D O I: 1 0 .1 5 0 1 / C o m m u a 1 _ 0 0 0 0 0 0 0 7 6 4 IS S N 1 3 0 3 –5 9 9 1
QUANTITATIVE ESTIMATES FOR JAIN-KANTOROVICH OPERATORS
EMRE DEN·IZ
Abstract. By using given arbitrary sequences, n > 0, n 2 N with the
property that limn!1n n= 0and limn!1 n= 0, we give a Kantorovich
type generalization of Jain operator based on the a Poisson disrtibition. Fristly we give the quantitative Voronovskaya type theorem. Then we also obtain the Grüss Voronovskaya type theorem in quantitative form .We show that they have an arbitrary good order of weighted approximation.
1. Introduction With the help of a Poisson type distribution;
! (k; ) = ( + k )
k 1
k! e
( +k ); k = 0; 1; 2; :: (1.1)
for 0 < < 1; j j < 1; in 1970, G. C. Jain [14] introduced a positive linear operator de…ned for f 2 C (R+) as
Pn[ ](f ) (x) = 1 X k=0 ! (k; nx) f k n ; (1.2) where 2 [0; 1) and 1 X k=0 ! (k; ) = 1:
As a particular case = 0; we obtain the well-known Szasz-Mirakyan operators studied in [6], [11] and [15]; Pn[0](f ) (x) Sn(f ) (x) = 1 X k=0 pn;k(x) f k n ; (1.3) where pn;k(x) = e nx (nx) k k! :
Received by the editors: March 24, 2016, Accepted: May 22, 2016.
2010 Mathematics Subject Classi…cation. Primary 41A36; Secondary 41A25.
Key words and phrases. Jain operators, Kantorovich operators, Voronovskaya type theorem, Grüss-Voronovskaya type theorem, Weighted approximation.
c 2 0 1 6 A n ka ra U n ive rsity 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 th e m a t ic s a n d S t a tis t ic s .
Recently, Agratini [3] studied class of integral type positive linear operators of Pn[ ]and obtained some approximation properties of them in weighted spaces. Also,
some authors studied generalizations of Jain’s operators ([10], [13],[16] and [17]). Now, we de…ne and investigate Kantorovich variant of Pn[ ] operator , in order
to obtain an approximation process for spaces of locally integrable functions on unbounded interval, replacing the sample values f (k=n) by the mean values of f in the intervals k n; k+1 n as follows: De…nition 1. For 2 [0; 1) S[ ]n (f ) (x) = n 1 X k=0 ! (k; nx) k+1 n Z k n f (t) dt (1.4) where ! (k; nx) = nx k! (nx + k ) k 1 e (nx+k ):
The Kantorovich method was applied to many generalizations of the Bernstein polynomials like for example Szász-Mirakyan, Baskakov and other operators. A recent contribution in this direction was given in [4]. We note that, P. L. Butzer [5] introduced and studied Szasz-Mirakyan-Kantorovich operators de…ned by
Sn[0](f ) (x) Kn(f ) (x) = n 1 X k=0 pn;k(x) k+1 n Z k n f (t) dt
for f 2 L1(0; 1), the space of integrable functions on unbounded interval [0; 1):
In this paper we study some approximation properties of the sequence of linear positive operators given by (1.4) in a weighted space.
The structure of the paper is as follows. In the second section, we calculate some moment of our operator in De…nition 1. In the third section, a Voronovskaya type theorem in quantitative form is obtained as well. In the fourth section, we also give a Grüss Voronovskaya type theorem in quantitative form. In the last section, some weighted approximation theorems are presented.
2. Moments of the Operators Sn[ ]
We begin with the following lemma which is necessary to prove the main result. Taking in view Lemma 1 in [2] has been established the following moments: Lemma 1. Let ej; j 2 N[f0g ; be the j-th monomial, ej(t) = tj: For the operators
de…ned by (1.2) (see also [14, Eq.(2.11)]) the moments are as follows: Pn[ ](e0) (x) = 1; Pn[ ](e1) (x) = x 1 ; P [ ] n (e2) (x) = x2 (1 )2 + x n (1 )3;
Pn[ ](e3) (x) = x3 (1 )3+ 3x2 n (1 )4 + x (2 + 1) n2(1 )5; Pn[ ](e4) (x) = x4 (1 )4+ 6x3 n (1 )5 + x2(8 + 7) n2(1 )6 + x 6 2+ 8 + 1 n3(1 )7 ; Pn[ ](e5) (x) = x5 (1 )5 + x 4 10 n (1 )6 + x 3 5 (5 + 4 ) n2(1 )7 +x215 1 + 4 + 2 2 n3(1 )8 + x 1 + 22 + 58 2+ 24 3 n4(1 )9 ; Pn[ ](e6) (x) = x6 (1 )6 + 15x5 n (1 )7 + x 45 (13 + 8 ) n2(1 )8+ x 330 3 + 8 + 3 2 n3(1 )9 +x231 + 292 + 478 2+ 144 3 n4(1 )10 +x1 + 4 13 + 82 + 111 2+ 30 3 n5(1 )11 :
Lemma 2. The operators Sn[ ]; de…ned by (1.4) the moments are as follows:
Sn[ ](e0) (x) = 1; Sn[ ](e1) (x) = x 1 + 1 2n; Sn[ ](e2) (x) = x2 (1 )2 + 2 2 + 2 x n (1 )3 + 1 3n2; Sn[ ](e3) (x) = x3 (1 )3 + 9 6 + 3 2 x2 2n (1 )4 + 7 10 + 15 2 8 3+ 2 4 x 2n2(1 )5 + 1 4n3; Sn[ ](e4) (x) = x4 (1 )4 + 2 4 2 + 2 x3 n (1 )5 + 15 12 + 18 2 8 3+ 2 4 x2 n2(1 )6 + 6 6 + 27 2 24 3+ 17 4 6 5+ 6 x n3(1 )7 + 1 5n4;
Sn[ ](e5) (x) = x5 (1 )5 + 5 5 2 + 2 x4 2n (1 )6 + 5 26 14 + 21 2 8 3+ 2 4 x3 3n2(1 )7 +5 18 4 + 42 2 28 3+ 19 4 6 5+ 6 x2 2n3(1 )8 + 62 + 44 + 566 2 392 3+ 595 4 386 5 6n4(1 )9 + 183 6 48 7+ 6 8 6n4(1 )9 ! x + 1 6n5; Sn[ ](e6) (x) = x 6 (1 )6 + 3 6 2 + 2 x5 n (1 )7 + 5 2 + 2 10 4 + 2 x4 n2(1 )8 +5 40 + 60 2 32 3+ 21 4 6 5+ 6 x3 n3(1 )9 + 129 + 168 + 612 2 224 3+ 400 4 218 5+ 99 6 24 7+ 3 8 x2 n4(1 )10 + 18 + 78 + 417 2 + 96 3+ 470 4 308 5 n5(1 )11 +269 6 134 7+ 48 8 10 9+ 10 n5(1 )11 ! x + 1 7n6:
Proof. Obviously by (1.4), we obtain S[ ]n (e0) (x) = 1: With a simple calculation,
we obtain that Sn[ ](e1) = Pn[ ](e1) (x)+ 1 2nP [ ] n (e0) (x) = x 1 + 1 2n; Sn[ ](e2) (x) = Pn[ ](e2) (x) + 1 nP [ ] n (e1) (x) + 1 3n2P [ ] n (e0) (x) = x 2 (1 )2 + x 2 2 + 2 n (1 )3 + 1 3n2; Sn[ ](e3) (x) = Pn[ ](e3) (x) + 3 2nP [ ] n (e2) (x) + 1 n2P [ ] n (e1) (x) + 1 4n3P [ ] n (e0) (x) = x 3 (1 )3 + x23 3 2 + 2 2n (1 )4 + x 7 10 + 15 2 8 3+ 2 4 2n2(1 )5 + 1 4n3
Similarly, for j 4; the proof can be done.
Lemma 3. The j-th order central moment of the operators Sn[ ] are as following
Sn[ ] '0x(t) (x) = 1; Sn[ ]'1x(t) (x) = x 1 + 1 2n; Sn[ ] '2x(t) (x) = x2 2 (1 )2+ x 1 + 2 2+ 3 n (1 )3 + 1 3n2; Sn[ ] '3x(t) (x) = x3 3 (1 )3 +x 23 2 + 2 2+ 3 2n (1 )4 +x5 5 2+ 12 3 8 4+ 2 5 2n2(1 )5 + 1 4n3; Sn[ ] '4x(t) (x) = x4 4 (1 )4+ x 32 2 3 + 2 2+ 3 n (1 )5 +x2 3 + 10 2 2 2 3+ 12 4 8 5+ 2 6 n2(1 )6 +x 5 + + 6 2+ 11 3 18 4+ 15 5 6 6+ 7 n3(1 )7 + 1 5n4; Sn[ ] '6x(t) (x) = x6 6 (1 )6 + x 53 4 5 + 2 2+ 3 n (1 )7 +x45 2 9 + 10 3 2+ 2 3+ 6 4 4 5+ 6 n2(1 )8 +x35 3 + 21 + 21 2+ 10 3+ 30 4+ 3 5 14 6+ 15 7 6 8+ 9 n3(1 )9 +x2 70 + 156 + 225 2 + 374 3+ 43 4 n4(1 )10 +7 5+ 160 6 153 7+ 84 8 24 9+ 3 10 n4(1 )10 ! +x 17 + 89 + 362 2+ 261 3+ 140 4+ 154 5 193 6 n5(1 )11 +196 7 117 8+ 45 9 10 10+ 11 n5(1 )11 ! + 1 7n6
where 'j
x(t) = (t x) j
; j = 0; 1; 2; ::: .
3. Voronovskaya Theorems
Let Bx2[0; 1) be the set of all functions f de…ned on [0; 1) satisfying the
con-dition jf (x)j Mf 1 + x2 with some constant Mf, depending only on f; but
independent of x. Bx2[0; 1) is called weighted space and it is a Banach space
endowed with the norm
kfkx2 = sup x2[0;1) f (x) 1 + x2: Let Cx2[0; 1) = C [0; 1) \ Bx2[0; 1) and by Ck x2[0; 1), we denote subspace of
all continuous functions f 2 Bx2[0; 1) for which limx!1 f (x)
1+x2 is …nite.
We know that usual …rst modulus of continuity ! ( ) does not tend to zero, as ! 0; on in…nite interval. Thus we use weighted modulus of continuity (f; ) de…ned on in…nite interval [0; 1) (see [12]). Let
(f; ) = sup
jhj< ; x2[0;1)
jf (x + h) f (x)j
(1 + h2) (1 + x2) for each f 2Cx2[0; 1) :
Now some elementary properties of (f; ) are collected in the following Lemma. Lemma 4. Let f 2 Ck
x2[0; 1) : Then,
i) (f; ) is a monotonically increasing function of ; 0: ii) For every f 2 Ck
x2[0; 1) ; lim
!0 (f; ) = 0:
iii) For each > 0;
(f; ) 2 (1 + ) 1 + 2 (f; ) : (3.1)
From the inequality (3.1) and de…nition of (f; ) we get
jf (t) f (x)j 2 1 + x2 1 + (t x)2 1 + jt xj 1 + 2 (f; ) (3.2) for every f 2 Cx2[0; 1) and x; t 2 [0; 1) :
Next, we give the quantitative Voronovskaya type theorem in weighted spaces, which states the following:
Theorem 1. Let f002 Ck
x2[0; 1) and 0 n< 1: Then, we have
sup x 0 nhS[ n] n (f ) (x) f (x) i 1 2[f0(x) + xf00(x)] (1 + x2)4 f 0 x2 n n 1 n + f00 x2 2 n + C n f00; 1 p n ; where n! 0; depending on n; as n ! 1 and C is a positive constant.
Proof. By the local Taylor’s formula there exist lying between x and y such that f (y) = f (x) + f0(x) (y x) +f 00 (x) 2 (y x) 2 + h (y; x) (y x)2; where h ( ; x) := f00( ) f00(x) 2
and h is a continuous function which vanishes at 0: Applying the operator S[ n]
n to
above equality, we obtain the equality S[ n] n (f ) (x) f (x) = f 0 (x) S[ n] n '1x(t) (x) + f00(x) 2 S [ n] n '2x(t) (x) +S[ n] n h ( ; x) (y x) 2 (x) ; also we can write that
S[ n] n (f ) (x) f (x) f0(x) 2n f00(x) 2n x f0(x) S[ n] n '1x(t) (x) 1 2n + f00(x) 2 h S[ n] n '2x(t) (x) x n i +S[ n] n h ( ; x) (y x) 2 (x) :
To estimate last inequality using the inequality (3.2) and the inequality j xj jy xj ; we can write that
jh ( ; x)j 1 + (y x)2 1 + x2 1 +jy xj 1 + 2 (f00; ) : Since jh ( ; x)j ( 2 1 + x2 1 + 2 2 (f00; ) ; jy xj < 8 1 + x2 (y x)4 4 1 + 2 2 (f00; ) ; jy xj choosing < 1, we have jh ( ; x)j 2 1 + x2 1 +(y x) 4 4 ! 1 + 2 2 (f00; ) 8 1 + x2 1 +(y x) 4 4 ! (f00; ) :
We deduce that S[ n] n jh ( ; x)j (y x) 2 (x) = 8 1 + x2 (f00; )nS[ n] n '2x(t) (x) +14S[ n] n '6x(t) (x) = 8 1 + x2 (f00; ) S[ n] n '2x(t) (x) ( 1 + 14S [ n] n '6x(t) (x) S[ n] n ('2x(t)) (x) ) : (3.3) Using Lemma 1 and calculating with simple, we have
S[ n] n '6x(t) (x) S[ n] n ('2x(t)) (x) x4 4 n (1 n)4 + x 3 21 2n n (1 n)5 + x2 660 n2(1 n) 6 + x 1122 n3(1 n) 7 + 1268 n4(1 n) 8: In (3.3), choosing = p1 n then, we have Sn[ ] jh ( ; x)j (y x)2 (x) C 1 + x2 x4+ x3+ x2+ x + 1 S[ n] n '2x(t) (x) f00; 1 pn ; where C is a positive constant. Thus we have desired result.
Remark 1. It is seen that Sn[ ] does not form an approximation process. In order
to transform it into an approximation process, the constant will be replaced by a number n2 [0; 1) and also
lim
n!1n n= 0:
The following estimate is Voronovskaya type asymptotic formula. Corollary 1. Let f002 Ck
x2[0; 1) , x > 0 be …xed and 0 n < 1: Then, we have
lim n!1n h S[ n] n (f ) (x) f (x) i = 1 2[f 0(x) + xf00(x)] : 4. Grüss Type Approximation
Let us to prove the following result called by us Grüss-Voronovskaya type theo-rem in quantitative form (see [9]).
Theorem 2. Suppose that the …rst and second derivative f0; g0; f00; g00 and (f g)00
exist at a point x 2 [0; 1) ; we have sup x 0 n S[ n] n (f g; x) Sn[ n](f ; x) Sn[ n](g; x) xf0(x) g0(x) (1 + x2)6 (f g)0 x2 n n 1 n + (f g)00 x2 2 n+ C n (f g) 00;p1 n + kfkx2 8 < : g 0 x2 n n 1 n + g00 x2 2 n+ C n g 00;p1 n 9 = ; + kgkx2 8 < : f 0 x2 n n 1 n + f00 x2 2 n+ C n f 00;p1 n 9 = ; +nAn(f ) An(g) ; where An(f ) = kf0kx2 n 1 n + 1 2n + kf00k x2 2 n and n! 0; depending on n; n ! 1: Proof. For x 2 [0; 1) ; we have
S[ n] n (f g; x) Sn[ n](f ; x) Sn[ n](g; x) x nf 0(x) g0(x) = S[ n] n (f g; x) f (x) g (x) 1 2n(f (x) g (x)) 0 x 2n(f (x) g (x)) 00 f (x) S[ n] n (g; x) g (x) 1 2ng 0(x) x 2ng 00(x) g (x) S[ n] n (f ; x) f (x) 1 2nf 0(x) x 2nf 00(x) + g (x) S[ n] n (g; x) Sn[ n](f ; x) f (x) = A1+ A2+ A3+ A4:
Thus we can write nS[ n]
n (f g; x) Sn[ n](f ; x) Sn[ n](g; x) xf0(x) g0(x)
(1 + x2)4
n
where n (1 + x2)6jA1j nS[ n] n ((f g; x) f (x) g (x)) 12(f (x) g (x))0 x2(f (x) g (x))00 (1 + x2)4 ; n (1 + x2)6jA2j jf (x)j (1 + x2) n S[ n] n (g; x) g (x) 12g0(x) x2g00(x) (1 + x2)4 ; n (1 + x2)6jA3j jg (x)j (1 + x2) n S[ n] n (f ; x) f (x) 12f0(x) x2f00(x) (1 + x2)4 ; n (1 + x2)6jA4j n S[ n] n (g; x) g (x) (1 + x2)3 S[ n] n (f ; x) f (x) (1 + x2)3 :
From Theorem 1, we have sup x 0 n jA1j (1 + x2)6 (f g) 0 x2 n n 1 n + (f g)00 x2 2 n+ C n (f g) 00;p1 n ; sup x 0 n jA2j (1 + x2)6 kfkx2 8 < : g 0 x2 n n 1 n + g00 x2 2 n +C n g 00;p1 n ; sup x 0 n jA3j (1 + x2)6 kgkx2 8 < : f 0 x2 n n 1 n + f00 x2 2 n +C n f 00;p1 n :
On the other hand, we can write S[ n] n (f ; x) f (x) = f0(x) S[ ]n '1x(t) (x) + 1 2S [ n] n f00( ) (t x) 2 ; x : Therefore, we have S[ n] n (f ; x) f (x) (1 + x2)3 jf0(x)j Sn[ ] '1x(t) (x) (1 + x2)3 + S[ n] n jf00( )j (t x)2; x 2 (1 + x2)3 kf0kx2 S[ ]n '1x(t) (x) (1 + x2)2 +1 2kf 00k x2 S[ n] n 1 + 2 (t x)2; x (1 + x2)2
where is a number between t and x. Case 1: t < < x; S[ n] n (f ; x) f (x) (1 + x2)3 kf 0k x2 Sn[ ] '1x(t) (x) (1 + x2)2 + 1 2kf 00k x2 Sn[ ] '2x(t) (x) (1 + x2) :
Case 2: x < < t; S[ n] n (f ; x) f (x) (1 + x2)3 kf0kx2 Sn[ ] '1x(t) (x) (1 + x2)2 + 1 2kf 00k x2 S[ n] n 1 + t2 (t x)2; x (1 + x2)2 = kf0kx2 Sn[ ]'1x(t) (x) (1 + x2)2 + 1 2kf 00k x2 ( Sn[ ] '2x(t) (x) 1 + x2 +2xS [ ] n '3x(t) (x) (1 + x2)2 + Sn[ ] '4x(t) (x) (1 + x2)2 ) : Thus, we obtain for two cases of that
sup x 0 S[ n] n (f ; x) f (x) (1 + x2)3 kf0kx2sup x 0 Sn[ ]'1x(t) (x) (1 + x2)2 + 1 2kf 00k x2 8 < :supx 0 Sn[ ]'2x(t) (x) 1 + x2 + 2 sup x 0 x Sn[ ]'3x(t) (x) (1 + x2)2 + supx 0 Sn[ ]'4x(t) (x) (1 + x2)2 9 = ; = kf0kx2 n 1 n + 1 2n + kf00k x2 2 n:= An(f ) : Thus the proof is completed.
5. Weight Approximation
Now, in this section we give some weight approximation theorems for the func-tions which belong to weighted space Ck
x2[0; 1) by S [ ]
n operators. For details of
proofs see [2] and [8]. Theorem 3. If f 2 Ck
x2[0; 1). then the inequality
sup x 0 S[ n] n (f ) (x) f (x) (1 + x2)52 K f ; r 1 n n !
is satis…ed for a su¢ ciently large n; where K is a constant. Theorem 4. For each f 2 Cxk2[0; 1), we have
lim
n!1 S
[ n]
n (f ) f
Now, we give the following theorem to approximation all functions in Cx2[0; 1) :
This type of results is given in Gadjiev et al. [7] for locally integrable functions. Theorem 5. For each f 2 Cx2[0; 1) and > 0; we have
lim n!1x2[0;1)sup S[ n] n (f ) (x) f (x) (1 + x2)1+ = 0: References
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Current address : Department of Mathematics, Faculty of Science and Arts, Kirikkale Univer-sity, 71450 Yahsihan, Kirikkale, Turkey
