polon. Math. 64(1996), No. 2, 153–159.

[8] X. M. Zeng, Bounds for Bernstein basis functions and Meyer–Konig and Zeller basis functions, J. Math. Anal. Appl. 219(1998), No. 2, 364–376.

Paper (preprint) in HTML form

gmj13024

RATE OF APPROXIMATION FOR CERTAIN DURRMEYER OPERATORS

VIJAY GUPTA, TENGIZ SHERVASHIDZE, AND MARIA CRACIUN

Abstract

In the present note, we study a certain Durrmeyer type integral modification of Bernstein polynomials. We investigate simultaneous approximation and estimate the rate of convergence in simultaneous approximation.

2000 Mathematics Subject Classification: 41A30, 41A36.
Key words and phrases: Lebesgue integrable functions, Bernstein polynomials, functions of bounded variation.

1. Introduction

Durrmeyer [3] introduced the integral modification of Bernstein polynomials to approximate Lebesgue integrable functions on the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1]. The operators introduced by Durrmeyer are defined by
(1) D n ( f , x ) = ( n + 1 ) ∑ k = 0 n p n , k ( x ) ∫ 0 1 p n , k ( t ) f ( t ) d t , x ∈ [ 0 , 1 ] (1) D n ( f , x ) = ( n + 1 ) ∑ k = 0 n   p n , k ( x ) ∫ 0 1   p n , k ( t ) f ( t ) d t , x ∈ [ 0 , 1 ] {:(1)D_(n)(f","x)=(n+1)sum_(k=0)^(n)p_(n,k)(x)int_(0)^(1)p_(n,k)(t)f(t)dt","quad x in[0","1]:}\begin{equation*} D_{n}(f, x)=(n+1) \sum_{k=0}^{n} p_{n, k}(x) \int_{0}^{1} p_{n, k}(t) f(t) d t, \quad x \in[0,1] \tag{1} \end{equation*}(1)Dn(f,x)=(n+1)∑k=0npn,k(x)∫01pn,k(t)f(t)dt,x∈[0,1]
where p n , k = ( n k ) x k ( 1 − x ) n − k p n , k = ( n k ) x k ( 1 − x ) n − k p_(n,k)=((n)/(k))x^(k)(1-x)^(n-k)p_{n, k}=\binom{n}{k} x^{k}(1-x)^{n-k}pn,k=(nk)xk(1−x)n−k.
Gupta [5] introduced a different Durrmeyer type modification of Bernstein polynomials and estimated the rate of convergence for functions of bounded variation. The operators introduced in [5] are defined by
(2) B n ( f , x ) = ∑ k = 0 n p n , k ( x ) ∫ 0 1 b n , k ( t ) f ( t ) d t , x ∈ [ 0 , 1 ] (2) B n ( f , x ) = ∑ k = 0 n   p n , k ( x ) ∫ 0 1   b n , k ( t ) f ( t ) d t , x ∈ [ 0 , 1 ] {:(2)B_(n)(f","x)=sum_(k=0)^(n)p_(n,k)(x)int_(0)^(1)b_(n,k)(t)f(t)dt","quad x in[0","1]:}\begin{equation*} B_{n}(f, x)=\sum_{k=0}^{n} p_{n, k}(x) \int_{0}^{1} b_{n, k}(t) f(t) d t, \quad x \in[0,1] \tag{2} \end{equation*}(2)Bn(f,x)=∑k=0npn,k(x)∫01bn,k(t)f(t)dt,x∈[0,1]
where
p n , k ( x ) = ( − 1 ) k x k k ! ϕ n ( k ) ( x ) , b n , k ( x ) = ( − 1 ) k + 1 x k k ! ϕ n ( k + 1 ) ( x ) p n , k ( x ) = ( − 1 ) k x k k ! ϕ n ( k ) ( x ) , b n , k ( x ) = ( − 1 ) k + 1 x k k ! ϕ n ( k + 1 ) ( x ) p_(n,k)(x)=(-1)^(k)(x^(k))/(k!)phi_(n)^((k))(x),quadb_(n,k)(x)=(-1)^(k+1)(x^(k))/(k!)phi_(n)^((k+1))(x)p_{n, k}(x)=(-1)^{k} \frac{x^{k}}{k!} \phi_{n}^{(k)}(x), \quad b_{n, k}(x)=(-1)^{k+1} \frac{x^{k}}{k!} \phi_{n}^{(k+1)}(x)pn,k(x)=(−1)kxkk!ϕn(k)(x),bn,k(x)=(−1)k+1xkk!ϕn(k+1)(x)
and ϕ n ( x ) = ( 1 − x ) n ϕ n ( x ) = ( 1 − x ) n phi_(n)(x)=(1-x)^(n)\phi_{n}(x)=(1-x)^{n}ϕn(x)=(1−x)n. It is easily verified that the values of p n , k ( x ) p n , k ( x ) p_(n,k)(x)p_{n, k}(x)pn,k(x) used in (1) and (2) are the same. Also, ∑ k = 0 n p n , k ( x ) = 1 , ∫ 0 1 b n , k ( t ) d t = 1 ∑ k = 0 n   p n , k ( x ) = 1 , ∫ 0 1   b n , k ( t ) d t = 1 sum_(k=0)^(n)p_(n,k)(x)=1,int_(0)^(1)b_(n,k)(t)dt=1\sum_{k=0}^{n} p_{n, k}(x)=1, \int_{0}^{1} b_{n, k}(t) d t=1∑k=0npn,k(x)=1,∫01bn,k(t)dt=1 and b n , n = 0 b n , n = 0 b_(n,n)=0b_{n, n}=0bn,n=0. Guo [4] estimated the rate of convergence for bounded variation functions for the usual Bernstein-Durrmeyer operators defined by (1). By considering the integral modification of Bernstein polynomials in form (2) some approximation properties become simpler in the analysis. Therefore it is important to carry
out a further study of different integral modifications of Bernstein polynomials B n ( f , x ) B n ( f , x ) B_(n)(f,x)B_{n}(f, x)Bn(f,x). Recently, Agratini [1] also estimated the rate of convergence for functions of bounded variation for some integral operators which include operators (2). He obtained the results on ordinary approximation. In the present paper, we estimate the rate of convergence in simultaneous approximation for operators B n ( f , x ) B n ( f , x ) B_(n)(f,x)B_{n}(f, x)Bn(f,x).

2. Auxiliary Results

In this section we give the results which are necessary to prove the main result.
Lemma 1 ([7]). For m , r ∈ N 0 ≡ N ∪ { 0 } m , r ∈ N 0 ≡ N ∪ { 0 } m,r inN^(0)-=N uu{0}m, r \in N^{0} \equiv N \cup\{0\}m,r∈N0≡N∪{0} (the set of non-negative integers), r ≤ n r ≤ n r <= nr \leq nr≤n, if we define
V r , n , m ( x ) = ∑ k = 0 n − r p n − r , x ( x ) ∫ 0 1 b n + r , k + r ( t ) ( t − x ) m d t V r , n , m ( x ) = ∑ k = 0 n − r   p n − r , x ( x ) ∫ 0 1   b n + r , k + r ( t ) ( t − x ) m d t V_(r,n,m)(x)=sum_(k=0)^(n-r)p_(n-r,x)(x)int_(0)^(1)b_(n+r,k+r)(t)(t-x)^(m)dtV_{r, n, m}(x)=\sum_{k=0}^{n-r} p_{n-r, x}(x) \int_{0}^{1} b_{n+r, k+r}(t)(t-x)^{m} d tVr,n,m(x)=∑k=0n−rpn−r,x(x)∫01bn+r,k+r(t)(t−x)mdt
then
V r , n , 0 ( x ) = 1 , V r , n , 1 ( x ) = ( 1 + r ) − x ( 1 + 2 r ) n + r + 1 V r , n , 2 ( x ) = ( r 2 + 3 r + 2 ) + 2 x ( n − 2 r 2 − 5 r − 2 ) − 2 x 2 ( n − 2 r 2 − 4 r − 1 ) n + r + 1 , V r , n , 0 ( x ) = 1 , V r , n , 1 ( x ) = ( 1 + r ) − x ( 1 + 2 r ) n + r + 1 V r , n , 2 ( x ) = r 2 + 3 r + 2 + 2 x n − 2 r 2 − 5 r − 2 − 2 x 2 n − 2 r 2 − 4 r − 1 n + r + 1 , {:[V_(r,n,0)(x)=1","quadV_(r,n,1)(x)=((1+r)-x(1+2r))/(n+r+1)],[V_(r,n,2)(x)=((r^(2)+3r+2)+2x(n-2r^(2)-5r-2)-2x^(2)(n-2r^(2)-4r-1))/(n+r+1)","]:}\begin{gathered} V_{r, n, 0}(x)=1, \quad V_{r, n, 1}(x)=\frac{(1+r)-x(1+2 r)}{n+r+1} \\ V_{r, n, 2}(x)=\frac{\left(r^{2}+3 r+2\right)+2 x\left(n-2 r^{2}-5 r-2\right)-2 x^{2}\left(n-2 r^{2}-4 r-1\right)}{n+r+1}, \end{gathered}Vr,n,0(x)=1,Vr,n,1(x)=(1+r)−x(1+2r)n+r+1Vr,n,2(x)=(r2+3r+2)+2x(n−2r2−5r−2)−2x2(n−2r2−4r−1)n+r+1,
and we have the recurrence relation
[ n + r + m + 1 ] V r , n , m + 1 ( x ) = x ( 1 − x ) [ V r , n , m ( 1 ) ( x ) + 2 m V r , n , m − 1 ( x ) ] + [ ( 1 − 2 x ) ( m + r + 1 ) + x ] V r , n , m ( x ) [ n + r + m + 1 ] V r , n , m + 1 ( x ) = x ( 1 − x ) V r , n , m ( 1 ) ( x ) + 2 m V r , n , m − 1 ( x ) + [ ( 1 − 2 x ) ( m + r + 1 ) + x ] V r , n , m ( x ) {:[{:[n+r+m+1]V_(r,n,m+1)(x)=:}x(1-x)[V_(r,n,m)^((1))(x)+2mV_(r,n,m-1)(x)]],[+[(1-2x)(m+r+1)+x]V_(r,n,m)(x)]:}\begin{aligned} {[n+r+m+1] V_{r, n, m+1}(x)=} & x(1-x)\left[V_{r, n, m}^{(1)}(x)+2 m V_{r, n, m-1}(x)\right] \\ & +[(1-2 x)(m+r+1)+x] V_{r, n, m}(x) \end{aligned}[n+r+m+1]Vr,n,m+1(x)=x(1−x)[Vr,n,m(1)(x)+2mVr,n,m−1(x)]+[(1−2x)(m+r+1)+x]Vr,n,m(x)
Remark 2. If n n nnn is sufficiently large, then by Lemma 1 it is easy to verift that
x ( 1 − x ) n ≤ V r , n , 2 ( x ) ≤ 2 x ( 1 − x ) n . x ( 1 − x ) n ≤ V r , n , 2 ( x ) ≤ 2 x ( 1 − x ) n . (x(1-x))/(n) <= V_(r,n,2)(x) <= (2x(1-x))/(n).\frac{x(1-x)}{n} \leq V_{r, n, 2}(x) \leq \frac{2 x(1-x)}{n} .x(1−x)n≤Vr,n,2(x)≤2x(1−x)n.
Lemma 3 ([8]). For every 0 ≤ k ≤ n , x ∈ ( 0 , 1 ) 0 ≤ k ≤ n , x ∈ ( 0 , 1 ) 0 <= k <= n,x in(0,1)0 \leq k \leq n, x \in(0,1)0≤k≤n,x∈(0,1) and for all n ∈ N n ∈ N n in Nn \in Nn∈N, we have
p n , k ( x ) ≤ 1 2 e n x ( 1 − x ) p n , k ( x ) ≤ 1 2 e n x ( 1 − x ) p_(n,k)(x) <= (1)/(sqrt(2enx(1-x)))p_{n, k}(x) \leq \frac{1}{\sqrt{2 e n x(1-x)}}pn,k(x)≤12enx(1−x)
Lemma 4 ([7]). If f ∈ L 1 [ 0 , 1 ] , f ( r − 1 ) ∈ A . C f ∈ L 1 [ 0 , 1 ] , f ( r − 1 ) ∈ A . C f inL_(1)[0,1],f^((r-1))in A.Cf \in L_{1}[0,1], f^{(r-1)} \in A . Cf∈L1[0,1],f(r−1)∈A.C. loc, f ( r ) ∈ L 1 [ 0 , 1 ] f ( r ) ∈ L 1 [ 0 , 1 ] f^((r))inL_(1)[0,1]f^{(r)} \in L_{1}[0,1]f(r)∈L1[0,1] and 1 ≤ r < n 1 ≤ r < n 1 <= r < n1 \leq r<n1≤r<n, then
B n ( r ) ( f , x ) = ( n ! ) 2 ( n − r ) ! ( n + r ) ! ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 1 b n + r , k + r ( t ) f ( r ) ( t ) d t B n ( r ) ( f , x ) = ( n ! ) 2 ( n − r ) ! ( n + r ) ! ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 1   b n + r , k + r ( t ) f ( r ) ( t ) d t B_(n)^((r))(f,x)=((n!)^(2))/((n-r)!(n+r)!)sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(1)b_(n+r,k+r)(t)f^((r))(t)dtB_{n}^{(r)}(f, x)=\frac{(n!)^{2}}{(n-r)!(n+r)!} \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{1} b_{n+r, k+r}(t) f^{(r)}(t) d tBn(r)(f,x)=(n!)2(n−r)!(n+r)!∑k=0n−rpn−r,k(x)∫01bn+r,k+r(t)f(r)(t)dt
Lemma 5. Let x ∈ ( 0 , 1 ) x ∈ ( 0 , 1 ) x in(0,1)x \in(0,1)x∈(0,1) and K n , r ( x , t ) = ∑ k = 0 n − r p n − r , k ( x ) b n + r , k + r ( t ) K n , r ( x , t ) = ∑ k = 0 n − r   p n − r , k ( x ) b n + r , k + r ( t ) K_(n,r)(x,t)=sum_(k=0)^(n-r)p_(n-r,k)(x)b_(n+r,k+r)(t)K_{n, r}(x, t)=\sum_{k=0}^{n-r} p_{n-r, k}(x) b_{n+r, k+r}(t)Kn,r(x,t)=∑k=0n−rpn−r,k(x)bn+r,k+r(t), then for n n nnn sufficiently large, we have
(3) λ n , γ ( x , y ) := ∫ 0 y K n , γ ( x , t ) d t ≤ 2 x ( 1 − x ) n ( x − y ) 2 , 0 ≤ y < x (3) λ n , γ ( x , y ) := ∫ 0 y   K n , γ ( x , t ) d t ≤ 2 x ( 1 − x ) n ( x − y ) 2 , 0 ≤ y < x {:(3)lambda_(n,gamma)(x","y):=int_(0)^(y)K_(n,gamma)(x","t)dt <= (2x(1-x))/(n(x-y)^(2))","quad0 <= y < x:}\begin{equation*} \lambda_{n, \gamma}(x, y):=\int_{0}^{y} K_{n, \gamma}(x, t) d t \leq \frac{2 x(1-x)}{n(x-y)^{2}}, \quad 0 \leq y<x \tag{3} \end{equation*}(3)λn,γ(x,y):=∫0yKn,γ(x,t)dt≤2x(1−x)n(x−y)2,0≤y<x
(4) 1 − λ n , γ ( x , z ) := ∫ z 1 K n , γ ( x , t ) d t ≤ 2 x ( 1 − x ) n ( z − x ) 2 , x < z < 1 (4) 1 − λ n , γ ( x , z ) := ∫ z 1   K n , γ ( x , t ) d t ≤ 2 x ( 1 − x ) n ( z − x ) 2 , x < z < 1 {:(4)1-lambda_(n,gamma)(x","z):=int_(z)^(1)K_(n,gamma)(x","t)dt <= (2x(1-x))/(n(z-x)^(2))","quad x < z < 1:}\begin{equation*} 1-\lambda_{n, \gamma}(x, z):=\int_{z}^{1} K_{n, \gamma}(x, t) d t \leq \frac{2 x(1-x)}{n(z-x)^{2}}, \quad x<z<1 \tag{4} \end{equation*}(4)1−λn,γ(x,z):=∫z1Kn,γ(x,t)dt≤2x(1−x)n(z−x)2,x<z<1
Proof. We first prove (3) by Lemma 1 as follows:
∫ 0 y K n , γ ( x , t ) d t ≤ ∫ 0 y K n , γ ( x , t ) ( x − t ) 2 ( x − y ) 2 d t ≤ 1 ( x − y ) 2 ∫ 0 1 K n , γ ( x , t ) ( t − x ) 2 d t ≤ V r , n , 2 ( x ) ( x − y ) 2 ≤ 2 x ( 1 − x ) n ( x − y ) 2 ∫ 0 y   K n , γ ( x , t ) d t ≤ ∫ 0 y   K n , γ ( x , t ) ( x − t ) 2 ( x − y ) 2 d t ≤ 1 ( x − y ) 2 ∫ 0 1   K n , γ ( x , t ) ( t − x ) 2 d t ≤ V r , n , 2 ( x ) ( x − y ) 2 ≤ 2 x ( 1 − x ) n ( x − y ) 2 {:[int_(0)^(y)K_(n,gamma)(x","t)dt <= int_(0)^(y)K_(n,gamma)(x","t)((x-t)^(2))/((x-y)^(2))dt],[ <= (1)/((x-y)^(2))int_(0)^(1)K_(n,gamma)(x","t)(t-x)^(2)dt <= (V_(r,n,2)(x))/((x-y)^(2)) <= (2x(1-x))/(n(x-y)^(2))]:}\begin{aligned} \int_{0}^{y} K_{n, \gamma}(x, t) d t & \leq \int_{0}^{y} K_{n, \gamma}(x, t) \frac{(x-t)^{2}}{(x-y)^{2}} d t \\ & \leq \frac{1}{(x-y)^{2}} \int_{0}^{1} K_{n, \gamma}(x, t)(t-x)^{2} d t \leq \frac{V_{r, n, 2}(x)}{(x-y)^{2}} \leq \frac{2 x(1-x)}{n(x-y)^{2}} \end{aligned}∫0yKn,γ(x,t)dt≤∫0yKn,γ(x,t)(x−t)2(x−y)2dt≤1(x−y)2∫01Kn,γ(x,t)(t−x)2dt≤Vr,n,2(x)(x−y)2≤2x(1−x)n(x−y)2
The proof of (4) is similar.

3. Main Result

Theorem. Let f ∈ H r , r ∈ N 0 f ∈ H r , r ∈ N 0 f inH_(r),r inN^(0)f \in H_{r}, r \in N^{0}f∈Hr,r∈N0. Then for every x ∈ ( 0 , 1 ) x ∈ ( 0 , 1 ) x in(0,1)x \in(0,1)x∈(0,1) and n n nnn sufficiently large, we have
| B n ( r ) ( f , x ) − 1 2 [ f + ( r ) ( x ) + f − ( r ) ( x ) ] | ≤ ( 2 + 2 r + 1 8 e ) | f + ( r ) ( x ) − f − ( r ) ( x ) | ( n − r ) x ( 1 − x ) (5) + 5 n x ( 1 − x ) ∑ k = 1 n ( ω x ( g x , r , x k ) + ω x ( g x , r , 1 − x k ) ) B n ( r ) ( f , x ) − 1 2 f + ( r ) ( x ) + f − ( r ) ( x ) ≤ 2 + 2 r + 1 8 e f + ( r ) ( x ) − f − ( r ) ( x ) ( n − r ) x ( 1 − x ) (5) + 5 n x ( 1 − x ) ∑ k = 1 n   ω x g x , r , x k + ω x g x , r , 1 − x k {:[|B_(n)^((r))(f,x)-(1)/(2)[f_(+)^((r))(x)+f_(-)^((r))(x)]| <= (2+2r+(1)/(sqrt(8e)))(|f_(+)^((r))(x)-f_(-)^((r))(x)|)/(sqrt((n-r)x(1-x)))],[(5)+(5)/(nx(1-x))sum_(k=1)^(n)(omega_(x)(g_(x,r),(x)/(sqrtk))+omega_(x)(g_(x,r),(1-x)/(sqrtk)))]:}\begin{gather*} \left|B_{n}^{(r)}(f, x)-\frac{1}{2}\left[f_{+}^{(r)}(x)+f_{-}^{(r)}(x)\right]\right| \leq\left(2+2 r+\frac{1}{\sqrt{8 e}}\right) \frac{\left|f_{+}^{(r)}(x)-f_{-}^{(r)}(x)\right|}{\sqrt{(n-r) x(1-x)}} \\ +\frac{5}{n x(1-x)} \sum_{k=1}^{n}\left(\omega_{x}\left(g_{x, r}, \frac{x}{\sqrt{k}}\right)+\omega_{x}\left(g_{x, r}, \frac{1-x}{\sqrt{k}}\right)\right) \tag{5} \end{gather*}|Bn(r)(f,x)−12[f+(r)(x)+f−(r)(x)]|≤(2+2r+18e)|f+(r)(x)−f−(r)(x)|(n−r)x(1−x)(5)+5nx(1−x)∑k=1n(ωx(gx,r,xk)+ωx(gx,r,1−xk))
where H r = { f : f ( r − 1 ) ∈ C [ 0 , 1 ] , f ± ( r ) ( x ) ∈ [ 0 , 1 ] , r > 0 } , H 0 = { f : f ± ( x ) ∈ [ 0 , 1 ] } , ω x ( f , t ) = sup u { ∣ f ( x + u − f ( x ) | : | u | ≤ t , x , x + u ∈ [ 0 , 1 ] } H r = f : f ( r − 1 ) ∈ C [ 0 , 1 ] , f ± ( r ) ( x ) ∈ [ 0 , 1 ] , r > 0 , H 0 = f : f ± ( x ) ∈ [ 0 , 1 ] } , ω x ( f , t ) = sup u   { ∣ f ( x + u − f ( x ) | : | u | ≤ t , x , x + u ∈ [ 0 , 1 ] } H_(r)={f:f^((r-1))in C[0,1],f_(+-)^((r))(x)in[0,1],r > 0},H_(0)={f:f_(+-)(x)in:}[0,1]},omega_(x)(f,t)=s u p_(u){∣f(x+u-f(x)|:|u| <= t,x,x+u in[0,1]}H_{r}=\left\{f: f^{(r-1)} \in C[0,1], f_{ \pm}^{(r)}(x) \in[0,1], r>0\right\}, H_{0}=\left\{f: f_{ \pm}(x) \in\right. [0,1]\}, \omega_{x}(f, t)=\sup _{u}\{\mid f(x+u-f(x)|:|u| \leq t, x, x+u \in[0,1]\}Hr={f:f(r−1)∈C[0,1],f±(r)(x)∈[0,1],r>0},H0={f:f±(x)∈[0,1]},ωx(f,t)=supu{∣f(x+u−f(x)|:|u|≤t,x,x+u∈[0,1]} and
g x , y = { f − ( r ) ( t ) − f − ( r ) ( x ) , 0 ≤ t < x 0 , t = x f + ( r ) ( t ) − f + ( r ) ( x ) , x < t ≤ 1 g x , y = f − ( r ) ( t ) − f − ( r ) ( x ) ,      0 ≤ t < x 0 ,      t = x f + ( r ) ( t ) − f + ( r ) ( x ) ,      x < t ≤ 1 g_(x,y)={[f_(-)^((r))(t)-f_(-)^((r))(x)",",0 <= t < x],[0",",t=x],[f_(+)^((r))(t)-f_(+)^((r))(x)",",x < t <= 1]:}g_{x, y}= \begin{cases}f_{-}^{(r)}(t)-f_{-}^{(r)}(x), & 0 \leq t<x \\ 0, & t=x \\ f_{+}^{(r)}(t)-f_{+}^{(r)}(x), & x<t \leq 1\end{cases}gx,y={f−(r)(t)−f−(r)(x),0≤t<x0,t=xf+(r)(t)−f+(r)(x),x<t≤1
Proof. Using Lemma 3, we can write
| B n ( r ) ( f , x ) − 1 2 { f + ( r ) ( t ) + f + ( r ) ( x ) } | ≤ | ( n ! ) 2 ( n − r ) ! ( n + r ) ! ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 x b n + r , k + r ( t ) f − ( r ) ( t ) d t − ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 x b n + r , k + r ( t ) f − ( r ) ( t ) d t ∣ + | ( n ! ) 2 ( n − r ) ! ( n + r ) ! ∑ k = 0 n − r p n − r , k ( x ) ∫ x 1 b n + r , k + r ( t ) f + ( r ) ( t ) d t B n ( r ) ( f , x ) − 1 2 f + ( r ) ( t ) + f + ( r ) ( x ) ≤ ( n ! ) 2 ( n − r ) ! ( n + r ) ! ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x   b n + r , k + r ( t ) f − ( r ) ( t ) d t − ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x   b n + r , k + r ( t ) f − ( r ) ( t ) d t ∣ + ( n ! ) 2 ( n − r ) ! ( n + r ) ! ∑ k = 0 n − r   p n − r , k ( x ) ∫ x 1   b n + r , k + r ( t ) f + ( r ) ( t ) d t {:[|B_(n)^((r))(f,x)-(1)/(2){f_(+)^((r))(t)+f_(+)^((r))(x)}|],[quad <= |((n!)^(2))/((n-r)!(n+r)!)sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(x)b_(n+r,k+r)(t)f_(-)^((r))(t)dt:}],[quad-sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(x)b_(n+r,k+r)(t)f_(-)^((r))(t)dt∣],[quad+|((n!)^(2))/((n-r)!(n+r)!)sum_(k=0)^(n-r)p_(n-r,k)(x)int_(x)^(1)b_(n+r,k+r)(t)f_(+)^((r))(t)dt:}]:}\begin{aligned} & \left|B_{n}^{(r)}(f, x)-\frac{1}{2}\left\{f_{+}^{(r)}(t)+f_{+}^{(r)}(x)\right\}\right| \\ & \quad \leq \left\lvert\, \frac{(n!)^{2}}{(n-r)!(n+r)!} \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{x} b_{n+r, k+r}(t) f_{-}^{(r)}(t) d t\right. \\ & \quad-\sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{x} b_{n+r, k+r}(t) f_{-}^{(r)}(t) d t \mid \\ & \quad+\left\lvert\, \frac{(n!)^{2}}{(n-r)!(n+r)!} \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{x}^{1} b_{n+r, k+r}(t) f_{+}^{(r)}(t) d t\right. \end{aligned}|Bn(r)(f,x)−12{f+(r)(t)+f+(r)(x)}|≤|(n!)2(n−r)!(n+r)!∑k=0n−rpn−r,k(x)∫0xbn+r,k+r(t)f−(r)(t)dt−∑k=0n−rpn−r,k(x)∫0xbn+r,k+r(t)f−(r)(t)dt∣+|(n!)2(n−r)!(n+r)!∑k=0n−rpn−r,k(x)∫x1bn+r,k+r(t)f+(r)(t)dt
− ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 x b n + r , k + r ( t ) f + ( r ) ( t ) d t ∣ + 1 2 | B n ( r ) ( sign ( t − x ) , x ) | | f + ( r ) ( x ) − f − ( r ) ( x ) | (6) = I 1 + I 1 + I 3 , say. − ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x   b n + r , k + r ( t ) f + ( r ) ( t ) d t ∣ + 1 2 B n ( r ) ( sign ( t − x ) , x ) f + ( r ) ( x ) − f − ( r ) ( x ) (6) = I 1 + I 1 + I 3 ,  say.  {:[-sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(x)b_(n+r,k+r)(t)f_(+)^((r))(t)dt∣],[+(1)/(2)|B_(n)^((r))(sign(t-x),x)||f_(+)(r)(x)-f_(-)^((r))(x)|],[(6)=I_(1)+I_(1)+I_(3)","quad" say. "]:}\begin{align*} & -\sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{x} b_{n+r, k+r}(t) f_{+}^{(r)}(t) d t \mid \\ & +\frac{1}{2}\left|B_{n}^{(r)}(\operatorname{sign}(t-x), x)\right|\left|f_{+}(r)(x)-f_{-}^{(r)}(x)\right| \\ = & I_{1}+I_{1}+I_{3}, \quad \text { say. } \tag{6} \end{align*}−∑k=0n−rpn−r,k(x)∫0xbn+r,k+r(t)f+(r)(t)dt∣+12|Bn(r)(sign(t−x),x)||f+(r)(x)−f−(r)(x)|(6)=I1+I1+I3, say. 
Using Lemma 4, we have
I 1 ≤ ( n ! ) 2 ( n − r ) ! ( n + r ) ! ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 x b n + r , k + r ( t ) | f − ( r ) ( t ) − f − ( r ) ( x ) | d t + | f − ( r ) ( x ) | | ( n + r ) ! ( n − r ) ! ( n ! ) 2 − 1 | ≤ ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 x b n + r , k + r ( t ) ω x ( x − t ) d t + O ( n − 1 ) ≤ ∑ k = 0 n − r p n − r , k ( x ) ( ∫ 0 x − δ + ∫ x − δ x ) b n + r , k + r ( t ) ω x ( x − t ) d t + O ( n − 1 ) ≤ ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 x − δ ω x ( x − t ) [ b n + r , k + r ( t ) d t ] + ω x ( δ ) ∑ k = 0 n − r p n − r , k ( x ) ∫ x − δ x b n + r , k + r ( t ) d t + O ( n − 1 ) ≤ ω x ( δ ) ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 x b n + r , k + r ( t ) d t + ∫ 0 x − δ 2 x ( 1 − x ) n ( t − x ) 2 d ( ω x ( x − t ) ) + O ( n − 1 ) I 1 ≤ ( n ! ) 2 ( n − r ) ! ( n + r ) ! ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x   b n + r , k + r ( t ) f − ( r ) ( t ) − f − ( r ) ( x ) d t + f − ( r ) ( x ) ( n + r ) ! ( n − r ) ! ( n ! ) 2 − 1 ≤ ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x   b n + r , k + r ( t ) ω x ( x − t ) d t + O n − 1 ≤ ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x − δ   + ∫ x − δ x   b n + r , k + r ( t ) ω x ( x − t ) d t + O n − 1 ≤ ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x − δ   ω x ( x − t ) b n + r , k + r ( t ) d t + ω x ( δ ) ∑ k = 0 n − r   p n − r , k ( x ) ∫ x − δ x   b n + r , k + r ( t ) d t + O n − 1 ≤ ω x ( δ ) ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x   b n + r , k + r ( t ) d t + ∫ 0 x − δ   2 x ( 1 − x ) n ( t − x ) 2 d ω x ( x − t ) + O n − 1 {:[I_(1) <= ((n!)^(2))/((n-r)!(n+r)!)sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(x)b_(n+r,k+r)(t)|f_(-)^((r))(t)-f_(-)^((r))(x)|dt],[+|f_(-)^((r))(x)||((n+r)!(n-r)!)/((n!)^(2))-1|],[ <= sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(x)b_(n+r,k+r)(t)omega_(x)(x-t)dt+O(n^(-1))],[ <= sum_(k=0)^(n-r)p_(n-r,k)(x)(int_(0)^(x-delta)+int_(x-delta)^(x))b_(n+r,k+r)(t)omega_(x)(x-t)dt+O(n^(-1))],[ <= sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(x-delta)omega_(x)(x-t)[b_(n+r,k+r)(t)dt]],[+omega_(x)(delta)sum_(k=0)^(n-r)p_(n-r,k)(x)int_(x-delta)^(x)b_(n+r,k+r)(t)dt+O(n^(-1))],[ <= omega_(x)(delta)sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(x)b_(n+r,k+r)(t)dt],[+int_(0)^(x-delta)(2x(1-x))/(n(t-x)^(2))d(omega_(x)(x-t))+O(n^(-1))]:}\begin{aligned} I_{1} \leq & \frac{(n!)^{2}}{(n-r)!(n+r)!} \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{x} b_{n+r, k+r}(t)\left|f_{-}^{(r)}(t)-f_{-}^{(r)}(x)\right| d t \\ & +\left|f_{-}^{(r)}(x)\right|\left|\frac{(n+r)!(n-r)!}{(n!)^{2}}-1\right| \\ \leq & \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{x} b_{n+r, k+r}(t) \omega_{x}(x-t) d t+O\left(n^{-1}\right) \\ \leq & \sum_{k=0}^{n-r} p_{n-r, k}(x)\left(\int_{0}^{x-\delta}+\int_{x-\delta}^{x}\right) b_{n+r, k+r}(t) \omega_{x}(x-t) d t+O\left(n^{-1}\right) \\ \leq & \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{x-\delta} \omega_{x}(x-t)\left[b_{n+r, k+r}(t) d t\right] \\ & +\omega_{x}(\delta) \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{x-\delta}^{x} b_{n+r, k+r}(t) d t+O\left(n^{-1}\right) \\ \leq & \omega_{x}(\delta) \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{x} b_{n+r, k+r}(t) d t \\ & +\int_{0}^{x-\delta} \frac{2 x(1-x)}{n(t-x)^{2}} d\left(\omega_{x}(x-t)\right)+O\left(n^{-1}\right) \end{aligned}I1≤(n!)2(n−r)!(n+r)!∑k=0n−rpn−r,k(x)∫0xbn+r,k+r(t)|f−(r)(t)−f−(r)(x)|dt+|f−(r)(x)||(n+r)!(n−r)!(n!)2−1|≤∑k=0n−rpn−r,k(x)∫0xbn+r,k+r(t)ωx(x−t)dt+O(n−1)≤∑k=0n−rpn−r,k(x)(∫0x−δ+∫x−δx)bn+r,k+r(t)ωx(x−t)dt+O(n−1)≤∑k=0n−rpn−r,k(x)∫0x−δωx(x−t)[bn+r,k+r(t)dt]+ωx(δ)∑k=0n−rpn−r,k(x)∫x−δxbn+r,k+r(t)dt+O(n−1)≤ωx(δ)∑k=0n−rpn−r,k(x)∫0xbn+r,k+r(t)dt+∫0x−δ2x(1−x)n(t−x)2d(ωx(x−t))+O(n−1)
Therefore
∫ 0 x − δ 2 x ( 1 − x ) n ( t − x ) 2 d ( ω x ( x − t ) ) ≤ 2 x ( 1 − x ) n [ − ω x ( δ ) δ 2 + ω x ( x ) x 2 ] + 2 x ( 1 − x ) n ∫ 0 x ω x ( t ) t − 3 d t ≤ 2 x ( 1 − x ) n [ ω x ( δ ) + ∑ k = 1 n ω x ( x k ) ] ∫ 0 x − δ   2 x ( 1 − x ) n ( t − x ) 2 d ω x ( x − t ) ≤ 2 x ( 1 − x ) n − ω x ( δ ) δ 2 + ω x ( x ) x 2 + 2 x ( 1 − x ) n ∫ 0 x   ω x ( t ) t − 3 d t ≤ 2 x ( 1 − x ) n ω x ( δ ) + ∑ k = 1 n   ω x x k {:[int_(0)^(x-delta)(2x(1-x))/(n(t-x)^(2))d(omega_(x)(x-t)) <= (2x(1-x))/(n)[(-omega_(x)(delta))/(delta^(2))+(omega_(x)(x))/(x^(2))]],[+(2x(1-x))/(n)int_(0)^(x)omega_(x)(t)t^(-3)dt],[ <= (2x(1-x))/(n)[omega_(x)(delta)+sum_(k=1)^(n)omega_(x)((x)/(sqrtk))]]:}\begin{aligned} \int_{0}^{x-\delta} \frac{2 x(1-x)}{n(t-x)^{2}} d\left(\omega_{x}(x-t)\right) \leq & \frac{2 x(1-x)}{n}\left[\frac{-\omega_{x}(\delta)}{\delta^{2}}+\frac{\omega_{x}(x)}{x^{2}}\right] \\ & +\frac{2 x(1-x)}{n} \int_{0}^{x} \omega_{x}(t) t^{-3} d t \\ \leq & \frac{2 x(1-x)}{n}\left[\omega_{x}(\delta)+\sum_{k=1}^{n} \omega_{x}\left(\frac{x}{\sqrt{k}}\right)\right] \end{aligned}∫0x−δ2x(1−x)n(t−x)2d(ωx(x−t))≤2x(1−x)n[−ωx(δ)δ2+ωx(x)x2]+2x(1−x)n∫0xωx(t)t−3dt≤2x(1−x)n[ωx(δ)+∑k=1nωx(xk)]
≤ 4 n x ( 1 − x ) ∑ k = 1 n ω x ( x k ) ≤ 4 n x ( 1 − x ) ∑ k = 1 n   ω x x k <= (4)/(nx(1-x))sum_(k=1)^(n)omega_(x)((x)/(sqrtk))\leq \frac{4}{n x(1-x)} \sum_{k=1}^{n} \omega_{x}\left(\frac{x}{\sqrt{k}}\right)≤4nx(1−x)∑k=1nωx(xk)
and
ω x ( δ ) ∑ k = 0 n − r p n − r , k ( x ) ∫ 0 x b n + r , k + r ( t ) d t ≤ ω x ( x n ) ≤ 1 n x ( 1 − x ) ∑ k = 1 n ω x ( x k ) ω x ( δ ) ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x   b n + r , k + r ( t ) d t ≤ ω x x n ≤ 1 n x ( 1 − x ) ∑ k = 1 n   ω x x k omega_(x)(delta)sum_(k=0)^(n-r)p_(n-r,k)(x)int_(0)^(x)b_(n+r,k+r)(t)dt <= omega_(x)((x)/(sqrtn)) <= (1)/(nx(1-x))sum_(k=1)^(n)omega_(x)((x)/(sqrtk))\omega_{x}(\delta) \sum_{k=0}^{n-r} p_{n-r, k}(x) \int_{0}^{x} b_{n+r, k+r}(t) d t \leq \omega_{x}\left(\frac{x}{\sqrt{n}}\right) \leq \frac{1}{n x(1-x)} \sum_{k=1}^{n} \omega_{x}\left(\frac{x}{\sqrt{k}}\right)ωx(δ)∑k=0n−rpn−r,k(x)∫0xbn+r,k+r(t)dt≤ωx(xn)≤1nx(1−x)∑k=1nωx(xk)
Thus
(7) I 1 ≤ 5 n x ( 1 − x ) ∑ k = 1 n ω x ( g x , r , x k ) + O ( n − 1 ) (7) I 1 ≤ 5 n x ( 1 − x ) ∑ k = 1 n   ω x g x , r , x k + O n − 1 {:(7)I_(1) <= (5)/(nx(1-x))sum_(k=1)^(n)omega_(x)(g_(x,r),(x)/(sqrtk))+O(n^(-1)):}\begin{equation*} I_{1} \leq \frac{5}{n x(1-x)} \sum_{k=1}^{n} \omega_{x}\left(g_{x, r}, \frac{x}{\sqrt{k}}\right)+O\left(n^{-1}\right) \tag{7} \end{equation*}(7)I1≤5nx(1−x)∑k=1nωx(gx,r,xk)+O(n−1)
Choosing δ 1 = 1 − x n δ 1 = 1 − x n delta_(1)=(1-x)/(sqrtn)\delta_{1}=\frac{1-x}{\sqrt{n}}δ1=1−xn and proceeding similar to I 1 I 1 I_(1)I_{1}I1, we get
I 2 ≤ ∑ k = 0 n − r p n − r , k ( x ) ( ∫ 0 x − δ 1 + ∫ x − δ 1 x ) b n + r , k + r ( t ) ω x ( g x , t , t − x ) d t + O ( n − 1 ) (8) ≤ 5 n x ( 1 − x ) ∑ k = 1 n ω x ( g x , r , 1 − x k ) + O ( n − 1 ) I 2 ≤ ∑ k = 0 n − r   p n − r , k ( x ) ∫ 0 x − δ 1   + ∫ x − δ 1 x   b n + r , k + r ( t ) ω x g x , t , t − x d t + O n − 1 (8) ≤ 5 n x ( 1 − x ) ∑ k = 1 n   ω x g x , r , 1 − x k + O n − 1 {:[I_(2) <= sum_(k=0)^(n-r)p_(n-r,k)(x)(int_(0)^(x-delta_(1))+int_(x-delta_(1))^(x))b_(n+r,k+r)(t)omega_(x)(g_(x,t),t-x)dt+O(n^(-1))],[(8) <= (5)/(nx(1-x))sum_(k=1)^(n)omega_(x)(g_(x,r),(1-x)/(sqrtk))+O(n^(-1))]:}\begin{align*} I_{2} & \leq \sum_{k=0}^{n-r} p_{n-r, k}(x)\left(\int_{0}^{x-\delta_{1}}+\int_{x-\delta_{1}}^{x}\right) b_{n+r, k+r}(t) \omega_{x}\left(g_{x, t}, t-x\right) d t+O\left(n^{-1}\right) \\ & \leq \frac{5}{n x(1-x)} \sum_{k=1}^{n} \omega_{x}\left(g_{x, r}, \frac{1-x}{\sqrt{k}}\right)+O\left(n^{-1}\right) \tag{8} \end{align*}I2≤∑k=0n−rpn−r,k(x)(∫0x−δ1+∫x−δ1x)bn+r,k+r(t)ωx(gx,t,t−x)dt+O(n−1)(8)≤5nx(1−x)∑k=1nωx(gx,r,1−xk)+O(n−1)
Next, we estimate I 3 I 3 I_(3)I_{3}I3 as follows:
B n ( r ) ( sign ( t − x ) , x ) = ∫ x 1 K n , γ ( x , t ) d t − ∫ 0 x K n , γ ( x , t ) d t = A n , γ − B n , γ , say. B n ( r ) ( sign ( t − x ) , x ) = ∫ x 1   K n , γ ( x , t ) d t − ∫ 0 x   K n , γ ( x , t ) d t = A n , γ − B n , γ ,  say.  {:[B_(n)^((r))(sign(t-x)","x)=int_(x)^(1)K_(n,gamma)(x","t)dt-int_(0)^(x)K_(n,gamma)(x","t)dt],[=A_(n,gamma)-B_(n,gamma)","quad" say. "]:}\begin{aligned} B_{n}^{(r)}(\operatorname{sign}(t-x), x) & =\int_{x}^{1} K_{n, \gamma}(x, t) d t-\int_{0}^{x} K_{n, \gamma}(x, t) d t \\ & =A_{n, \gamma}-B_{n, \gamma}, \quad \text { say. } \end{aligned}Bn(r)(sign(t−x),x)=∫x1Kn,γ(x,t)dt−∫0xKn,γ(x,t)dt=An,γ−Bn,γ, say. 
It is easy to verify that A n , γ ( x ) + B n , γ ( x ) = 1 A n , γ ( x ) + B n , γ ( x ) = 1 A_(n,gamma)(x)+B_(n,gamma)(x)=1A_{n, \gamma}(x)+B_{n, \gamma}(x)=1An,γ(x)+Bn,γ(x)=1. Thus B n ( r ) ( sign ( t − x ) , x ) = 1 − 2 A n , γ ( x ) B n ( r ) ( sign ( t − x ) , x ) = 1 − 2 A n , γ ( x ) B_(n)^((r))(sign(t-x),x)=1-2A_(n,gamma)(x)B_{n}^{(r)}(\operatorname{sign}(t-x), x)=1- 2 A_{n, \gamma}(x)Bn(r)(sign(t−x),x)=1−2An,γ(x). Using Lemma 1, Lemma 2 and the fact that ∑ j = 0 k p n , j ( x ) = ∫ x 1 b n , k ( t ) d t ∑ j = 0 k   p n , j ( x ) = ∫ x 1   b n , k ( t ) d t sum_(j=0)^(k)p_(n,j)(x)=int_(x)^(1)b_(n,k)(t)dt\sum_{j=0}^{k} p_{n, j}(x)=\int_{x}^{1} b_{n, k}(t) d t∑j=0kpn,j(x)=∫x1bn,k(t)dt, we have
A n , γ ( x ) = ∑ k = 0 n − r p n − r , k ( x ) ∑ j = 0 k + r p n + r , j ( x ) = ∑ k = 0 n − r p n − r , k ( x ) ∑ j = 0 k p n + r , j ( x ) + ∑ k = 0 n − r p n − r , k ( x ) ∑ j = k + 1 k + r p n + r , j ( x ) ≤ ∑ k = 0 n − r p n − r , k ( x ) ∑ j = 0 k p n + r , j ( x ) + r 2 e ( n + r ) x ( 1 − x ) . A n , γ ( x ) = ∑ k = 0 n − r   p n − r , k ( x ) ∑ j = 0 k + r   p n + r , j ( x ) = ∑ k = 0 n − r   p n − r , k ( x ) ∑ j = 0 k   p n + r , j ( x ) + ∑ k = 0 n − r   p n − r , k ( x ) ∑ j = k + 1 k + r   p n + r , j ( x ) ≤ ∑ k = 0 n − r   p n − r , k ( x ) ∑ j = 0 k   p n + r , j ( x ) + r 2 e ( n + r ) x ( 1 − x ) . {:[A_(n,gamma)(x)=sum_(k=0)^(n-r)p_(n-r,k)(x)sum_(j=0)^(k+r)p_(n+r,j)(x)],[=sum_(k=0)^(n-r)p_(n-r,k)(x)sum_(j=0)^(k)p_(n+r,j)(x)+sum_(k=0)^(n-r)p_(n-r,k)(x)sum_(j=k+1)^(k+r)p_(n+r,j)(x)],[ <= sum_(k=0)^(n-r)p_(n-r,k)(x)sum_(j=0)^(k)p_(n+r,j)(x)+(r)/(sqrt(2e(n+r)x(1-x))).]:}\begin{aligned} A_{n, \gamma}(x) & =\sum_{k=0}^{n-r} p_{n-r, k}(x) \sum_{j=0}^{k+r} p_{n+r, j}(x) \\ & =\sum_{k=0}^{n-r} p_{n-r, k}(x) \sum_{j=0}^{k} p_{n+r, j}(x)+\sum_{k=0}^{n-r} p_{n-r, k}(x) \sum_{j=k+1}^{k+r} p_{n+r, j}(x) \\ & \leq \sum_{k=0}^{n-r} p_{n-r, k}(x) \sum_{j=0}^{k} p_{n+r, j}(x)+\frac{r}{\sqrt{2 e(n+r) x(1-x)}} . \end{aligned}An,γ(x)=∑k=0n−rpn−r,k(x)∑j=0k+rpn+r,j(x)=∑k=0n−rpn−r,k(x)∑j=0kpn+r,j(x)+∑k=0n−rpn−r,k(x)∑j=k+1k+rpn+r,j(x)≤∑k=0n−rpn−r,k(x)∑j=0kpn+r,j(x)+r2e(n+r)x(1−x).
By the Berry-Esseen theorem [2] we readily obtain
(9) | ∑ k = 0 j p n − r , k ( x ) − 1 2 π ∫ − ∞ ( j − ( n − r ) x / ( n − r ) x ( 1 − x ) e − t 2 / 2 d t | < 1 ( n − r ) x ( 1 − x ) (9) ∑ k = 0 j   p n − r , k ( x ) − 1 2 π ∫ − ∞ ( j − ( n − r ) x / ( n − r ) x ( 1 − x )   e − t 2 / 2 d t < 1 ( n − r ) x ( 1 − x ) {:(9)|sum_(k=0)^(j)p_(n-r,k)(x)-(1)/(sqrt(2pi))int_(-oo)^((j-(n-r)x//sqrt((n-r)x(1-x)))e^(-t^(2)//2)dt| < (1)/(sqrt((n-r)x(1-x))):}\begin{equation*} \left|\sum_{k=0}^{j} p_{n-r, k}(x)-\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{(j-(n-r) x / \sqrt{(n-r) x(1-x)}} e^{-t^{2} / 2} d t\right|<\frac{1}{\sqrt{(n-r) x(1-x)}} \tag{9} \end{equation*}(9)|∑k=0jpn−r,k(x)−12π∫−∞(j−(n−r)x/(n−r)x(1−x)e−t2/2dt|<1(n−r)x(1−x)
and
(10) | ∑ k = 0 j p n + r , k ( x ) − 1 2 π ∫ − ∞ ( j − ( n + r ) x / ( n + r ) x ( 1 − x ) e − t 2 / 2 d t | < 1 ( n + r ) x ( 1 − x ) (10) ∑ k = 0 j   p n + r , k ( x ) − 1 2 π ∫ − ∞ ( j − ( n + r ) x / ( n + r ) x ( 1 − x )   e − t 2 / 2 d t < 1 ( n + r ) x ( 1 − x ) {:(10)|sum_(k=0)^(j)p_(n+r,k)(x)-(1)/(sqrt(2pi))int_(-oo)^((j-(n+r)x//sqrt((n+r)x(1-x)))e^(-t^(2)//2)dt| < (1)/(sqrt((n+r)x(1-x))):}\begin{equation*} \left|\sum_{k=0}^{j} p_{n+r, k}(x)-\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{(j-(n+r) x / \sqrt{(n+r) x(1-x)}} e^{-t^{2} / 2} d t\right|<\frac{1}{\sqrt{(n+r) x(1-x)}} \tag{10} \end{equation*}(10)|∑k=0jpn+r,k(x)−12π∫−∞(j−(n+r)x/(n+r)x(1−x)e−t2/2dt|<1(n+r)x(1−x)
Hence by (9) and (10), we have
(11) | ∑ k = 0 j p n − r , k ( x ) − ∑ k = 0 j p n + r , k ( x ) | < 2 + r ( n − r ) x ( 1 − x ) (11) ∑ k = 0 j   p n − r , k ( x ) − ∑ k = 0 j   p n + r , k ( x ) < 2 + r ( n − r ) x ( 1 − x ) {:(11)|sum_(k=0)^(j)p_(n-r,k)(x)-sum_(k=0)^(j)p_(n+r,k)(x)| < (2+r)/(sqrt((n-r)x(1-x))):}\begin{equation*} \left|\sum_{k=0}^{j} p_{n-r, k}(x)-\sum_{k=0}^{j} p_{n+r, k}(x)\right|<\frac{2+r}{\sqrt{(n-r) x(1-x)}} \tag{11} \end{equation*}(11)|∑k=0jpn−r,k(x)−∑k=0jpn+r,k(x)|<2+r(n−r)x(1−x)
By (11) it follows that
(12) | ∑ k = 0 n − r p n − r , k ( x ) ( ∑ j = 0 k p n + r , j ( x ) − ∑ j = 0 k p n − r , j ( x ) ) | < 2 + r ( n − r ) x ( 1 − x ) . (12) ∑ k = 0 n − r   p n − r , k ( x ) ∑ j = 0 k   p n + r , j ( x ) − ∑ j = 0 k   p n − r , j ( x ) < 2 + r ( n − r ) x ( 1 − x ) . {:(12)|sum_(k=0)^(n-r)p_(n-r,k)(x)(sum_(j=0)^(k)p_(n+r,j)(x)-sum_(j=0)^(k)p_(n-r,j)(x))| < (2+r)/(sqrt((n-r)x(1-x))).:}\begin{equation*} \left|\sum_{k=0}^{n-r} p_{n-r, k}(x)\left(\sum_{j=0}^{k} p_{n+r, j}(x)-\sum_{j=0}^{k} p_{n-r, j}(x)\right)\right|<\frac{2+r}{\sqrt{(n-r) x(1-x)}} . \tag{12} \end{equation*}(12)|∑k=0n−rpn−r,k(x)(∑j=0kpn+r,j(x)−∑j=0kpn−r,j(x))|<2+r(n−r)x(1−x).
Let
S = ∑ k = 0 n − r p n − r , k ( x ) ∑ j = 0 k p n − r , j ( x ) . S = ∑ k = 0 n − r   p n − r , k ( x ) ∑ j = 0 k   p n − r , j ( x ) . S=sum_(k=0)^(n-r)p_(n-r,k)(x)sum_(j=0)^(k)p_(n-r,j)(x).S=\sum_{k=0}^{n-r} p_{n-r, k}(x) \sum_{j=0}^{k} p_{n-r, j}(x) .S=∑k=0n−rpn−r,k(x)∑j=0kpn−r,j(x).
If ξ ξ xi\xiξ and η η eta\etaη are independent random variables with the same distribution P P P\mathcal{P}P assigning the probability p k , ∑ k = 1 ∞ p k = 1 p k , ∑ k = 1 ∞   p k = 1 p_(k),sum_(k=1)^(oo)p_(k)=1p_{k}, \sum_{k=1}^{\infty} p_{k}=1pk,∑k=1∞pk=1, to the number b k , k = 1 , 2 , … b k , k = 1 , 2 , … b_(k),k=1,2,dotsb_{k}, k=1,2, \ldotsbk,k=1,2,…, such that b 1 < b + 2 < ⋯ b 1 < b + 2 < ⋯ b_(1) < b+2 < cdotsb_{1}<b+2<\cdotsb1<b+2<⋯, then
P ( ξ ≤ η ) = ∑ k = 1 ∞ P ( η = b k ) P ( ξ ≤ b k ) = ∑ k = 1 ∞ p k ∑ i = 1 k p i P ( ξ = η ) = ∑ k = 1 ∞ P ( η = b k ) P ( ξ = b k ) = ∑ k = 1 ∞ p k 2 P ( ξ ≤ η ) = ∑ k = 1 ∞   P η = b k P ξ ≤ b k = ∑ k = 1 ∞   p k ∑ i = 1 k   p i P ( ξ = η ) = ∑ k = 1 ∞   P η = b k P ξ = b k = ∑ k = 1 ∞   p k 2 {:[P(xi <= eta)=sum_(k=1)^(oo)P(eta=b_(k))P(xi <= b_(k))=sum_(k=1)^(oo)p_(k)sum_(i=1)^(k)p_(i)],[P(xi=eta)=sum_(k=1)^(oo)P(eta=b_(k))P(xi=b_(k))=sum_(k=1)^(oo)p_(k)^(2)]:}\begin{aligned} & P(\xi \leq \eta)=\sum_{k=1}^{\infty} P\left(\eta=b_{k}\right) P\left(\xi \leq b_{k}\right)=\sum_{k=1}^{\infty} p_{k} \sum_{i=1}^{k} p_{i} \\ & P(\xi=\eta)=\sum_{k=1}^{\infty} P\left(\eta=b_{k}\right) P\left(\xi=b_{k}\right)=\sum_{k=1}^{\infty} p_{k}^{2} \end{aligned}P(ξ≤η)=∑k=1∞P(η=bk)P(ξ≤bk)=∑k=1∞pk∑i=1kpiP(ξ=η)=∑k=1∞P(η=bk)P(ξ=bk)=∑k=1∞pk2
and we obtain
(13) 0 < ∑ k = 1 ∞ p k ∑ t = 1 k p t − 1 2 = 1 2 ∑ k = 1 ∞ p k 2 (13) 0 < ∑ k = 1 ∞   p k ∑ t = 1 k   p t − 1 2 = 1 2 ∑ k = 1 ∞   p k 2 {:(13)0 < sum_(k=1)^(oo)p_(k)sum_(t=1)^(k)p_(t)-(1)/(2)=(1)/(2)sum_(k=1)^(oo)p_(k)^(2):}\begin{equation*} 0<\sum_{k=1}^{\infty} p_{k} \sum_{t=1}^{k} p_{t}-\frac{1}{2}=\frac{1}{2} \sum_{k=1}^{\infty} p_{k}^{2} \tag{13} \end{equation*}(13)0<∑k=1∞pk∑t=1kpt−12=12∑k=1∞pk2
If now P P P\mathcal{P}P is the binomial distribution with the parameters n − r n − r n-rn-rn−r (number of independent trials) and x x xxx (success probability in each trial), then we have S = P ( ξ ≤ η ) S = P ( ξ ≤ η ) S=P(xi <= eta)S= P(\xi \leq \eta)S=P(ξ≤η) and due to (13) and Lemma 2, we have
0 < S − 1 2 = 1 2 ∑ k = 0 n − r p n − r , k 2 ( x ) ≤ 1 2 2 e ( n − r ) x ( 1 − x ) ∑ k = 0 n − r p n − r , k ( x ) (14) = 1 2 2 e ( n − r ) x ( 1 − x ) 0 < S − 1 2 = 1 2 ∑ k = 0 n − r   p n − r , k 2 ( x ) ≤ 1 2 2 e ( n − r ) x ( 1 − x ) ∑ k = 0 n − r   p n − r , k ( x ) (14) = 1 2 2 e ( n − r ) x ( 1 − x ) {:[0 < S-(1)/(2)=(1)/(2)sum_(k=0)^(n-r)p_(n-r,k)^(2)(x) <= (1)/(2sqrt(2e(n-r)x(1-x)))sum_(k=0)^(n-r)p_(n-r,k)(x)],[(14)=(1)/(2sqrt(2e(n-r)x(1-x)))]:}\begin{align*} 0<S-\frac{1}{2} & =\frac{1}{2} \sum_{k=0}^{n-r} p_{n-r, k}^{2}(x) \leq \frac{1}{2 \sqrt{2 e(n-r) x(1-x)}} \sum_{k=0}^{n-r} p_{n-r, k}(x) \\ & =\frac{1}{2 \sqrt{2 e(n-r) x(1-x)}} \tag{14} \end{align*}0<S−12=12∑k=0n−rpn−r,k2(x)≤122e(n−r)x(1−x)∑k=0n−rpn−r,k(x)(14)=122e(n−r)x(1−x)
Combining (12) and (14), we obtain
| ∑ k = 0 n − r p n − r , k ( x ) ∑ j = 0 k p n + r , j ( x ) − 1 2 | ≤ ( 2 + 2 r + 1 8 e ) 1 ( n − r ) x ( 1 − x ) B n ( r ) ( sign ( t − x ) , x ) = | 2 A n , γ ( x ) − 1 | (15) ≤ 2 ( 2 + 2 r + 1 8 e ) 1 ( n − r ) x ( 1 − x ) ∑ k = 0 n − r   p n − r , k ( x ) ∑ j = 0 k   p n + r , j ( x ) − 1 2 ≤ 2 + 2 r + 1 8 e 1 ( n − r ) x ( 1 − x ) B n ( r ) ( sign ( t − x ) , x ) = 2 A n , γ ( x ) − 1 (15) ≤ 2 2 + 2 r + 1 8 e 1 ( n − r ) x ( 1 − x ) {:[|sum_(k=0)^(n-r)p_(n-r,k)(x)sum_(j=0)^(k)p_(n+r,j)(x)-(1)/(2)| <= (2+2r+(1)/(sqrt(8e)))(1)/(sqrt((n-r)x(1-x)))],[B_(n)^((r))(sign(t-x)","x)=|2A_(n,gamma)(x)-1|],[(15) <= 2(2+2r+(1)/(sqrt(8e)))(1)/(sqrt((n-r)x(1-x)))]:}\begin{align*} \left|\sum_{k=0}^{n-r} p_{n-r, k}(x) \sum_{j=0}^{k} p_{n+r, j}(x)-\frac{1}{2}\right| & \leq\left(2+2 r+\frac{1}{\sqrt{8 e}}\right) \frac{1}{\sqrt{(n-r) x(1-x)}} \\ B_{n}^{(r)}(\operatorname{sign}(t-x), x) & =\left|2 A_{n, \gamma}(x)-1\right| \\ & \leq 2\left(2+2 r+\frac{1}{\sqrt{8 e}}\right) \frac{1}{\sqrt{(n-r) x(1-x)}} \tag{15} \end{align*}|∑k=0n−rpn−r,k(x)∑j=0kpn+r,j(x)−12|≤(2+2r+18e)1(n−r)x(1−x)Bn(r)(sign(t−x),x)=|2An,γ(x)−1|(15)≤2(2+2r+18e)1(n−r)x(1−x)
Combining estimates of (6), (7), (8) and (15), our theorem follows.

Acknowledgement

This article was completed in the framework of a cooperation with the Department of Mathematics of the Rome University "La Sapienza".

References

  1. O. Agratini, On the rate of convergence of some integral operators for functions of bounded variation. Studia Sci. Math. Hungar. 42(2005), No. 2, 235-252
  2. R. N. Bhattacharya and R. Ranga Rao, Normal approximation and asymptotic expansions. Wiley Series in Probability and Mathematical Statistics. John Wiley & Sons, New York-London-Sydney, 1976.
  3. J. L. Durrmeyer, Une formule d'inversion de la transformee de Laplace: Application a la Theorie des Moments. These de 3e cycles, Faculte des Sciences de l' Universite de Paris, 1967.
  4. S. S. Guo, On the rate of convergence of the Durrmeyer operator for functions of bounded variation. J. Approx. Theory 51(1987), No. 2, 183-192.
  5. V. Gupta, A note on the rate of convergence of Durrmeyer type operators for function of bounded variation. Soochow J. Math. 23(1997), No. 1, 115-118.
  6. V. Gupta and P. Maheshwari, Bezier variant of a new Durrmeyer type operators. Riv. Mat. Univ. Parma (7) 2(2003), 9-21.
  7. V. Gupta and G. S. Srivastava, Approximation by Durrmeyer-type operators. Ann. Polon. Math. 64(1996), No. 2, 153-159.
  8. X. M. Zeng, Bounds for Bernstein basis functions and Meyer-König and Zeller basis functions. J. Math. Anal. Appl. 219(1998), No. 2, 364-376.
    (Received 19.09.2005; revised 22.04.2006)
    Authors' addresses:
    V. Gupta
School of Applied Sciences
Netaji Subhas Institute of Technology
Sector 3 Dwarka, New Delhi 110075,
India
E-mail: vijaygupta2001@hotmail.com

T. ShervashidzeA. Razmadze Mathematical InstituteGeorgian Academy of Science1, M. Aleksidze St., Tbilisi 0193GeorgiaE-mail: sher@rmi.acnet.geM. CraciunTiberiu Popoviciu, Institute of Numerical AnalysisP. O. Box 68-1, 3400 Cluj-NapocaRomaniaE-Mail: craciun@ictp.acad.ro

2006

Related Posts