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Başlık: Blending type approximation by bézier-summation-integral type operatorsYazar(lar):ACAR, Tuncer; KAJLA, ArunCilt: 67 Sayı: 2 Sayfa: 195-208 DOI: 10.1501/Commua1_0000000874 Yayın Tarihi: 2018 PDF

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C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat. Volum e 67, N umb er 2, Pages 195–208 (2018) D O I: 10.1501/C om mua1_ 0000000874 ISSN 1303–5991

http://com munications.science.ankara.edu.tr/index.php?series= A 1

BLENDING TYPE APPROXIMATION BY

BÉZIER-SUMMATION-INTEGRAL TYPE OPERATORS

TUNCER ACAR AND ARUN KAJLA

Abstract. In this note we construct the Bézier variant of summation integral type operators based on a non-negative real parameter. We present a direct approximation theorem by means of the …rst order modulus of smoothness and the rate of convergence for absolutely continuous functions having a derivative equivalent to a function of bounded variation. In the last section, we study the quantitative Voronovskaja type theorem.

1. Introduction

In 1912 Bernstein introduced the most famous algebraic polynomials Bn(f ; x) in

approximation theory in order to give a constructive proof of Weierstrass’s theorem which is given by Bn(f ; x) = n X k=0 pn;k(x)f k n ; x 2 [0; 1]; where pn;k(x) = n k x k(1 x)n k

and he proved that if f 2 C[0; 1] then Bn(f ; x)

converges uniformly to f (x) in [0; 1]:

The Bernstein operators have been used in many branches of mathematics and computer science. Since their useful structure, Bernstein polynomials and their modi…cations have been intensively studied. Among the other we refer the readers to (cf. [2, 3, 12, 8, 23, 27]).

Received by the editors: June 12, 2017, Accepted: July 21, 2017. 2010 Mathematics Subject Classi…cation. Primary 41A25, 26A15.

Key words and phrases. Blending type approximation, Bézier-Summation-Integral type oper-ators, weighted approximation, rate of convergence.

c 2 0 1 8 A n ka ra U n ive rsity. C o m m u n ic a tio n s Fa c u lty o f S c ie n c e s U n ive rs ity o f A n ka ra -S e rie s A 1 M a t h e m a tic s a n d S t a tis tic s . C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra -S é rie s A 1 M a t h e m a tic s a n d S t a tis t ic s .

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For f 2 C[0; 1]; Chen et al. [10] introduced a generalization of the Bernstein operators based on a non-negative parameter (0 1) as follows:

Tn( )(f ; x) = n X k=0 p( )n;k(x)f k n ; x 2 [0; 1] (1.1) where p( )n;k(x) = n 2 k (1 )x + n 2 k 2 (1 )(1 x) + n k x(1 x) x k 1(1 x)n k 1

and n 2: They proved the rate of convergence, Voronovskaja type asymptotic formula and Shape preserving properties for these operators. For the special case,

= 1; these operators reduce the well-known Bernstein operators.

In [19], Kajla and Acar introduced a sequence of summation-integral type operators as follows: D( )n (f ; x) = (n + 1) n X k=0 p( )n;k(x) Z 1 0 pn;k(t) f (t)dt; (1.2)

where f 2 L1[0; 1] (the space of all Lebesgue integrable functions on [0; 1]),

pn;k(t) =

n k t

k(1 t)n k and p( )

n;k(x) is de…ned as above. In [19], Voronoskaja

type asymptotic formula, rate of convergence, local and global convergence results were established for these operators (1.2).

The aim of this paper is to introduce Bézier variant of the operators (1.2) and obtain the direct approximation results. Furthermore we study the rate of con-vergence for an absolutely continuous function f having a derivative f0 equivalent

with a function of bounded variation on [0; 1] and quantitative Voronovskaja type theorem.

A Bézier curve is a parametric curve frequently used in computer graphics and image processing. These are mainly used in approximation, interpolation, curve …tting etc. Bézier-Bernstein type operators were established by many mathemati-cians. The pioneer works in this direction are due to [3, 5, 9, 13, 24, 26, 28, 29, 30]. In these works, the direct approximation results were obtained and the rate of convergence for functions of bounded variation were established. The order of ap-proximation of the summation-integral type operators for functions with derivatives of bounded variation is estimated in [1, 4, 6, 7, 14, 15, 16, 17, 18, 21, 20, 22, 25].

For f 2 L1[0; 1]; we de…ne the Bézier variant of the operators Dn( )(f ; x) as

Sn;( )(f ; x) = (n + 1) n X k=0 Q( )n;k; (x) Z 1 0 pn;k(t)f (t)dt; x 2 [0; 1]; (1.3)

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where 1; Q( )n;k; (x) = [Jn;k; (x)] [Jn;k+1; (x)] and Jn;k; (x) = n

X

j=k

p( )n;j(x); when k n and 0 otherwise.

Alternatively we may rewrite the operators (1.3) as

Sn;( )(f ; x) = Z 1 0 M n; ; (x; t)f (t)dt; x 2 [0; 1]; (1.4) where Mn; ; (x; t) = (n + 1) n X k=0 Q( )n;k; (x)pn;k(t):

If = 1 then the operators Sn;( )(f ; x) reduce to the operators D( )n (f ; x):

Throughout this article, C denotes a positive constant independent of n and x, not necessarily the same at each occurrence.

To express our results we give the following auxiliary results. Lemma 1. [19] Let ei(t) = ti; i = 0; 4, then we have

(1) Dn( )(e0; x) = 1; (2) Dn( )(e1; x) = x + 1 2x (n + 2); (3) Dn( )(e2; x) = x2+ 2x2( 3n 4) (n + 2)(n + 3) + 2x(2n + 1) (n + 2)(n + 3) + 2 (n + 2)(n + 3); (4) Dn( )(e3; x) = x3+ 6x3( n(5 + 2n ) 2(1 + )) (n + 2)(n + 3)(n + 4) + 3x2(n(3n 2 1) + 10( 1)) (n + 2)(n + 3)(n + 4) + 18x(n + 1) (n + 2)(n + 3)(n + 4)+ 6 (n + 2)(n + 3)(n + 4); (5) Dn( )(e4; x) = x4+ x4( 4(n + 3)(16 + n(3 + 5n)) + 12 (n 3)(n 2)) (n + 2)(n + 3)(n + 4)(n + 5) +4x 3(n 2) (n(4n 3 1) + 33( 1)) (n + 2)(n + 3)(n + 4)(n + 5) + 24x2 n + 3n2+ 14( 1) 4n (n + 2)(n + 3)(n + 4)(n + 5) + 48x(2n 3 + 3) (n + 2)(n + 3)(n + 4)(n + 5)+ 24 (n + 2)(n + 3)(n + 4)(n + 5): Hence we get D( )n ((t x); x) = 1 2x n + 2 < 1(1 x) n + 1 ; 8x 2 [0; 1] and 8n 2 N (1.5) with 1 2 and Dn( )((t x)2; x) = 2x(1 x)(n 2) (n + 2)(n + 3) + 2 (n + 2)(n + 3): From [19], we have D( )n ((t x)2; x) < 2 (n + 2) 2 n(x); 8x 2 [0; 1] and 8n 2 N

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where 2

n(x) = '2(x) + (n+2)1 and '

2(x) = x(1 x): Then we can write

Dn( )((t x)2; x) < 2 2 n(x) n + 2 ; 2 2: (1.6) Dn( )((t x)4; x) = 12x3(x 2) (n(n 2 19) + 46 36) (n + 2)(n + 3)(n + 4)(n + 5) + 12x2(n(n 2 25) + 58 38) (n + 2)(n + 3)(n + 4)(n + 5) + 24x(3n 6 + 1) (n + 2)(n + 3)(n + 4)(n + 5)+ 24 (n + 2)(n + 3)(n + 4)(n + 5) (1.7) Remark 1. We have Sn;( )(e0; x) = n X k=0 Q( )n;k; (x) = [Jn;0; (x)] = 2 4 n X j=0 p( )n;j(x) 3 5 = 1; since n X j=0 p( )n;j(x) = 1:

Lemma 2. [19] Let f 2 C[0; 1]. Then for x 2 [0; 1] we have k D( )n (f ) k k f k :

2. Direct Estimates

To describe our results, we recall the de…nitions of the …rst order modulus of smoothness and the K-functional [11]. Let '(x) = px(1 x); f 2 C[0; 1]: The …rst order modulus of smoothness is given by

!'(f ; t) = sup 0<h t f x +h'(x) 2 f x h'(x) 2 ; x h'(x) 2 2 [0; 1] ; and the appropriate Peetre’s K-functional is de…ned by

K'(f ; t) = inf g2W'fjjf gjj + tjj'g 0jj + t2 jjg0jjg (t > 0); where W' = fg : g 2 ACloc; jj'g 0

jj < 1; jjg0jj < 1g and jj:jj is the uniform norm on C[0; 1]: It is well known that ([11], Thm. 3.1.2 ) K'(f ; t) !'(f ; t) which

means that there exists a constant M > 0 such that

M 1!'(f ; t) K'(f ; t) M !'(f ; t): (2.1)

Lemma 3. Let f 2 C[0; 1]. Then, for x 2 [0; 1]; we have k Sn;( )(f ) k k f k :

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Proof. Applying the inequality j a b j j a b j with 0 a; b 1; 1 and from de…nition of Q( )n;k; (x); we may write

0 < [Jn;k; (x)] [Jn;k+1; (x)] (Jn;k; (x) Jn;k+1; (x))

= p( )n;k(x):

Hence from the de…nition Sn;( )(f ; x) and Lemma 2, we obtain

k Sn;( )(f ) k k D( )n (f ) k k f k :

Now we study a direct approximation theorem for the operators Sn;( ).

Theorem 1. Suppose that f be in C[0; 1] and '(x) =px(1 x) then for every x 2 [0; 1); we have

j Sn;( )(f ; x) f (x) j< C!' f ;

1 p

n + 2 ; (2.2)

where C is a constant independent of n and x:

Proof. By the de…nition of K'(f ; t), for …xed n; x; we can choose g = gn;x 2 W'

such that jjf gjj +p 1 n + 2jj'g 0jj + 1 n + 2jjg 0jj 2K ' f ; 1 p n + 2 : (2.3)

Using Remark 1, we can write

j Sn;( )(f ; x) f (x) j j Sn;( )(f g; x) j +jf gj+ j Sn;( )(g; x) g(x) j (2.4)

2jjf gjj+ j Sn;( )(g; x) g(x) j :

We only need to compute the second term in the above equation. We will have to split the estimate into two domains, i.e. x 2 Fnc= [0; 1=n] and x 2 Fn= (1=n; 1):

Using the representation g(t) = g(x) + Z t x g0(u)du; we get Sn;( )(g; x) g(x) = Sn;( ) Z t x g0(u)du; x : (2.5) If x 2 Fn = (1=n; 1) then n(x) '(x): We have Z t x g0(u)du jj'g0jj Z t x 1 '(u)du : (2.6)

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For any x; t 2 (0; 1); we …nd that Z t x 1 '(u)du = Z t x 1 p u(1 u)du Z t x 1 p u+ 1 p 1 u du 2 jpt px j + jp1 t p1 x j = 2jt xj p 1 t +px+ 1 p 1 t +p1 x < 2jt xj p1 x+ 1 p 1 x 2p2 jt xj '(x) : (2.7)

Combining (2.5)-(2.7) and using Cauchy-Schwarz inequality, we obtain jSn;( )(g; x) g(x)j < 2 p 2jj'g0 1(x)Sn;( )(jt xj; x) 2p2jj'g0 1(x) Sn;( )((t x)2; x) 1=2 2p2jj'g0 1(x) Dn( )((t x)2; x) 1=2: From (1.6), we get jSn;( )(g; x) g(x)j < Cjj'g 0jj p n + 2: (2.8) For x 2 Fc n = [0; 1=n]; n(x) 1 p n + 2 and Z t x g0(u)du jjg0jj jt xj: Therefore, using Cauchy-Schwarz inequality we have

jSn;( )(g; x) g(x)j jjg0jjSn;( )(jt xj; x) Cjjg0jjpn(x) n + 2 < C n + 2jjg 0jj: (2.9)

From (2.8) and (2.9), we have

jSn;( )(g; x) g(x)j < C jj'g 0jj p n + 2+ 1 n + 2jjg 0jj : (2.10)

Using K'(f ; t) !'(f ; t) and (2.3), (2.4), (2.10), we get the desired relation (2.2).

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3. Rate of Convergence

In this section we would like to obtain the rate of convergence of the operators Sn;( )(f ; x) for an absolutely continuous function f having a derivative f0 equivalent

to a function of bounded variation on [0; 1]:

Throughout this section DBV [0; 1] will denote the class of all absolutely continuous functions f de…ned on [0; 1] and having on (0; 1); a derivative f0 equivalent with a function of bounded variation on [0; 1]: We notice that the functions f 2 DBV [0; 1] possess a representation

f (x) = Z x

0

g(t)dt + f (0);

where g 2 BV [0; 1]; i.e., g is a function of bounded variation on [0; 1]:

Lemma 4. Let x 2 (0; 1]; then for 1; 2 2 and su¢ ciently large n; we have

(1) #n; ; (x; y) = Z y 0 M n; ; (x; t)dt < 2 (n + 2) 2 n(x) (x y)2; 0 y < x; (2) 1 #n; ; (x; z) = Z 1 z M n; ; (x; t)dt < 2 (n + 2) 2 n(x) (z x)2; x < z < 1:

Proof. (i) From Lemmas 1 and 2, we get

#n; ; (x; y) = Z y 0 M n; ; (x; t)dt Z y 0 x t x y 2 Mn; ; (x; t)dt = Sn;( )((t x)2; x) (x y) 2 Dn( )((t x)2; x) (x y) 2 < 2 (n + 2) 2 n(x) (x y)2:

The proof of (ii) is similar to the proof of (i). Hence it is omitted.

Theorem 2. Let f 2 DBV (0; 1); 1 and let b

a(fx0) be the total variation of fx0

on [a; b] [0; 1]: Then for every x 2 (0; 1) and for su¢ ciently large n, we have Sn;( )(f ; x) f (x) < 1 + 1 f 0(x+) + f0(x ) s 2 (n + 2) n(x) + s 2 (n + 2) n(x) + 1 f 0(x+) f0(x ) + 2 2 n(x) (n + 2)x 1 [pn ] X k=1 x x (x=k)(fx0) + x pn xx (x=pn)(fx0) + 2 2 n(x) (n + 2)(1 x) 1 [pn ] X k=1 x+((1 x)=k) x (fx0) + 1 x p n x+((1 x)=pn) x (fx0);

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where 2 2 and the auxiliary function and fx0 is de…ned by fx0(t) = 8 < : f0(t) f0(x ); 0 t < x 0; t = x f0(t) f0(x+) x < t 1:

Proof. Using the fact that Z 1 0 M n; ; (x; t)dt = Sn;( )(e0; x) = 1, we have Sn;( )(f ; x) f (x) = Z 1 0 [f (t) f (x)]Mn; ; (x; t)dt = Z 1 0 Z t x f0(u)du Mn; ; (x; t)dt: (3.1)

From de…nition of the function f0

x; for any f 2 DBV (0; 1); we can write

f0(t) = 1 + 1 f 0(x+) + f0(x ) + f0 x(t) +1 2 f 0(x+) f0(x ) sgn(t x) + 1 + 1 + x(t) f0(x) 1 2 f 0(x+) + f0(x ) ; (3.2) where x(t) = 1 ; x = t 0 ; x 6= t: It is clear that Z 1 0 M n; ; (x; t) Z t x f0(x) 1 2 f 0(x+) + f0(x ) x(t)dudt = 0:

By (1.4) and simple computations, we have

P1 = Z 1 0 Z t x 1 + 1 f 0(x+) + f0(x ) du M n; ; (x; t)dt = 1 + 1 f 0(x+) + f0(x ) Z 1 0 jt xjM n; ; (x; t)dt 1 + 1 f 0(x+) + f0(x ) S( ) n; ((t x)2; x) 1=2

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and P2 = Z 1 0 Zt x 1 2 f 0(x+) f0(x ) sgn(u x) + 1 + 1 du Mn; ; (x; t)dt = 1 2 f 0(x+) f0(x ) Z x 0 Zx t sgn(u x) + 1 + 1 du Mn; ; (x; t)dt + Z 1 x Zt x sgn(u x) + 1 + 1 du Mn; ; (x; t)dt + 1 f 0(x+) f0(x ) Z 1 0 jt xj M n; ; (x; t)dt = + 1 f 0(x+) f0(x ) S( ) n; (jt xj ; x) + 1 f 0(x+) f0(x ) S( ) n; ((t x)2; x) 1=2 :

By using (1.6) and considering (3.1), (3.2) we obtain the following estimate Sn;( )(f ; x) f (x) < jEn; ; (fx0; x) + Fn; ; (fx0; x)j + 1 + 1jf 0(x+) + f0(x )j s 2 (n + 2) n(x) + + 1jf 0(x+) f0(x )j s 2 (n + 2) n(x); (3.3) where En; ; (fx0; x) = Z x 0 Z t x fx0(u)du Mn; ; (x; t)dt and Fn; ; (fx0; x) = Z 1 x Z t x fx0(u)du Mn; ; (x; t)dt:

To complete the proof, it is su¢ cient to estimate the terms En; ; (fx0; x); Fn; ; (fx0; x):

Since Rabdt#n; ; (x; t) 1 for all [a; b] [0; 1]; using integration by parts and

ap-plying Lemma 4 with y = x (x=pn); we have jEn; ; (fx0; x)j = Z x 0 Z t x fx0(u)du dt#n; ; (x; t) = Z x 0 #n; ; (x; t)fx0(t)dt Z y 0 + Z x y jf 0 x(t)j j#n; ; (x; t)j dt < 2 2 n(x) (n + 2) Z y 0 x t(fx0)(x t) 2dt + Z x y x t(fx0)dt 2 2n(x) (n + 2) Z y 0 x t(fx0)(x t) 2dt + x p n x x (x=pn)(fx0):

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By the substitution of u = x=(x t); we get 2 2n(x) (n + 2) Z x (x=pn) 0 (x t) 2 xt(fx0)dt = 2 2 n(x) (n + 2)x 1 Z p n 1 x x (x=u)(fx0)du 2 2n(x) (n + 2)x 1 [pXn ] k=1 Z k+1 k x x (x=u)(fx0)du < 2 2 n(x) (n + 2)x 1 [pXn ] k=1 x x (x=k)(fx0):

Hence we reach the following result

jEn; ; (fx0; x)j < 2 2n(x) (n + 2)x 1 [pXn ] k=1 x x (x=k)(fx0) + x pn xx (x=pn)(fx0):

Using integration by parts and applying Lemma 4 with z = x + ((1 x)=pn); we have jFn; ; (fx0; x)j = Z 1 x Z t x fx0(u)du Mn; ; (x; t)dt = Z z x Z t x fx0(u)du dt(1 #n; ; (x; t)) + Z1 z Z t x fx0(u)du dt(1 #n; ; (x; t)) = Z t x fx0(u)du (1 #n; ; (x; t)) z x Z z x fx0(t)(1 #n; ; (x; t))dt + Z 1 z Z t x fx0(u)du dt(1 #n; ; (x; t)) = Z z x fx0(u)du(1 #n; ; (x; z)) Z z x fx0(t)(1 #n; ; (x; t))dt + Zt x fx0(u)du(1 #n; ; (x; t)) 1 z Z 1 z fx0(t)(1 #n; ; (x; t))dt = Z z x fx0(t)(1 #n; ; (x; t))dt + Z 1 z fx0(t)(1 #n; ; (x; t))dt < 2 2 n(x) (n + 2) Z 1 z t x(fx0)(t x) 2dt + Z z x t x(fx0)dt 2 2n(x) (n + 2) Z 1 x+((1 x)=pn) t x(fx0)(t x) 2dt + (1 x) pn x+((1 x)= pn) x (fx0):

By the substitution of u = (1 x)=(t x); we get

2 2n(x) (n + 2) Z1 x+((1 x)=pn) t x(fx0)(t x) 2dt = 2 2n(x) (1 x)(n + 2) Z pn 1 x+((1 x)=u) x (fx0)du < 2 2 n(x) (1 x)(n + 2) [pn ] X k=1 Z k+1 k x+((1 x)=u) x (fx0)du 2 2n(x) (1 x)(n + 2) [pn ] X k=1 x+((1 x)=k) x (fx0):

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Thus, we get jFn; ; (fx0; x)j < 2 2n(x) (1 x)(n + 2) [pXn ] k=1 x+((1 x)=k) x (fx0) +1p x n x+((1 x)=pn) x (fx0): (3.4)

Collecting the estimates (3.3)-(3.4), we get the required result. This completes the proof of theorem.

4. quantitative Voronovskaja-type theorem

Now we are going to study a quantitative Voronovskaja-type result for the op-erators Sn;( ): This result is given using the …rst order Ditzian-Totik modulus of

smoothness.

Theorem 3. Let f 2 C2[0; 1]: Then there hold

p n Sn;( )(f ; x) f (x) s 2 '2(x) + 1 n + 2 jjf 00jj + jjf00jj p n' 2(x) +pC n!'(x) f 00;2 p 3 p n'(x) ! + (n 1); p n Sn;( )(f ; x) f (x) s 2 '2(x) + 1 n + 2 jjf 00jj + jjf00jj p n' 2(x) +pC n!'(x)'(x) f 00;2 p 3 p n ! + (n 1)

Proof. Let f 2 C2[0; 1] and x; t 2 [0; 1]: Then Taylor’s expansion, we may write

f (t) f (x) = (t x)f0(x) + Z t x (t u)f00(u)du: Thus, f (t) f (x) = f0(x)(t x) 1 2(t x) 2f00(x) + Z t x (t u)f00(u)du Z t x (t u)f00(u)du: Operating Sn;( )( ; x) to both sides of the above relation, we get

jSn;( )(f ; x) f (x)j = jf0(x)jSn;( )(jt xj; x) + 1 2jf 00(x)jS( ) n; ((t x)2; x) +Sn;( ) Z t x jt uj jf 00(u) f00(x)jdu ; x : (4.1)

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Therefore, g 2 W' we have

Z t

x jt uj jf

00(u) f00(x)jdu jjf00 gjj(t x)2

+ 2jj'g0jj' 1(x)jt xj3:

Thus, in view of (4.1), (??), (1.7) and using Cauchy-Schwarz inequality, we may write Sn;( )(f ; x) f (x) jf0(x)jS( )n; (jt xj; x) + 1 2jf 00(x)jS( ) n; ((t x)2; x) +jjf00 gjjS( ) n; ((t x)2; x) + 2jj'g0jj' 1(x)Sn;( )(jt xj3; x) jjf0jj Sn;( )((t x)2; x) 1=2 +1 2jjf 00jjS( ) n; ((t x)2; x) + jjf00 gjjSn;( )((t x)2; x) +2jj'g0jj' 1(x) Sn;( )((t x)2; x) 1=2 Sn;( )((t x)4; x) 1=2 = s 2 (n + 2) ' 2(x) + 1 n + 2 jjf 00jj + jjf00jj n + 2 ' 2(x) + 1 n + 2 + 2 n + 2 ' 2(x) + 1 n + 2 jjf 000jj' 1(x) '2(x) + 1 n + 2 12x3(x 2) (n(n 2 19) + 46 36) (n + 2)(n + 3)(n + 4)(n + 5) +12x 2(n(n 2 25) + 58 38) (n + 2)(n + 3)(n + 4)(n + 5) + 24x(3n 6 + 1) (n + 2)(n + 3)(n + 4)(n + 5) + 24 (n + 2)(n + 3)(n + 4)(n + 5) 1=2 s 2 (n + 2) ' 2(x) + 1 n + 2 jjf 00jj + jjf00jj n + 2' 2(x) + 2 n + 2 ( '2(x)jjf000jj'(x)2 p 3 p n ) + (n 3=2): Since '2(x) '(x) 1; x 2 [0; 1]; we have Sn;( )(f ; x) f (x) s 2 (n + 2) ' 2(x) + 1 n + 2 jjf 00jj + jjf00jj n + 2' 2(x) + 2 n + 2 ( jjf000jj'(x)2 p 3 pn ) + (n 3=2): (4.2) Sn;( )(f ; x) f (x) s 2 (n + 2) ' 2(x) + 1 n + 2 jjf 00jj + jjf00jj n + 2' 2(x) + 2 n + 2'(x) ( jjf000jj2 p 3 p n ) + (n 3=2): (4.3)

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By taking the in…mum on the right hand side of the above relations over g 2 W , we get p n Sn;( )(f ; x) f (x) s 2 '2(x) + 1 n + 2 jjf 00jj + jjf00jj p n' 2(x) +pC nK' f 00;2 p 3 p n'(x) ! + (n 1); p n Sn;( )(f ; x) f (x) s 2 '2(x) + 1 n + 2 jjf 00jj + jjf00jj p n' 2(x) +pC n'(x)K' f 00;2 p 3 p n ! + (n 1):

Applying K'(f; t) !'(f; t), the theorem is proved.

The …rst author is partially supported by Research Project of Kirikkale Univer-sity, BAP, 2017/014 (Turkey).

References

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Current address : Tuncer Acar: Kirikkale University, Faculty of Science and Arts, Department of Mathematics, Yahsihan, 71450, Kirikkale, Turkey

E-mail address : tunceracar@ymail.com

ORCID Address: http://orcid.org/0000-0003-0982-9459

Current address : Arun Kajla: Department of Mathematics, Central University of Haryana, Haryana-123031, India

E-mail address : rachitkajla47@gmail.com

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