On simultaneous approximation of the Bernstein Durrmeyer operators

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On simultaneous approximation of the Bernstein Durrmeyer operators
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  On simultaneous approximation of the Bernstein Durrmeyer operators q Vijay Gupta a, * , Antonio-Jesús López-Moreno b , José-Manuel Latorre-Palacios c a School of Applied Sciences, Netaji Subhas Institute of Technology, Sector 3 Dwarka, New Delhi 110 078, India b Departamento de Matemáticas, Universidad de Jaén, Campus Las Lagunillas, 23071-Jaen, Spain c IES Los Cerros, C/Cronista Juan de la Torre s/n, 23400-Úbeda, Jaén, Spain a r t i c l e i n f o Keywords: Bernstein polynomialsDurrmeyer operators Bézier   variant a b s t r a c t The present paper deals with the study of the rate of convergence of the  Bézier   variant of certain Bernstein Durrmeyer type operators in simultaneous approximation.   2009 Elsevier Inc. All rights reserved. 1. Introduction In the year 1967 Durrmeyer [2] introduced an integral modification of the Bernstein polynomials with the purpose of approximating Lebesgue integrable functions on [0,1] as D n ð  f  ;  x Þ ¼ ð n  þ 1 Þ X nk ¼ 0  p n ; k ð  x Þ Z   10  p n ; k ð t  Þ  f  ð t  Þ dt  ;  x  2 ½ 0 ; 1  ; where  p n ; k ð  x Þ¼  nk    x k ð 1   x Þ n  k .Derriennic [1] first studied these operators in details and she obtained some direct results in ordinary and simultaneousapproximation. Zeng and Chen [7] estimated the rate of convergence of the generalized Durrmeyer type operators for func-tions of bounded variation. Since then, many authors have defined different modifications of the Durrmeyer operators. Wepresent in this paper the family of Durrmeyer type operators P  n ; r  ð  f  ;  x Þ ¼ n  P nk ¼ 1  p n ; k ð  x Þ R  10  p n  1 ; k  1 ð t  Þ  f  ð t  Þ dt   þ  p n ; 0 ð  x Þ  f  ð 0 Þ ;  r   ¼  0 ; n r  ð n þ r   1 Þ r   1 P n  r k ¼ 0  p n  r  ; k ð  x Þ R  10  p n þ r   1 ; k þ r   1 ð t  Þ  f  ð t  Þ dt  ;  r   >  0 ; 8>>><>>>: where  r  ; n 2 N 0  with  r  6 n  and for any  a ; b 2 N 0 ,  a b ¼ a ð a  1 Þð a  b þ 1 Þ ,  a 0 ¼ 1 is the falling factorial. The operators of this family are linear and positive and they all are related by means of the following differentiation formula that we willprove later: D s P  n ; r  ð  f  ;  x Þ ¼  P  n ; r  þ s ð D s  f  ;  x Þ ;  8 r  ; s  2 N 0 :  ð 1 Þ 0096-3003/$ - see front matter   2009 Elsevier Inc. All rights reserved.doi:10.1016/j.amc.2009.02.052 q The present work carried out in joint collaboration, when the first author visited Departamento de Matemáticas, Universidad de Jaén during 25June–10 July 2008. *  Corresponding author. E-mail addresses:  vijaygupta2001@hotmail.com (V. Gupta), ajlopez@ujaen.es (A.-J. López-Moreno), jomalapa@ujaen.es (J.-M. Latorre-Palacios). Applied Mathematics and Computation 213 (2009) 112–120 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc  Notice that the family  P  n ; a , as a particular case ( r   ¼ 0), contains the sequence  P  n ; 0  introducedby Srivastava and Gupta [4].Inseveral recent papers manyauthors consider the socalled  Bézier   modifications of different sequences of operators thatare obtained by changing the classical basis functions by certain  Bézier   type basis functions. The  Bézier   modification of thefamily  P  n ; r   is given, for any  a P 1, by P  n ; r  ; a ð  f  ;  x Þ ¼ n  P nk ¼ 1 Q  ð a Þ n ; k ð  x Þ R  10  p n  1 ; k  1 ð t  Þ  f  ð t  Þ dt   þ  Q  ð a Þ n ; 0 ð  x Þ  f  ð 0 Þ ;  r   ¼  0 ; n r  ð n þ r   1 Þ r   1 P n  r k ¼ 0 Q  ð a Þ n  r  ; k ð  x Þ R  10  p n þ r   1 ; k þ r   1 ð t  Þ  f  ð t  Þ dt  ;  r   >  0 ; 8>>><>>>: where  Q  ð a Þ n ; k ð  x Þ¼  J  a n ; k ð  x Þ  J  a n ; k þ 1 ð  x Þ  and  J  n ; k ð  x Þ¼ P n j ¼ k  p n ;  j ð  x Þ .This new family of   Bézier   type operators, for  a ¼ 1, yields again the family  P  n ; r  . In particular,  P  n ; 0 ; a  is the sequence intro-duced by Gupta and Maheshwari [3].The aim of this paper is to study the approximation properties of the operators  P  n ; r  ; a  in the space of functions H   ¼ f  f   :  ½ 0 ; 1  ! R :  f   ð  x Þ  exists everywhere and be bounded on  ½ 0 ; 1 g : Usually the rate of approximation for this kind of linear positive operators is analyzed for the class of continuous functionsor, in a more general setting, for the class of functions of bounded variation (see, for example [6,3]). We would like to stressthat the space  H   is larger than the space of bounded variation functions on [0,1]. For instance, the function  f  ð  x Þ ¼ 0 ;  if   x  ¼  0 ;  x sin  1  x 2   ;  if 0  <  x 6 1 ( is not of bounded variation but  f   2 H  . Moreover, as a consequence of our results for  P  n ; r  ; a  we will also establish estimates of the simultaneous approximation error for the operators  P  n ; r  , that is to say, we also study the rate of convergence for thederivatives of   P  n ; r  .The operators  P  n ; r  ; a  also admit the integral representation P  n ; r  ; a ð  f  ;  x Þ ¼ Z   10 K  r  ; n ; a ð  x ; t  Þ  f  ð t  Þ dt  ; where the kernel  K  r  ; n ; a  is given by K  r  ; n ; a ð  x ; t  Þ ¼ n  P nk ¼ 1 Q  ð a Þ n ; k ð  x Þ  p n  1 ; k  1 ð t  Þþ  Q  ð a Þ n ; 0 ð  x Þ d ð t  Þ ;  r   ¼  0 ; n r  ð n þ r   1 Þ r   1 P n  r k ¼ 0 Q  ð a Þ n  r  ; k ð  x Þ  p n þ r   1 ; k þ r   1 ð t  Þ ;  r   >  0 : 8>>><>>>: Here,  d ð t  Þ  denotes the Dirac delta function.We will give our estimates in terms of the local moduli of continuity  X  x  ; X  x ; þ  and  X  x . For the class of bounded functionson the interval [0,1], we consider the following quantities: X  x  ð  f  ; d 1 Þ ¼  sup t  2½  x  d 1 ;  x  j  f  ð t  Þ  f  ð  x Þj ; X  x þ ð  f  ; d 2 Þ ¼  sup t  2½  x ;  x þ d 2  j  f  ð t  Þ  f  ð  x Þj ; X  x ð  f  ; k Þ ¼  sup t  2½  x   x = k ;  x þð 1   x Þ = k  j  f  ð t  Þ  f  ð  x Þj ; where  x 2½ 0 ; 1  and 0 6 d 1  6  x ; 0 6 d 2  6 1   x  and  k P 1. We summarize several properties of these moduli in the followingitems:(i)  X  x  ð  f  ; d 1 Þ and X  x þ ð  f  ; d 2 Þ aremonotonenon-decreasingwithrespect to d 1  and d 2  respectively. Also X  x ð  f  ; k Þ ismonotonenon-increasing with respect to  k .(ii) lim d 1 ! 0 þ X  x  ð  f  ; d 1 Þ¼ 0 ; lim d 2 ! 0 þ X  x þ ð  f  ; d 2 Þ¼ 0 and lim k !1 X  x ð  f  ; k Þ¼ 0 if   f   is continuous on the left, continuous on theright, or continuous at the point  x  respectively.(iii)  X  x  ð  f  ; d 1 Þ 6 X  x ð  f  ;  x = d 1 Þ  and  X  x þ ð  f  ; d 2 Þ 6 X  x ð  f  ; ð 1   x Þ = d 2 Þ .(iv)  X  x  ð  f  ; d 1 Þ 6 V   x x  d 1 ð  f  Þ ; X  x þ ð  f  ; d 2 Þ 6 V   x þ d 2  x  ð  f  Þ  and  X  x ð  f  ; k Þ 6 V   x þð 1   x Þ = k  x   x = k  ð  f  Þ .All these properties can be found in [6]. 2. Auxiliary results In the sequel we shall need several lemmas. First we will study the moments of the operators  P  n ; r  . For this purpose wedefine for any  m 2 N 0 V. Gupta et al./Applied Mathematics and Computation 213 (2009) 112–120  113  T  n ; r  ; m ð  x Þ ¼ ð n  þ  r  Þ r  n r   P  n ; r  ðð t     x Þ m ;  x Þ ¼ n  P nk ¼ 1  p n ; k ð  x Þ R  10  p n  1 ; k  1 ð t  Þð t     x Þ m dt   þ ð  x Þ m  p n ; 0 ð  x Þ ;  r   ¼  0 ; ð n  þ  r  Þ  P n  r k ¼ 0  p n  r  ; k ð  x Þ R  10  p n þ r   1 ; k þ r   1 ð t  Þð t     x Þ m dt  ;  r   >  0 : 8>>><>>>: Lemma 1.  The following claims hold: (i)  For any r  ; m 2 N 0  and x 2½ 0 ; 1  , it is satisfied the recurrence relation ð n  þ  m  þ  r   þ 1 Þ T  n ; r  ; m þ 1 ð  x Þ ¼  x ð 1   x Þ½ DT  n ; r  ; m ð  x Þþ 2 mT  r  ; n ; m  1 ð  x Þ þ ½ð m  þ  r  Þ  x ð 1 þ 2 m  þ 2 r  Þ T  n ; r  ; m ð  x Þ ; where for m ¼ 0 , we denote T  n ; r  ;  1 ð  x Þ¼ 0 . (ii)  For all r   2 N 0  and x 2½ 0 ; 1  , T  n ; r  ; 0 ð  x Þ ¼  1 ;  T  n ; r  ; 1 ð  x Þ ¼  r     x ð 1 þ 2 r  Þ n  þ  r   þ 1  ; T  n ; r  ; 2 ð  x Þ ¼  2 nx ð 1   x Þ þ  r  ð 1 þ  r  Þ  2 rx ð 2 r   þ 3 Þþ 2  x 2 ð 2 r  2 þ 4 r   þ 1 Þð n  þ  r   þ 1 Þð n þ  r   þ 2 Þ  : (iii)  For all r  ; m 2 N 0  and x 2½ 0 ; 1  , T  n ; r  ; m ð  x Þ ¼  O n ½ð m þ 1 Þ = 2    : Proof.  We first prove the recurrence relation of claim (i). For a start we tackle the case  r   >  0. By definition of   T  n ; r  ; m ð  x Þ , wehave  x ð 1   x Þ DT  n ; r  ; m ð  x Þ ¼ ð n þ  r  Þ X n  r k ¼ 0  x ð 1   x Þ Dp n  r  ; k ð  x Þ Z   10  p n þ r   1 ; k þ r   1 ð t  Þð t     x Þ m dt     m ð n  þ  r  Þ X n  r k ¼ 0  x ð 1   x Þ  p n  r  ; k ð  x Þ Z   10  p n þ r   1 ; k þ r   1 ð t  Þð t     x Þ m  1 dt  : Hence, using the identity  x ð 1   x Þ Dp n ; k ð  x Þ¼ð k  nx Þ  p n ; k ð  x Þ , we have  x ð 1   x Þ½ DT  n ; r  ; m ð  x Þþ mT  n ; r  ; m  1 ð  x Þ¼ð n þ r  Þ X n  r k ¼ 0 ½ k ð n  r  Þ  x   p n  r  ; k ð  x Þ Z   10  p n þ r   1 ; k þ r   1 ð t  Þð t    x Þ m dt  ¼ð n þ r  Þ X n  r k ¼ 0  p n  r  ; k ð  x Þ Z   10 ½ð k þ r   1 Þð n þ r   1 Þ t  þð n þ r   1 Þð t    x Þþ½ð 1  r  Þ  x ð 1  2 r  Þf g  p n þ r   1 ; k þ r   1 ð t  Þð t    x Þ m dt  ¼ð n þ r  Þ X n  r k ¼ 0  p n  r  ; k ð  x Þ Z   10 t  ð 1  t  Þ Dp n þ r   1 ; k þ r   1 ð t  Þð t    x Þ m dt  þð n þ r   1 Þ T  n ; r  ; m þ 1 ð  x Þþ ð 1  r  Þ  x ð 1  2 r  Þ½  T  n ; r  ; m ð  x Þ¼ð n þ r  Þ X n  r k ¼ 0  p n  r  ; k ð  x Þ Z   10 ½ð 1  2  x Þð t    x Þð t    x Þ 2 þ  x ð 1   x Þ Dp n þ r   1 ; k þ r   1 ð t  Þð t    x Þ m dt  þð n þ r   1 Þ T  n ; r  ; m þ 1 ð  x Þþ½ð 1  r  Þ  x ð 1  2 r  Þ T  n ; r  ; m ð  x Þ¼ð m þ 1 Þð 1  2  x Þ T  n ; r  ; m ð  x Þþð m þ 2 Þ T  n ; r  ; m þ 1 ð  x Þ mx ð 1   x Þ T  n ; r  ; m  1 ð  x Þþð n þ r   1 Þ T  n ; r  ; m þ 1 ð  x Þþ½ð 1  r  Þ  x ð 1  2 r  Þ T  n ; r  ; m ð  x Þ : Thus, we get  x ð 1   x Þ½ DT  n ; r  ; m ð  x Þþ 2 mT  n ; r  ; m  1 ð  x Þ ¼ ½ð 1   r  Þ  x ð 1  2 r  Þ ð m  þ 1 Þð 1  2  x Þ T  n ; r  ; m ð  x Þ þð n  þ  m þ  r   þ 1 Þ T  n ; r  ; m þ 1 ð  x Þ : For  r  ¼ 0, we have that T  n ; 0 ; m ð  x Þ ¼  n X nk ¼ 1  p n ; k ð  x Þ Z   10  p n  1 ; k  1 ð t  Þð t     x Þ m dt   |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}  ¼ e T  n ; 0 ; m ð  x Þ þð  x Þ m  p n ; 0 ð  x Þ  |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}  ¼  b T  n ; 0 ; m ð  x Þ : We can repeate the proof of the case  r   >  0 taking the lower limit of the sums as  k ¼ 1. In this way we prove that  e T  n ; 0 ; m ð  x Þ satisfy the identity of claim (i). Besides, by direct substitution it can be checked that for  b T  n ; 0 ; m ð  x Þ  claim (i) also holds.Once completed the proof of the recurrence relation, claims (ii) and (iii) can be derived easily.  h In Section 1, we asserted that the family of operators  P  n ; r   satisfy the differentiation formula (1). Let us prove it now. 114  V. Gupta et al./Applied Mathematics and Computation 213 (2009) 112–120  Lemma 2.  For any r  ; s 2 N 0  and x 2½ 0 ; 1  , D s P  n ; r  ð  f  ;  x Þ ¼  P  n ; r  þ s ð D s  f  ;  x Þ : Proof.  We will first prove the identity D s P  n ; 0 ð  f  ;  x Þ ¼  P  n ; s ð D s  f  ;  x Þ ð 2 Þ by induction on  s .Using the identity Dp n ; k ð  x Þ ¼  n ½  p n  1 ; k  1 ð  x Þ  p n  1 ; k ð  x Þ ;  ð 3 Þ we have DP  n ; 0 ð  f  ;  x Þ ¼  n X nk ¼ 1 Dp n ; k ð  x Þ Z   10  p n  1 ; k  1 ð t  Þ  f  ð t  Þ dt     n ð 1   x Þ n  1  f  ð 0 Þ¼  n 2 X nk ¼ 1 ½  p n  1 ; k  1 ð  x Þ  p n  1 ; k ð  x Þ Z   10  p n  1 ; k  1 ð t  Þ  f  ð t  Þ dt     n ð 1   x Þ n  1  f  ð 0 Þ¼  n 2  p n  1 ; 0 ð  x Þ Z   10  p n  1 ; 0 ð t  Þ  f  ð t  Þ dt     n ð 1   x Þ n  1  f  ð 0 Þþ  n 2 X n  1 k ¼ 1  p n  1 ; k ð  x Þ Z   10 ½  p n  1 ; k ð t  Þ  p n  1 ; k  1 ð t  Þ  f  ð t  Þ dt  ¼  n 2  p n  1 ; 0 ð  x Þ Z   10 ð 1   t  Þ n  1  f  ð t  Þ dt     n ð 1   x Þ n  1  f  ð 0 Þþ  n 2 X n  1 k ¼ 1  p n  1 ; k ð  x Þ Z   10  Dp n ; k ð t  Þ n f  ð t  Þ dt  : Integrating by parts the above integrals in the last equality, we get DP  n ; 0 ð  f  ;  x Þ ¼  n X n  1 k ¼ 0  p n  1 ; k ð  x Þ Z   10  p n ; k ð t  Þ Df  ð t  Þ dt   ¼  P  n ; 1 ð Df  ;  x Þ ; which means that the identity is satisfied for  s ¼ 1. Let us suppose now that the result holds for  s . Then D s P  n ; 0 ð  f  ;  x Þ ¼  P  n ; s ð D s  f  ;  x Þ ¼  n s ð n  þ  s   1 Þ s  1 X n  sk ¼ 0  p n  s ; k ð  x Þ Z   10  p n þ s  1 ; k þ s  1 ð t  Þ D s  f  ð t  Þ dt  : Again differentiating and making use of  (3), in a similar way we have D s þ 1 P  n ; 0 ð  f  ;  x Þ ¼  n s ð n  þ  s  1 Þ s  1 X n  sk ¼ 0 ð n    s Þ½  p n  s  1 ; k  1 ð  x Þ   p n  s  1 ; k ð  x Þ  Z   10  p n þ s  1 ; k  s  1 ð t  Þ D s  f  ð t  Þ dt  ¼  n s þ 1 ð n  þ  s  1 Þ s  1 X n  s  1 k ¼ 0  p n  s  1 ; k ð  x Þ Z   10 ½  p n þ s  1 ; k  s ð t  Þ  p n þ s  1 ; k  s  1 ð t  Þ D s  f  ð t  Þ dt  ¼  n s þ 1 ð n  þ  s  1 Þ s  1 X n  s  1 k ¼ 0  p n  s  1 ; k ð  x Þ Z   10  Dp n þ s ; k þ s ð t  Þ n  þ  s D s  f  ð t  Þ dt  ¼  n s þ 1 ð n  þ  s  1 Þ s X n ð s þ 1 Þ k ¼ 0  p n ð s þ 1 Þ ; k ð  x Þ Z   10  p n þ s ; k þ s ð t  Þ D s þ 1  f  ð t  Þ dt   ¼  P  n ; s þ 1 ð D s þ 1  f  ;  x Þ : Therefore we obtain the result for  s þ 1 which close the induction process.For the general statement of the lemma take  g   : ½ 0 ; 1 ! R  in such a way that  D r   g   ¼  f  . Then, D s P  n ; r  ð  f  ;  x Þ ¼  D s P  n ; r  ð D r   g  ;  x Þ ¼  D r  þ s P  n ; 0 ð  g  ;  x Þ ¼  P  n ; r  þ s ð D r  þ s  g  ;  x Þ ¼  P  n ; r  þ s ð D s  f  ;  x Þ ; where we have used (2) in the intermediate equalities.  h We prove now several lemmas that will be used in the following section. Lemma 3.  The following claims hold: (i)  For every x 2ð 0 ; 1 Þ  and k ; n 1 ; n 2  2 N 0 , with  0  <  k 6 n 1  <  n 2  we have  J  n 1 ; k ð  x Þ  <  J  n 2 ; k ð  x Þ : (ii)  For every x 2½ 0 ; 1  , r  2 N , V. Gupta et al./Applied Mathematics and Computation 213 (2009) 112–120  115  X n j ¼ 0  J  a n þ r  ;  j ð  x Þ   J  a n ;  j ð  x Þ   6 a rx : Proof.  Let us prove first claim(i). It is clear that it is sufficient to prove the inequality for  J  n ; k  and  J  n þ 1 ; k . For this purposetakeinto account first that from the definition of   J  n ; k  we know that  J  n þ 1 ; k   J  n ; k   ð 0 Þ¼ 0 and  J  n þ 1 ; k   J  n ; k   ð 1 Þ¼ 0 and also that DJ  n ; k ð  x Þ ¼  np n  1 ; k  1 ð  x Þ  and  DJ  n þ 1 ; k ð  x Þ ¼ ð n  þ 1 Þ  p n ; k  1 ð  x Þ : In this case, D ð  J  n þ 1 ; k    J  n ; k Þð  x Þ ¼ ð n  þ 1 Þ  nk   1    x k  1 ð 1   x Þ n  k þ 1   n n   1 k   1    x k  1 ð 1   x Þ n  k ¼  nk   1    x k  1 ð 1   x Þ n  k k   ð n  þ 1 Þ  x ð Þ : Then it is immediate that(a) for  x 2ð 0 ; 1 Þ ;  D ð  J  n þ 1 ; k   J  n ; k Þð  x Þ¼ 0 ()  x ¼  kn þ 1 ,(b) for  x 2ð 0 ;  kn þ 1 Þ ;  D ð  J  n þ 1 ; k   J  n ; k Þð  x Þ >  0.From (b) it is clear that ð  J  n þ 1 ; k   J  n ; k Þð  kn þ 1 Þ >  0 and then (a) together with the mean value theorem lead us to the conclusion.To prove claim (ii), we know that, provided  a P 1,  b a  a a 6 a ð b  a Þ whenever 0 6 a 6 b 6 1. Then, if we use claim (1), X n j ¼ 0  J  a n þ r  ;  j ð  x Þ   J  a n ;  j ð  x Þ    ¼ X n j ¼ 0  J  a n þ r  ;  j ð  x Þ  J  a n ;  j ð  x Þ   6 a X n j ¼ 0  J  n þ r  ;  j ð  x Þ  J  n ;  j ð  x Þ   6 a X n þ r  j ¼ 0  J  n þ r  ;  j ð  x Þ X n j ¼ 0  J  n ;  j ð  x Þ  ! ¼  a  1 þ ð n  þ  r  Þ  x   ð 1 þ  nx Þð Þ ¼  a rx :   Since for all  n 2 N 0 ,  J  n ; 0 ð  x Þ¼ 1, we can concludefromthe precedinglemma that for  x 2½ 0 ; 1  and 0 6 k 6 n 1  6 n 2  we havethat  J  n 1 ; k ð  x Þ 6  J  n 2 ; k ð  x Þ . Remark 1.  For every0 6 k 6 n ;  x 2ð 0 ; 1 Þ ; a P 1 and for all  n , if we applythe meanvalue theoremwe knowthat there exists n 2½ 0 ; 1   with  J  n ; k þ 1 ð  x Þ 6 n 6  J  n ; k ð  x Þ  such that Q  ð a Þ n ; k ð  x Þ ¼  J  a n ; k ð  x Þ  J  a n ; k þ 1 ð  x Þ ¼  a n a  1  J  n ; k ð  x Þ  J  n ; k þ 1 ð  x Þ   6 a  p n ; k ð  x Þ : Then, considering the results in [5, Theorem 1], we can write the following inequalities: Q  ð a Þ n ; k ð  x Þ 6 a  p n ; k ð  x Þ 6  a  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 enx ð 1   x Þ p   : As a consequence, we also have  K  n ; r  ; a ð  x ; t  Þ 6 a K  n ; r  ; 1 ð  x ; t  Þ . Remark 2.  For a fixed  x 2ð 0 ; 1 Þ , when n sufficiently large, it can be seen from Lemma 1, that  x ð 1   x Þ n  6 T  n ; r  ; 2 ð  x Þ 6 Cx ð 1   x Þ n for any  C   >  2. Lemma 4.  Let x 2ð 0 ; 1 Þ  and C   >  2 , then for n sufficiently large, we have k n ; r  ; a ð  x ;  y Þ ¼ Z   y 0 K  n ; r  ; a ð  x ; t  Þ dt  6 C  a  x ð 1   x Þ n ð  x   y Þ 2  ;  0 6  y  <  x ; 1   k n ; r  ; a ð  x ;  z  Þ ¼ Z   1  z  K  n ; r  ; a ð  x ; t  Þ dt  6 C  a  x ð 1   x Þ n ð  z    x Þ 2  ;  x  <  z   <  1 : The proof of the above lemma follows easily by using Remark 2. For instance, for the first inequality we have k n ; r  ; a ð  x ;  y Þ ¼ Z   y 0 K  n ; r  ; a ð  x ; t  Þ dt  6 Z   y 0 K  n ; r  ; a ð  x ; t  Þ ð t     x Þ 2 ð  y    x Þ 2  dt   ¼  P  n ; r  ; a ðð t     x Þ 2 ;  x Þð  y    x Þ 2  6 a P  n ; r  ; 1 ðð t     x Þ 2 ;  x Þð  y    x Þ 2  ¼  n r  ð n  þ  r  Þ r  a T  n ; r  ; 2 ð  x Þð  y    x Þ 2  : 3. Rate of convergence In this section we give the rate of convergence for bounded functions on [0,1] for the operators  P  n ; r  ; a . At the end of thesection we also analyze the simultaneous approximation for  P  n ; r  . 116  V. Gupta et al./Applied Mathematics and Computation 213 (2009) 112–120
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