Some Approximation Properties Of King Type Generalization Of Modi…ed Positive Linear Operators

Some Approximation Properties Of King Type Generalization Of Modi…ed Positive Linear Operators

Prashantkumar Patel

Received 10 July 2019

Abstract

In manuscript, a variant of the operators de…ned in [1], which preserve the linear functions are inves- tigated. The rate of convergence of these operators with the help of K-functional is discussed. Also, the modi…cations of the operators which preservex² is de…ned. These modi…ed operators yield better error estimates than the operator de…ned in [1]. Asymptotic formula, Quantitative Voronovskaya theorems and Grüss-type approximation results for these King’s Type Generalized operators are established.

1 Introduction

In [1], the author constructed a class of linear and positive operators given by S^{[ ]}_n (f; x) =

X1 k=0

p (k; nx)f k

n ; n2N, (1)

where 0 <1 and p (k; nx) = (1 )1

k!(nx+ k)^ke ^{(nx+ k)}. For = 0, the operators Sn^{[ ]} reduce to Szàsz-Mirakjan operators [2] as

S_n^[0](f; x) =Sn(f; x) =e ^nx X1 k=0

(nx)^k k! f k

n ; x 0: (2)

In [3], King presented an example of operators of Bernstein type which preserve the test functionse0(x) = 1 and e2(x) = x² of the Bohman Korovkin theorem. Motivated by King’s work [3], the Szàsz-Mirakjan operators were modi…ed in [4] and [5]. Moreover, by letting vn(x) := x _2n¹, n 2 N; it has been shown in [5] that the operators de…ned by Vn(f;x) := Sn(f;vn(x))do not preserve the test functions e1(x) =x ande2(x) =x² but provide the best error estimation among all Szàsz-Mirakjan operators for all continuous bounded functionf on[0;1)and for eachx2[¹₂;1).

The aim of this manuscript is to investigate a variant of the operators (1) de…ned by Patel, which preserves the linear functions. The rate of convergence of these operators with the help of K-functional is disscussed.

The modi…cations of these operators which preservex²was introduced. These modi…ed operators yield better error estimates than the operators de…ned in [1]. Also, an asymptotic formula, quantitative Voronovskaya theorems and Grüss-type approximation results for these King’s type generalized operators are estabished.

2 Construction of The Operators

The following lemmas are needed for further discussion:

Mathematics Sub ject Classi…cations: primary 41A25, 41A30, 41A36.

yDepartment of Mathematics, St. Xavier’s College(Autonomous), Ahmedabad-380 009 (Gujarat), India

323

Lemma 1 ([1]) Let 0< <1,j j<1; r2N, P(r; ; ) =

X1 k=0

1

k!( + k)^k+re ^{( + k)} and

(1 )P(0; ; ) = 1:

Then

P(r; ; ) = X1 k=0

k( +k )P(r 1; +k ; ):

Whenj j<1 , one has P(1; ; ) =

(1 )² +

2

(1 )³; P(2; ; ) =

2

(1 )³ + 3 ² (1 )⁴ +

3(1 + 2 ) (1 )⁵ ; P(3; ; ) =

3

(1 )⁴ + 6 ^{2 2} (1 )⁵ +

3(4 + 11 ) (1 )⁶ +

4+ 8 ⁵+ 6 ⁶ (1 )⁷ ; P(4; ; ) =

4

(1 )⁵ + 10 ^{3 2}

(1 )⁶ +5 ² 2 ³+ 7 ⁴ (1 )⁷ +5 ⁴+ 10 ⁵+ 10 ⁶

(1 )⁸ +

5+ 22 ⁶+ 58 ⁷+ 24 ⁸

(1 )⁹ ;

P(5; ; ) =

5

(1 )⁶ + 15 ^{4 2}

(1 )⁷ +5 ³ 4 ³+ 17 ⁴

(1 )⁸ +15 ^{2 4}+ 12 ⁵+ 15 ⁶ (1 )⁹ + 6 ⁵+ 157 ⁶+ 508 ⁷+ 274 ⁸

(1 )¹⁰ +

6+ 52 ⁷+ 328 ⁸+ 444 ⁹+ 120 ¹⁰

(1 )¹¹ ;

P(6; ; ) =

6

(1 )⁷ + 21 ^{5 2}

(1 )⁸ +35 ^{4 3}+ 5 ⁴

(1 )⁹ +35 ^{3 4}+ 14 ⁵+ 21 ⁶ (1 )¹⁰ +7 ² 3 ⁵+ 91 ⁶+ 349 ⁷+ 232 ⁸

(1 )¹¹ +7 ⁶+ 60 ⁷+ 444 ⁸+ 728 ⁹+ 252 ¹⁰ (1 )¹²

+

7+ 114 ⁸+ 1452 ⁹+ 4400 ¹⁰+ 3708 ¹¹+ 720 ¹²

(1 )¹³ :

3 Estimation of Moments

The following lemmas are required to prove main results.

Lemma 2 ([1]) The operators Sn^{[ ]}; n 1, de…ned by (1) satis…es the following relations 1. S^{[ ]}n (1; x) = 1;

2. S^{[ ]}n (t; x) = ₍₁^x ₎+_{n(1 )}2;

3. S^{[ ]}n (t²; x) = ₍₁^x2₎2 +^{x(1+2 )}_{n(1 )}3 +_n2^{(1+2 )}(1 )⁴; 4. S^{[ ]}n (t³; x) = ₍₁^x3₎3 +^3x_n(1^{2(1+ )}₎4 +^x(^{1+8 +6 2})

n²(1 )⁵ + (^{1+8 +6 2})

n³(1 )⁶ ;

5. ^{[ ]}n (t⁴; x) =₍₁^x4₎4 +^2x_n(1^{3(3+2 )}₎5 +^x

2(^{7+26 +12 2})

n²(1 )⁶ +^x(^{1+22 +58 2+24 3})

n³(1 )⁷ + (^{1+22 +58 2+24 3})

n⁴(1 )⁸ : By a straightforward calculation, one obtain

S_n^{[ ]}(t⁵; x) = (1 )

n⁵ (P(5; nx+ 5 ; ) + 10P(4; nx+ 4 ; ) + 25P(3; nx+ 3 ; ) +15P(2; nx+ 2 ; ) +P(1; nx+ ; ))

= x⁵

(1 )⁵+5x⁴(2 + )

n(1 )⁶ +5x³ 5 + 12 + 4 ²

n²(1 )⁷ +15x² 1 + 9 + 14 ²+ 4 ³

n³(1 )⁸ :

+x 1 + 52 + 328 ²+ 444 ³+ 120 ⁴

n⁴(1 )⁹ + 1 + 52 + 328 ²+ 444 ³+ 120 ⁴

n⁵(1 )¹⁰ ;

S_n^{[ ]}(t⁶; x) = (1 )

n⁶ (P(6; nx+ 6 ; ) + 15P(5; nx+ 5 ; )

+65P(4; nx+ 4 ; ) + 90P(3; nx+ 3 ; ) + 31P(2; nx+ 2 ; ) +P(1; nx+ ; )): Hence, we have

S_n^{[ ]}(t⁶; x) = x⁶

(1 )⁶+3x⁵(5 + 2 )

n(1 )⁷ +5x⁴ 13 + 23 + 6 ²

n²(1 )⁸ +10x³ 9 + 50 + 55 ²+ 12 ³ n³(1 )⁹

+x² 31 + 562 + 1978 ²+ 1794 ³+ 360 ⁴ n⁴(1 )¹⁰

+x 1 + 114 + 1452 ²+ 4400 ³+ 3708 ⁴+ 720 ⁵ n⁵(1 )¹¹

+ 1 + 114 + 1452 ²+ 4400 ³+ 3708 ⁴+ 720 ⁵

n⁶(1 )¹² :

4 Approximation Properties

The modi…cation of the classical Bernstein operators proposed by King [3]. These type of Bernstein operators preserve test function e0 &e2 and give faster convergence. This approach was applied to some well known operators, details are found in [6, 7, 8, 9, 10]. One can observe that, the operator Sn^{[ ]} introduced in (1) preserve only the constant so further modi…cation of these operators is proposed to be made so that the modi…ed operators preserve the constant as well as linear functions, for this purpose the modi…cation ofS^{[ ]}n

is de…ned as follows:

S_n^{[ ]}(f; x) = X1 k=0

p (k; nrn(x))f k

n ; (3)

where

rn(x) =nx(1 )²

n(1 ) =x(1 )

n(1 )

andp (k; nx)is as de…ned in (1).

Lemma 3 The operators de…ned in (3) satis…es for eachx 0, the following identities S_n^{[ ]}(1; x) = 1;

S_n^{[ ]}(t; x) =x;

S_n^{[ ]}(t²; x) =x²+ x n(1 )² +

2

n²(1 )⁴;

S_n^{[ ]}(t³; x) =x³+ 3x²

n(1 )²+x 1 + 2 + 3 ² n²(1 )⁴ +

2(3 + 2 ) n³(1 )⁶; S_n^{[ ]}(t⁴; x) =x⁴+ 6x³

n(1 )²+x² 7 + 8 + 6 ²

n²(1 )⁴ +x 1 + 8 + 24 ²+ 8 ³

n³(1 )⁶ +

2 7 + 20 + 9 ² n⁴(1 )⁸ ; S_n^{[ ]}(t⁵; x) = x⁵+ 10x⁴

n(1 )² +5x³ 5 + 4 + 2 ²

n²(1 )⁴ +5x² 3 + 12 + 18 ²+ 4 ³ n³(1 )⁶ +x 1 + 22 + 133 ²+ 164 ³+ 45 ⁴

n⁴(1 )⁸ +

2 15 + 110 + 160 ²+ 44 ³

n⁵(1 )¹⁰ ;

S_n^{[ ]}(t⁶; x) = x⁶+ 15x⁵

n( 1 + )² +5x⁴ 13 + 8 + 3 ²

n²( 1 + )⁴ +10x³ 9 + 24 + 24 ²+ 4 ³ n³( 1 + )⁶ +x² 31 + 292 + 868 ²+ 684 ³+ 135 ⁴

n⁴( 1 + )⁸

+x 1 + 52 + 598 ²+ 1684 ³+ 1385 ⁴+ 264 ⁵ n⁵( 1 + )¹⁰

+31 ²+ 472 ³+ 1543 ⁴+ 1344 ⁵+ 265 ⁶

n⁶( 1 + )¹² :

By direct computation, the proof of the above identities can be obtained. By linear properties of the operatorsSn^{[ ]}, one has

S_n^{[ ]}(t x; x) = 0;

S_n^{[ ]}((t x)²; x) = x n(1 )² +

2

n²(1 )⁴; S_n^{[ ]}((t x)³; x) = x(1 + 2 )

n²(1 )⁴ +

2(3 + 2 ) n³(1 )⁶; S_n^{[ ]}((t x)⁴; x) = 3x²

n²(1 )⁴ +x 1 + 8 + 12 ² n³(1 )⁶ +

2 7 + 20 + 9 ² n⁴(1 )⁸ ; S_n^{[ ]}((t x)⁵; x) = 10x²(1 + 2 )

n³(1 )⁶ +x 1 + 22 + 98 ²+ 64 ³

n⁴(1 )⁸ +

2 15 + 110 + 160 ²+ 44 ³

n⁵(1 )¹⁰ ;

S_n^{[ ]}((t x)⁶; x) = 15x³

n³(1 )⁶ +5x² 5 + 32 + 35 ²

n⁴(1 )⁸ +x 1 + 52 + 508 ²+ 1024 ³+ 425 ⁴ n⁵(1 )¹⁰

+

2 31 + 472 + 1543 ²+ 1344 ³+ 265 ⁴

n⁶(1 )¹² :

Remark 1 Let n >1 be a given number. For every0< <1, one has S_n^{[ ]}((t x)²; x) 1

n(1 )² x+ 1 n(1 )² : Remark 2 Let n >1 be a given number. For every0< <1, one has

S_n^{[ ]}((t x)⁴; x) = 3n²x²(1 )⁴+ ² 7 + 20 + 9 ² +nx(1 )² 1 + 8 + 12 ² n⁴(1 s )⁸

3n²x²+ 36 + 21nx n⁴(1 )⁸

36(x²+x+ 1) n²(1 )⁸ :

Remark 3 Let n >1 be a given number. For every0< <1 andx2(0;1), one has Sn^{[ ]}((t x)⁶; x)

Sn^{[ ]}((t x)²; x) = 15n³x³(1 )⁶+ 5n²x²(1 )⁴(1 + 5 )(5 + 7 )+

n⁴(1 )⁸(nx(1 )²+n²x²(1 )⁴+ ²) +nx(1 )²(1 + (52 + (508 + (1024 + 425 ))))

n⁴(1 )⁸(nx(1 )²+n²x²(1 )⁴+ ²) +

2(31 + (472 + (1543 + (1344 + 265 )))) n⁴(1 )⁸(nx(1 )²+n²x²(1 )⁴+ ²)

15n³x³+ 360n²x²+ 2010nx+ 3655

n⁴(1 )⁸(nx(1 )²+n²x²(1 )⁴+ ²): (4) Using(1 )⁴ (1 )² and1< n < n², one get

nx(1 )²+n²x²(1 )⁴+ ² nx(1 )⁴+nx²(1 )⁴+ ² n(1 )⁴(x+x²):

Therefore,

1

nx(1 )²+n²x²(1 )⁴+ ²

1

n(1 )⁴(x+x²): (5)

Also,

15n³x³+ 360n²x²+ 2010nx+ 3655 360n³x²(x+ 1) + 3655n(x+ 1)

3655(x+ 1)n³(x²+ 1): (6) Using (5)and (6)in (4), one has

Sn^{[ ]}((t x)⁶; x) Sn^{[ ]}((t x)²; x)

3655(x+ 1)n³(x²+ 1)

n⁵(1 )¹²(x+x²) = 3655(x²+ 1) n²(1 )¹²x:

The general technique to construct sequences of operators of discrete type with the same property as given in [3] was obtained by Agratini [11]. With the help of this technique, the function is de…ned as

e r_n(x) =

p1 4 ²+ 4x²n²(1 )⁴ 1 2

2n(1 ) :

Using this function, the following linear and positive operators are de…ned Se_n^{[ ]}(f; x) =

X1 k=0

p (k; nern(x))f k

n : (7)

Direct computation gives

Se_n^{[ ]}(1; x) = 1;

Se_n^{[ ]}(t; x) =

p1 4 ²+ 4n²x²(1 )⁴ 1

2n(1 )² ;

Se_n^{[ ]}(t²; x) =x²: Also,

Se_n^{[ ]}(t x; x) =

p1 4 ²+ 4n²x²(1 )⁴ 1

2n(1 )² x;

Se_n^{[ ]}((t x)²; x) = 2x²+x 1 p

1 4 ²+ 4n²x²( 1 + )⁴ n(1 )²

! :

Now, by de…ning a function similar to the one used in [5],

b

rn(x) = (1 ) x 1

2n n(1 ): The linear and positive operators are de…ne, forn2Nandf 2C([0;1)),

Sb_n^{[ ]}(f; x) = X1 k=0

p (k; nbr_n(x))f k

n : (8)

The following identities, for the operatorsSbn^{[ ]},x ¹₂, can be obtained Se_n^{[ ]}(1; x) = 1;

Se_n^{[ ]}(t; x) =x 1 2n; Se_n^{[ ]}(t²; x) =x²+x(2 )

n(1 )² + 1 + 8 ² 4 ³+ ⁴ 4n²(1 )⁴ : and

Se_n^{[ ]}(t x; x) = 1 2n; Se_n^{[ ]}((t x)²; x) = x

n(1 )² + 1 + 8 ² 4 ³+ ⁴ 4n²(1 )⁴ :

The modi…cation of Jain operators [12], the author consider the restriction that n !0 as n! 1, in order to get convergence. Here, in (3),(7)and (8)need not to be take any restriction on . Because of this interesting property its worth to study these operators.

4.1 Asymptotic Formula

Theorem 4 Let f be a continuous function on [0;1)and 0 a < b <1. Then the sequence fSn^{[ ]}(f; )g converges uniformly to f asn! 1 in[a; b].

Proof. Using Lemma 3, observe that Sn^{[ ]}(1; x), Sn^{[ ]}(t; x) and Sn^{[ ]}(t²; x) converges uniformly to 1, x and x² as n! 1 respectively on every compact subset of [0;1). Thus, the required result follows from Bohman-Korovkin theorem.

Similarly, we have

Theorem 5 Let f be a continuous function on [0;1)and 0 a < b <1. Then the sequence fSen^{[ ]}(f; )g converges uniformly to f asn! 1 in[a; b].

Theorem 6 Let f be a continuous function on [0;1)and0 a < b <1. Then the sequencefSbn^{[ ]}(f; x)g converges uniformly to f(x)asn! 1 in[a; b].

Theorem 7 Let f be a bounded integrable function on [0;1) which has second derivative at a point x2 [0;1). Then

nlim!1n h

S_n^{[ ]}(f; x) f(x) i

= x

2(1 )²f⁰⁰(x):

Proof. By the Taylor’s expansion off, one has

f(t) =f(x) +f⁰(x)(t x) +1

2f⁰⁰²+r(t; x)(t x)²; (9)

wherer(t; x)is the remainder term andlim_t!xr(t; x) = 0. OperatingS_n to the equation (9), one obtains S_n^{[ ]}(f; x) f(x) =f⁰(x)S_n^{[ ]}((t x); x) +f⁰⁰(x)

2 S_n^{[ ]}((t x)²; x) +S_n^{[ ]}(r(t; x)(t x)²; x): (10) Using the Cauchy-Schwarz inequality, one has

S_n^{[ ]}(r(t; x)(t x)²; x) q

Sn^{[ ]}((r(t; x))²; x) q

Sn^{[ ]}((t x)⁴; x): (11) Asr(x; x) = 0andr²(t; x)2C₂[0;1), one has

nlim!1S_n^{[ ]}((r(t; x))²; x) = (r(x; x))²= 0 (12) uniformly with respect tox2[0; A]. From (11) and (12), one has

nlim!1nh

S_n^{[ ]}(f; x) f(x)i

= lim

n!1n f⁰(x)S_n^{[ ]}((t x); x) +f⁰⁰(x)

2 S_n^{[ ]}((t x)²; x) +S_n^{[ ]}(r(t; x)(t x)²; x)

i :

= f⁰(x) lim

n!1nS_n^{[ ]}((t x); x) +f⁰⁰(x) 2 lim

n!1nS_n^{[ ]}((t x)²; x)

= f⁰⁰(x) 2 lim

n!1

x (1 )² +

2

n(1 )⁴ = x

2(1 )²f⁰⁰(x):

Theorem 8 Let f be a bounded integrable function on [0;1) which has second derivative at a point x2 [0;1). Then

nlim!1n

hSb_n^{[ ]}(f; x) f(x) i

= x

2(1 )²f⁰⁰(x):

The proof of the above theorem follows along the lines of Theorem7, thus the details are omitted.

4.2 The Rate of Convergence

The setC_B([0;1))is the class of real valued continuous bounded functionsf onx2[0;1)with the norm kfk= sup_x2[0;1)jf(x)j. Forf 2C_B([0;1))and >0 them-th order modulus of continuity is de…ned as

!_m(f; ) = sup

0 h

sup

x2[0;1)j4^mhf(x)j;

where 4 is the forward di¤erence. For m = 1, !_m(f; )is the usual modulus of continuity. The Peetre’s K-functional is de…ned as

K₂(f; ) = inf

g2C_B²([0;1))fkf gk+ kg⁰⁰k:g2C_B² ([0;1))g; where

C_B² ([0;1)) =fg2CB([0;1)) :g⁰; g⁰⁰2CB([0;1))g: The following direct results are established in this section:

Theorem 9 Let f 2C_B([0;1))and0 <1. Then jS_n^{[ ]}(f; x) f(x)j 2K f; 1

2n(1 )² x+ 1

n(1 )² : (13)

Proof. Consider anyg 2 C_B² ([0;1)) and an arbitrarily …xedx 2[0;1). Using the Taylor formula, one obtain

g(t) g(x) = (t x)g⁰(x) + Z t

x

(t u)g⁰⁰(u)du; t 0:

SinceSn^{[ ]} is a linear and positive operator, one obtains

jS_n^{[ ]}(g; x) g(x)j jg⁰(x)S_n^{[ ]}((t x); x)j+ S_n^{[ ]} Z t

x

(t u)g⁰⁰(u)du; x S_n^{[ ]}

Z t x

(t u)g⁰⁰(u)du ; x kg⁰⁰kS_n^{[ ]}

Z t x

(t u)du ; x kg⁰⁰kS_n^{[ ]} (t x)²; x kg⁰⁰k

n(1 )² x+ 1 n(1 )² : Forf 2C_B([0;1)), one obtain

jS_n^{[ ]}(f; x) f(x)j jS_n^{[ ]}(f g; x)j+jS_n^{[ ]}(g; x) g(x)j+jg(x) f(x)j 2kf gk+ kg⁰⁰k

n(1 )² x+ 1 n(1 )² : By using theK-functional described by

K(f; ) = inf

g2CB([0;1))(kf gk+ kgk); forx 0,

jS_n^{[ ]}(f; x) f(x)j 2K f; 1

2n(1 )² x+ 1

n(1 )² :

Theorem 10 Letf 2C_B([0;1))and0 <1. Then

jSe_n^{[ ]}(f; x) f(x)j 2!(f;e(x));

where

e(x) = vu

ut2x²+x 1 p

1 4 ²+ 4n²x²( 1 + )⁴ n(1 )²

! :

Proof. Letf 2C_B([0;1))andx 0. Using linearity and monotonicity ofSe_n, one gets jSe_n^{[ ]}(f; x) f(x)j !(f; ) 1 +1q

Sen^{[ ]}((t x)²; x) :

Choosing =e(x), the proof follows.

Theorem 11 Letf 2C_B([0;1)),0 <1andx ¹₂. Then jSb_n^{[ ]}(f; x) f(x)j 2! f;b(x) ; where

b(x) = s

x

n(1 )² + 1 + 8 ² 4 ³+ ⁴ 4n²(1 )⁴ :

4.3 Quantitative Voronovskaya Theorem

The weighted modulus of continuity is denoted by (f; )and given by (f; ) = sup

0 h< ;x2[0;1)

jf(x+h) f(x)j (1 +h²)(1 +x²) forf 2Cx²([0;1)). For everyf 2C_x2([0;1));

lim!0 (f; ) = 0 and

(f; ) = 2(1 + )(1 + ²) (f; ); >0; (14)

which was proved in [13, p. 359–360]. Using the de…nition of the weighted modulus of continuity, one write jf(y) f(x)j (1 +x²)(1 + (y x)²) (f;jx yj):

Also, using the property (14), one has

jf(y) f(x)j (1 + (y x)²)(1 +x²) (f;jx yj)

2 1 + jy xj 1 + ² (f; ) 1 + (y x)² (1 +x²): (15) Now, we give the quantitative Voronovskaya-type theorem in weighted spaces, as follows:

Theorem 12 LetSn^{[ ]} be de…ned as in (3), where0 <1. Iff 2C([0;1))andf⁰⁰2C_x2([0;1)), then one has for x2(0;1)that

jS_n^{[ ]}(f; x) f(x) f⁰⁰(x)

2 S_n^{[ ]}((t x)²; x)j 16(1 +x²)

n(1 )² f⁽²⁾; ⁴ s

3655(x²+ 1) n²(1 )¹²x

!

x+ 1

n(1 )² : Proof. Using the Taylor’s expansion off, one obtain

f(t) =f(x) +f⁰(x)(t x) +f⁰⁰(x)

2 (t x)²+R2(f;t; x); (16) where

R₂(f;t; x) =(t x)²

2! f⁽²⁾( ) f⁽²⁾(x) for 2(t; x):

Applying the operatorsSn^{[ ]}to both sides of (16), we obtain S_n^{[ ]}(f; x) f(x) f⁰(x)S_n^{[ ]}(t x; x) f⁰⁰(x)

2 S_n^{[ ]}((t x)²; x) =S_n^{[ ]}(R₂(f;t; x); x): (17)

Hence, using the inequalityj xj jt xj, jR₂(f;t; x)j:= 1

2jf⁽²⁾( ) f⁽²⁾(x)j 1

2 (f⁽²⁾;j xj) 1 + ( x)² (1 +x²) 1

2 (f⁽²⁾;jt xj) 1 + (t x)² (1 +x²)

1 + jt xj 1 + ² (f⁽²⁾; ) 1 + (t x)² (1 +x²):

Ifjt xj< , one has

jR₂(f;t; x)j 2 1 + ^{2 2} (f⁽²⁾; )(1 +x²) and ifjt xj , one has

jR₂(f;t; x)j 2 1 + ^{2 2}(t x)⁴

4 (f⁽²⁾; )(1 +x²):

Therefore any fort2[0;1)and choosing <1, we get

jR₂(f;t; x)j 2 1 + ^{2 2} 1 + (t x)⁴

4 (f⁽²⁾; )(1 +x²) 8 1 +(t x)⁴

4 (1 +x²) (f⁽²⁾; ):

Using the above inequality, we deduce

S_n^{[ ]}(jR2(f;t; x)j; x) = S_n^{[ ]}((t x)²jR₂(f;t; x)j; x)

8(1 +x²) (f⁽²⁾; ) S_n^{[ ]}((t x)²; x) + 1

4S_n^{[ ]}((t x)⁶; x) : Using above inequality, Remarks1,3 and the equation (17), one has

jS_n^{[ ]}(f; x) f(x) f⁰⁰(x)

2 S_n^{[ ]}((t x)²; x)j

8(1 +x²) (f⁽²⁾; ) S_n^{[ ]}((t x)²; x) + 1

4S_n^{[ ]}((t x)⁶; x) 8(1 +x²) (f⁽²⁾; )S_n^{[ ]}((t x)²; x) 1 + 1

4

3655(x²+ 1) n²(1 )¹²x : Choosing = 3655(x²+ 1)

n²(1 )¹²x

1=4

, one has

jS_n^{[ ]}(f; x) f(x) f⁰⁰(x)

2 S_n^{[ ]}((t x)²; x)j 16(1 +x²)

n(1 )² (f⁽²⁾; ) x+ 1 n(1 )² ; as desired.

4.4 Grüss-Type Approximation Results

Theorem 13 Let Sn^{[ ]} be de…ned as in (3), where 0 < 1. Let E be subspace of C([0;1)). f; g 2 E[C_x2([0;1))be such thatf²; g²2E[C_x2([0;1))andf g2E:Then for …xedx2[0;1)the inequality

jS_n^{[ ]}(f g; x) S_n^{[ ]}(f; x)S_n^{[ ]}(g; x)j Af(x)Ag(x);

where

Af(x) =p

32(1 +x²) ( (f²; ) + (1 +M )(1 +x²)kfkx² (f; )) and

Ag(x) =p

32(1 +x²) ( (g²; ) + (1 +M )(1 +x²)kgk^x2 (g; )):

Proof. Denote

Dn(f; g;x) =S_n^{[ ]}(f g; x) S_n^{[ ]}(f; x)S_n^{[ ]}(g; x): (18) Applyig the Cauchy-Schwarz inequality, one has

jDn(f; g;x)j p

Dn(f; f;x)Dn(g; g;x):

On the other hand, using (15),

jS_n^{[ ]}(f; x) f(x)j 2(1 +x²)(1 + ²) (f; )S_n^{[ ]} 1 + jy xj 1 + (y x)² ; x : (19) LetA(x; y) := 1 +^{jy xj} 1 + (y x)² . Since

A(x; y)

( 2(1 + ²); jy xj ; 2(1 + ²)^(y 4^x)4; jy xj ; we obtain for allx; t2[0;1)that

A(x; y) 2(1 + ²) 1 +(y x)⁴

4 : (20)

Using the inequalities (19) and (20), one gets

jS_n^{[ ]}(f; x) f(x)j 4(1 +x²)(1 + ²)² (f; )S_n^{[ ]} 1 +(y x)⁴

4 ; x

16(1 +x²) (f; ) 1 + ⁴S_n^{[ ]} (y x)⁴; x : (21) By de…nition (18),

D_n(f; f;x) = S_n^{[ ]}(f²; x) f²(x) +f²(x) S_n^{[ ]}(f; x) ²

= S_n^{[ ]}(f²; x) f²(x) + f(x) S_n^{[ ]}(f; x) f(x) +S_n^{[ ]}(f; x) : Sincef 2C_x2[0;1),

Sn^{[ ]}(f; x) 1 +x²

kfkx²Sn^{[ ]}((1 +t²); x) 1 +x² : Now, using0< <1 andn 1, one gets

S_n^{[ ]}((1 +t²); x) = 1 +x²+ x n(1 )² +

2

n²(1 )⁴ 1 +x²+

2+nx(1 )² n²( 1 + )⁴ 1 +x²+ 1

n²(1 )⁴+ x n(1 )⁴ 1

(1 )⁴ 2 +x²+x :

Ifx2[0;1], thenx 1 and ifx2(1;1), thenx x². Hence, one has S_n^{[ ]}((1 +t²); x) 2 1 +x²

(1 )⁴ : Finally,

Sn^{[ ]}(f; x) 1 +x²

kfkx²M (1 +x²)

1 +x² =kfkx²M : Hence,

jD_n(f; f;x)j jS_n^{[ ]}(f²; x) f²(x)j+jf(x) S_n^{[ ]}(f; x)j(kfkx²+M kfkx²) (1 +x²):

Using the inequality (21), one deduce

jDn(f; f;x)j 16(1 +x²) (f²; ) 1 + ⁴S_n^{[ ]} (y x)⁴; x

+16(1 +x²)²(1 +M )kfkx² (f; ) 1 + ⁴S_n^{[ ]} (y x)⁴; x 16(1 +x²) (f²; ) 1 + ⁴36(x²+x+ 1)

n²(1 )⁸

+16(1 +x²)²(1 +M )kfkx² (f; ) 1 + ⁴36(x²+x+ 1) n²(1 )⁸ and choosing = ⁴

q36(x²+x+1)

n²(1 )⁸ , one has

jDn(f; f;x)j 32(1 +x²) (f²; ) + (1 +M )(1 +x²)kfkx² (f; ) : We may obtain similar estimate forjDn(g; g;x)j, which completes the proof.

Acknowledgments. The author would like to thank the referee for his/her valuable suggestions which improved the paper considerably.

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