Vo lu m e 6 5 , N u m b e r 2 , P a g e s 9 7 –1 1 9 (2 0 1 6 ) 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 3 IS S N 1 3 0 3 –5 9 9 1
SOME APPROXIMATION PROPERTIES OF KANTOROVICH VARIANT OF CHLODOWSKY OPERATORS BASED ON
q INTEGER
AL·I KARA·ISA AND AL·I ARAL
Abstract. In this paper, we introduce two di¤erent Kantorovich type gener-alization of the q Chlodowsky operators. For the …rst operators we give some weighted approximation theorems and a Voronovskaja type theorem. Also, we present the local approximation properties and the order of convergence for unbounded functions of these operators . For second operators, we obtain a weighted statistical approximation property.
1. Introduction
In 1997, G. Phillips [21] introduced the generalization of Bernstein polynomials based on q integers. The author estimated the rate of convergence and obtained a Voronovskaja-type theorem for the generalization of Bernstein operators. Recently, generalizations of positive linear operators based on q integers were de…ned and studied by several authors. For example; Karsli and Gupta [3] introduced the following q Chlodowsky polynomials de…ned as:
(Cn;qf ) (x) = n X k=0 f [k]q [n]qbn ! n k q x bn k n kY1 i=0 1 qi x bn n k ; 06 x 6 bn;
where bnis a positive increasing sequence with bn! 1. They investigated the rate
of convergence and the monotonicity property of these operators. For more works, see references [4, 5, 6, 7, 8, 9, 10].
In this study, we de…ne Kantorovich type generalization of q Chlodowsky op-erators. We examine the statical approximation properties of our new operator by the help of Korovkin-type theorem in weighted space. Further, we present the local approximation properties and the order of convergence for unbounded functions of
Received by the editors: March 21, 2016, Accepted: May 25, 2016.
2010 Mathematics Subject Classi…cation. Primary 41A36; Secondary 41A25.
Key words and phrases. q Chlodowsky operators, modulus of continuity, local approximation, Peetre’s K-functional, statistical convergence.
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 tic s .
these operators. Furthermore, we prove and state a Voronovskaja type theorem for our new operators.
The Kantorovich type generalization of q Chlodowsky operators as follows:
Cnq(f ; x) = [n]q n X k=0 q k n k qP q n;k(x) [k+1]q [n]q Z [k]q [n]q f (bnt)dqt; (1.1) where n> 1, q 2 (0; 1] and Pn;kq (x) = bx n k 1 bx n n k q :
Assume that f is a monotone increasing function on [0; bn], then using (1.2) one
can easily veri…ed that Cq
n(f ; x) are linear and positive operators for 0 < q6 1.
Let us recall some de…nitions and notations regarding the concept of q calculus. For any …xed real number q > 0 and non-negative integer, the q integer of the number n is de…ned by
[n]q := (1 q
n)=(1 q),
q 6= 1
1; q = 0 :
The q factorial [n]q! is de…ned as following
[n]q! := [n]q[n 1]q [1]q; n 2 N
1; n = 0 :
The q binomial coe¢ cients are also de…ned as n
k
q
= [n]q!
[k]q! [n k]q!; 06 k 6 n:
The q analogue of the integration in the interval [0; b] is de…ned as (see [11])
b Z 0 f (t)dqt = (1 q)b 1 X j=0 f (qjb)qj; 0 < q < 1: Over a general interval [a; b], one can write
b Z a f (t)dqt = b Z 0 f (t)dqt a Z 0 f (t)dqt: (1.2)
Further results related to q calculus can be found in [1, 2]. 2. Some Basic Results
Lemma 2.1. By the de…nition of q integral, we have [k+1]q [n]q Z [k]q [n]q dqt = qk [n]q ; [k+1]q [n]q Z [k]q [n]q bntdqt = bnqk [n]2 q[2]q ([2]q[k]q+ 1) ; [k+1]q [n]q Z [k]q [n]q b2nt2dqt = b2 nqk [n]3q[3]q [3]q[k] 2 q+ (2q + 1) [k]q+ 1 ; [k+1]q [n]q Z [k]q [n]q b3nt3dqt = b3 nqk [n]4q[4]q [4]q[k] 3 q+ 3q 2+ 2q + 1 [k]2 q+ (3q + 1) [k]q+ 1 ; [k+1]q [n]q Z [k]q [n]q b4nt4dqt = b4 nqk [n]5q[5]q n [5]q[k]4q+ 4q3+ 3q2+ 2q + 1 [k]3q + 6q2+ 3q + 1 [k]2q+ (4q + 1) [k]q+ 1o: Lemma 2.2. The following equalities hold.
n X k=0 bn [k]q [n]q n k qP q n;k(x) = x; n X k=0 b2n[k] 2 q [n]2q n k qP q n;k(x) = q [n 1]q [n]q x 2+ bn [n]qx; n X k=0 b3n[k] 3 q [n]3q n k qP q n;k(x) = q3[n 1]q[n 2]q [n]2q x 3+ q 2+ 2q [n 1] qbn [n]2q x 2 + b 2 n [n]2qx;
n X k=0 b4n[k] 4 q [n]4q n k qP q n;k(x) = q6[n 1] q[n 2]q[n 3]q [n]3q x 4 +q 3 q2+ 2q + 3 [n 1] q[n 2]qbn [n]3q x 3 +q q 2+ 3q + 3 [n 1] qb2n [n]3q x 2+ b3n [n]3qx: Proof. Using the equality
[n]q = [j]q+ qj[n j]q; 06 j 6 n; (2.1) we have n X k=0 bn [k]q [n]q n k q Pn;kq (x) = x; n X k=0 b2n[k] 2 q [n]2q n k qP q n;k(x) = n X k=1 b2n[k]q [n]q n 1 k 1 qP q n;k(x) = n X k=1 b2nq [k 1]q+ 1 [n]q n 1 k 1 qP q n;k(x) = qb 2 n [n]q n X k=1 [k 1]q n 1 k 1 qP q n;k(x) + bn [n]qx = qb 2 n[n 1]q [n]q n X k=2 n 2 k 2 qP q n;k(x) + bn [n]qx = qb 2 n[n 1]q [n]q n X k=0 n 2 k qP q n;k+2(x) + bn [n]qx = q [n 1]q [n]q x 2+ bn [n]qx and n X k=0 b3n[k] 3 q [n]3q n k qP q n;k(x) = n X k=1 b3n[k] 2 q [n]2q n 1 k 1 qP q n;k(x) = b 3 nq2 [n]2q n X k=1 [k 1]2q n 1 k 1 qP q n;k(x) +b 3 n2q [n]2q n X k=1 [k 1]q n 1 k 1 q Pn;kq (x) + b 3 n [n]2q n X k=1 n 1 k 1 q Pn;kq (x)
= q 2b3 n[n 1]q [n]2q n X k=2 [k 1]q n 2 k 2 qP q n;k(x) +2qb 3 n[n 1]q [n]2q n X k=2 n 2 k 2 qP q n;k(x) + b2 n [n]2qx = q 3b3 n[n 1]q [n]2q n X k=2 [k 2]q n 2 k 2 q Pn;kq (x) +q 2b3 n[n 1]q [n]2q n X k=2 n 2 k 2 q Pn;kq (x) +2qbn[n 1]q [n]2q x 2+ b2n [n]2qx = q 3[n 1] q[n 2]q [n]2q x 3+ q 2+ 2q b n[n 1]q [n]2q x 2+ b2n [n]2qx: Finally, we have n X k=0 b4n[k] 4 q [n]4q n k qP q n;k(x) = n X k=1 b4n[k] 3 q [n]3q n 1 k 1 qP q n;k(x) = q 3b4 n [n]3q n X k=1 [k 1]3q n 1 k 1 qP q n;k(x) +3q 2b4 n [n]3q n X k=1 [k 1]2q n 1 k 1 qP q n;k(x) +3qb 4 n [n]3q n X k=1 [k 1]q n 1 k 1 qP q n;k(x) + b3 n [n]3qx = q 3b4 n[n 1]q [n]3q n X k=2 [k 1]2q n 2 k 2 q Pn;kq (x) +3q 2b4 n[n 1]q [n]3q n X k=2 [k 1]q n 2 k 2 q Pn;kq (x) +3qb 2 n[n 1]q [n]3q x 2+ b3n [n]3qx = q 5b4 n[n 1]q [n]3q n X k=2 [k 2]2q n 2 k 2 qP q n;k(x) +2q 4b4 n[n 1]q [n]3q n X k=2 [k 2]q n 2 k 2 q Pn;kq (x)
+q 3b2 n[n 1]q [n]3q x 2+3q 3[n 1] q[n 2]qbn [n]3q x 3 +3q 2[n 1] qb2n [n]3q x 2+3q [n 1]qb2n [n]3q x 2+ b3n [n]3qx = q 5[n 1] q[n 2]qb4n [n]3q n X k=3 [k 2]q n 3 k 3 q Pn;kq (x) + 2q 4+ 3q3 b n[n 1]q[n 2]q [n]3q x 3+ 3q 2+ 3q b2 n[n 1]q [n]3q x 2+ b3n [n]3qx = q 6[n 1] q[n 2]q[n 3]q [n]3q x 4+q 3 q2+ 2q + 3 [n 1] q[n 2]qbn [n]3q x 3 +q q 2+ 3q + 3 [n 1] qb2n [n]3q x 2+ b3n [n]3qx:
Lemma 2.3. The operators de…ned by (1.1) satisfy the following properties: Cnq(1; x) = 1; Cnq(t; x) = x + K0bn [n]q ; (2.2) Cnq t2; x = q [n 1]q [n]q x 2+K1bn [n]q x + K2b2n [n]2q ; (2.3) Cnq t3; x = q 3[n 1] q[n 2]q [n]2q x 3+qK3[n 1]qbn [n]2q x 2+K4b2n [n]2q x + b3 n [4]q[n]3q; Cnq t4; x = q 6[n 1] q[n 2]q[n 2]q [n]3q x 4+q 3K 5[n 1]q[n 2]qbn [n]3q x 3 +qK6[n 1]qb 2 n [n]3q x 2+K7b3n [n]3q x + b4 n [5]q[n]4q; for all x 2 [0; bn], where
K0 = 1 [2]q; K1= q2+ 3q + 2 [3]q ; K2= 1 [3]q; K4 = q3+ 4q2+ 6q + 2 [4]q ; K3= q4+ 3q3+ 6q2+ 5q + 3 [4]q ; K5 = q6+ 3q5+ 6q4+ 10q3+ 9q2+ 7q + 4 [5]q ;
K6 = q6+ 4q5+ 11q4+ 18q3+ 15q2+ 11q + 5 [5]q ; K7 = q4+ 5q3+ 10q2+ 10q + 4 [5]q :
Proof. By using de…nition of Cnq(f; x), Lemma 2.1 and Lemma 2.2, we get
Cnq(1; x) = [n]q n X k=0 q k n k qP q n;k(x) qk [n]q = 1; Cnq(t; x) = [n]q n X k=0 q k n k qP q n;k(x) qkb n [2]q[n]2q [2]q[k]q+ 1 = n X k=0 bn [k]q [n]q n k qP q n;k(x) + bn [2]q[n]q = x +K0bn [n]q ; Cnq t2; x = [n]q n X k=0 q k n k q Pn;kq (x) q kb2 n [3]q[n]3q [3]q[k] 2 q+ (2q + 1) [k]q+ 1 = b2n n X k=0 n k qP q n;k(x) + (2q + 1) b2n [3]q[n]3q n X k=0 [k]q [n]q n k qP q n;k(x) + b2n [3]q[n]2q = q [n 1]q [n]q x 2+K1bn [n]q x + K2b2n [n]2q : For t3, we get Cnq t3; x = [n]q n X k=0 q k n k qP q n;k(x) qkb3n [4]q[n]4q n [4]q[k]3q+ 3q2+ 2q + 1 [k]2q + (3q + 1) [k]q+ 1o = b3n n X k=0 [k]3q [n]3q n k q Pn;kq (x) + 3q 2+ 2q + 1 b3 n [4]q[n]4q n X k=0 [k]2q [n]2q n k q Pn;kq (x) +(3q + 1) b 3 n [4]q[n]2q n X k=0 [k]q [n]q n k qP q n;k(x) + b3n [4]q[n]3q = q 3[n 1] q[n 2]q [n]2q x 3+qK3[n 1]qbn [n]2q x 2+K4b2n [n]2q x + b3n [4]q[n]3q
and …nally Cnq t4; x = [n]q n X k=0 q k n k qP q n;k(x) qkb4 n [5]q[n]5q([5]q[k] 4 q + 4q3+ 3q2+ 2q + 1 [k]3q + 6q2+ 3q + 1 [k]2q + (4q + 1) [k]q+ 1) = b4n n X k=0 [k]4q [n]4q n k q Pn;kq (x) +b 4 n 4q3+ 3q2+ 2q + 1 [5]q[n]4q n X k=0 [k]3q [n]3q n k q Pn;kq (x) +b 4 n 6q2+ 3q + 1 [5]q[n]2q n X k=0 [k]2q [n]2q n k q Pn;kq (x) +b 4 n(4q + 1) [5]q[n]3q n X k=0 [k]q [n]q n k q Pn;kq (x) + b 4 n [5]q[n]4q = q 6[n 1] q[n 2]q[n 2]q [n]3q x 4+q 3K 5[n 1]q[n 2]qbn [n]3q x 3 +qK6[n 1]qb 2 n [n]3q x 2+K7b3n [n]3q x + b4 n [5]q[n]4q: 3. Weighted approximation We consider the following class of functions:
Let Hx2[0; 1) be the set of all functions f de…ned on [0; 1) satisfying the condition jf (x)j 6 Mf 1 + x2 ; where Mf is a constant depending only on f:
By Cx2[0; 1), we denote the subspace of all continuous functions belonging to Hx2[0; 1) : Also, let C
x2[0; 1) be the subspace of all functions f 2 Cx2[0; 1) ; for which lim
x!1 f (x)
1+x2 is …nite. The norm on Cx2[0; 1) is kfkx2= supx2[0; 1)jf (x)j1+x2: Now, we shall discuss the weighted approximation theorem, where the approxi-mation formula holds true on the interval [0; 1) :
Theorem 3.1. Let q = qn satisfy 0 < qn 6 1 and for n su¢ ciently large qn ! 1
and bn
[n]qn ! 0 with bn ! 1 : Let f 2 Cx2[0; 1) and f be a monotone increasing function on [0; 1) : Then we have
lim
n!106x6bsupn jCqn
n (f ; x) f (x)j
Proof. Setting the operators Cqn n (f ; x) = Cqn n (f ; x) if 06 x 6 bn f (x) if x > bn
and using the theorem in [15] for the operators Cqn
n ; we see that it is su¢ cient to
verify the following three conditions
lim n!1k C qn n (t ; x) x kx2 = lim n!106x6bsupn jCqn n (t ; x) x j 1 + x2 = 0; 1; 2: (3.1) Since Cqn
n (1; x) = 1 the …rst condition of (3.1) is ful…lled for = 0 :
Using Lemma 2.3, we can write sup 06x6bn jCqn n (t; x) xj 1 + x2 = 1 1 + qn bn [n]qn and sup 06x6bn Cqn n t2; x x2 1 + x2 6 06x6bsup n x2 1 + x2 bn [n]qn + sup 06x6bn x 1 + x2 qn2+ 3qn+ 2 q2 n+ qn+ 1 bn [n]qn + sup 06x6bn 1 1 + x2 1 q2 n+ qn+ 1 b2 n [n]2 qn : which implies that
lim n!106x6bsupn jCqn n (t; x) xj 1 + x2 = 0 and lim n!106x6bsupn Cqn n t2; x x2 1 + x2 = 0:
Thus the proof is completed.
We know that for f 2 Cx2[0; 1) Theorem 3.1 is not true (see [15]). But we can give following property of Cnq:
Theorem 3.2. Let q = qn satisfy 0 < qn 6 1 and for n su¢ ciently large qn ! 1
and bn
[n]qn ! 0 with bn ! 1 : Let f 2 Cx2[0; 1) and f be a monotone increasing function on [0; 1) : Then we have
lim n!1 1 p bn sup 06x6bn jCqn n (f ; x) f (x)j 1 + x2 = 0
Proof. f is continuous function we can write jf (t) f (x)j < " if jt xj < and jt xj > we have
Thus we can write
jf (t) f (x)j < " + Cf( ) (t x)2+ 1 + x2 jt xj :
for x 2 [0; bn] and t 2 [0; 1) : Since Cnqn linear and positive operator we have
jCqn n (f ; x) f (x)j 6 " + Cf( ) Cnqn (t x) 2 ; x +Cf( ) 1 + x2 Cnqn(jt xj ; x) 6 " + Cf( ) Cnqn (t x) 2 ; x +Cf( ) 1 + x2 r Cqn n (t x)2; x :
From Lemma 2.3, we have jCqn n (f ; x) f (x)j 1 + x2 6 " 1 + x2 + Cf( ) x (3bn x) (1 + x2) [n] qn + b 2 n [n]2 qn + s x (3bn x) [n]qn + b 2 n [n]2 qn # and sup 06x6bn jCqn n (f ; x) f (x)j 1 + x2 6 " sup 06x6bn 1 1 + x2 + Cf( ) x (3bn x) (1 + x2) [n] qn + b 2 n [n]2 qn + s x (3bn x) [n]qn + b 2 n [n]2 qn # 6 " + Cf( ) " 3bn [n]qn + b 2 n [n]2 qn + s 3b2 n [n]qn + b 2 n [n]2 qn # : Therefore 1 p bn sup 06x6bn jCqn n (f ; x) f (x)j 1 + x2 6 " p bn +Cf( ) " 3pbn [n]qn + b 3=2 n [n]2 qn + s 3bn [n]qn + bn [n]2 qn #
which proves the theorem.
4. Voronovskaja type theorem Now, we give a Voronovskaja type theorem for Cnq(f; x).
Lemma 4.1. Let q := (qn), 0 < qn6 1, be sequence such that qn ! 1 as n ! 1.
Then, we have the following limits: (i) limn!1[n]bnqnC qn n ((t x)2; x) = x (ii) limn!1 [n]2qn b2 n C qn n ((t x)4; x) = 2x2.
Proof. (i) From Lemma 2.3, we have Cqn n ((t x)2; x) = x2 [n]qn + bn [n]qn q3 n+ 2qn2+ 3qn q3 n+ 2q2n+ 2qn+ 1 x + b 2 n (q2 n+ qn+ 1)[n]2qn : (4.1) Then, we get [n]qn bn Cqn n ((t x)2; x) = x2 bn + q 3 n+ 2q2n+ 3qn q3 n+ 2q2n+ 2qn+ 1 x + bn (q2 n+ qn+ 1)[n]qn :
Let us take the limit of both sides of the above equality as n ! 1, then we have
lim n!1 [n]qn bn Cqn n ((t x)2; x) = lim n!1 x2 bn + q 3 n+ 2q2n+ 3qn q3 n+ 2q2n+ 2qn+ 1 x + bn (q2 n+ qn+ 1)[n]qn = x:
(ii) Again from Lemma 2.3 and by the linearity of the operators Cqn
n (f; x), we get Cnq((t x)4; x) = D1;nx4+ D2;nx3+ D3;nx2+ D4;nx + D5;n; where D1;n = q6 n[n 1]qn[n 2]qn[n 3]qn [n]3 qn 4q3 n[n 1]qn[n 2]qn [n]2 qn +6qn[n 1]qn [n]qn 3; D2;n = q3 n[n 1]qn[n 2]qn [n]2 qn K5;qn 4K3;qnqn[n 1]qn [n]qn + 6K2;qn 4K0;qn bn [n]qn ; D3;n = qn[n 1]qn [n]qn K6;qn+ 6K1;qn 4K4;qn b2 n [n]2 qn ; D4;n = K7;qn 4 q3 n+ qn2+ qn+ 1 b3n [n]3 qn ; D5;n = 1 q4+ q3 n+ qn2+ qn+ 1 b4 n [n]4 qn :
and K0;qn = 1 1 + qn ; K1;qn = qn2+ 3qn+ 2 q2 n+ qn+ 1 ; K2;qn = 1 q2 n+ qn+ 1 K3;qn = q4 n+ 3q3n+ 6qn2+ 5qn+ 3 q3 n+ qn2+ qn+ 1 ; K4;qn = qn3+ 4q2n+ 6qn+ 2 q3 n+ q2n+ qn+ 1 ; K5;qn = q6 n+ 3q5n+ 6qn4+ 10q3n+ 9qn2+ 7qn+ 4 q4 n+ qn3+ qn2+ qn+ 1 ; K6;qn = q6 n+ 4q5n+ 11qn4+ 18qn3+ 15qn2+ 11qn+ 5 q4 n+ qn3+ qn2+ qn+ 1 ; K7;qn = q4 n+ 5q3n+ 10qn2+ 10qn+ 4 q4 n+ qn3+ qn2+ qn+ 1 : By (2.1), we get lim n!1 [n]2 qn b2 n fD 1;ng = lim n!1 [n]2 qn b2 n ( (1 qn)2[n]2qn+ [n]qn(q 3 n+ 3q2n 1) (qn3+ qn2+ 2qn+ 1) [n]3 qn ) = lim n!1 (1 qn)2[n]qn b2 n +q 3 n+ 3q2n 1 b2 n qn3+ qn2+ 2qn+ 1 [n]qnb2n = lim n!1 (qn 1)(1 qnn) b2 n +q 3 n+ 3qn2 1 b2 n q3 n+ q2n+ 2qn+ 1 [n]qnbn = 0: (4.2) Again, by using (2.1), we have
[n]2 qn b2 n fD 2;ng = [n]qn bn ( [n]2 qn(K5;qn 4K3;qn+ 6K2;qn 4K0;qn) + [n]qn(4K3;qn qn 2)(qn+ 1)K6;qn [n]2 qn )
Taking the limit of both sides of the above equality, we get lim n!1 [n]2 qn b2 n fD2;ng = lim n!1 [n]qn(K5;qn 4K3;qn+ 6K2;qn 4K0;qn) bn +4K3;qn qn 2 bn +(qn+ 1)K6;qn [n]qnbn
Since limn!14K3;qnbnqn 2 = 0 and limn!1 (qn+1)K6;qn [n]qnbn = 0, we have lim n!1 [n]2 qn b2 n fD 2;ng = lim n!1 [n]qn(K5;qn 4K3;qn+ 6K2;qn 4K0;qn) bn = lim n!1 q2 n(1 qn)(1 qnn) (qn+ 1)(qn3+ qn2+ qn) (q8 n+ 3q7n+ 6qn6+ 9q5n+ 12q4n+ 9qn3+ 8q2n+ 7qn+ 5) (q5 n+ qn4+ q3n+ q2n+ qn+ 1)bn = 0: (4.3)
Finally, using (2.1), we get lim n!1 [n]2 qn b2 n fD3;ng = lim n!1 K6;qn [n]qn + K6;qn+ 6K1;qn 4K4;qn = K6;qn+ 6K1;qn 4K4;qn= 2: (4.4) It is clear that lim n!1 [n]2qn b2 n fD 4;nx + D5;ng = 0:
By combining (4.2)-(??), we reach the desired the result.
Theorem 4.2. Let f 2 Cx2[0; 1) such that f0; f002 Cx2[0; 1). Then, we have lim n!1 [n]qn bn (Cqn n (f; x) f (x)) = 1 2f 0(x) + x 2f 00(x):
Proof. We write Taylor’s expansion of f as follows: f (t) = f (x) + f0(x)(t x) + 1
2f
00(x)(t x)2+ "(t; x)(t x)2;
where "(t; x) ! 0 as t ! x. By linearity of the operators Cqn
n (f; x) we get Cqn n (f; x) f (x) = f0(x)Cnqn((t x); x)+ 1 2f 00(x)Cqn n ((t x)2; x)+Cnqn "(t; x)(t x)2; x :
From Lemma 2.3, we have
Cqn n (f; x) f (x) = f0(x) bn (1 + qn)[n]qn +1 2f 00(x) x2 [n]qn + bn [n]qn q3 n+ 2qn2+ 3qn q3 n+ 2qn2+ 2qn+ 1 x + b 2 n (q2 n+ qn+ 1)[n]2qn +Cqn n "(t; x)(t x)2; x
For the last term on the right hand side, using Cauchy-Schwartz inequality, we get lim n!1 [n]qn bn Cqn n "(t; x)(t x)2; x 6 q lim n!1C qn n ("2(t; x); x) s lim n!1 [n]2 qn b2 n Cqn n ((t x)4; x):
Since limn!1Cnqn "2(t; x); x = 0 and by Lemma 4.1(ii) limn!1 [n]2 qn b2 n C qn n (t x)4; x
is …nite, we have limn!1[n]bnqnC qn n "(t; x)(t x)2; x = 0. Therefore, we obtain lim n!1 [n]qn bn (Cqn n (f; x) f (x)) = 1 2f 00(x) lim n!1 x2 bn + q 3 n+ 2q2n+ 3qn q3 n+ 2q2n+ 2qn+ 1 x + bn (q2 n+ qn+ 1)[n]qn + f0(x) 1 + qn = 1 2f 0(x) + x 2f 00(x):
This step completes the proof.
5. Local approximation
In this section, we give a local approximation theorem regarding the our opera-tors. The Peetre’s K-functional is de…ned by
K2(f ; ) := inf
g2C2[0;1)fkf gk + kg 00kg ;
where C2
B[0; 1) = fg 2 CB[0; 1) : g0; g002 CB[0; 1)g, CB[0; 1) denotes the space
of all real-valued bounded and continuous functions.
By using Devore-Lorentz theorem (see[19], Thm 2.4, pp.177), for f 2 CB[0; 1)
and C > 0 we have
K2(f ; )6 C!2 f ;
p
(5.1) where !2 is the second modulus of continuity of f .
In this section, we need the following lemmas for proving our main theorem. Lemma 5.1. Let g 2 C2
B[0; 1). The following inequality holds:
~ Cnq(g; x) g (x) n(x) kg00k ; where n(x) = x(3b[n]nqx) + b2n [n]2 q + bn [n]q. Proof. Let us de…ne auxiliary operators
~ Cnq(f ; x) := Cnq(f ; x) f x + 1 1 + q bn [n]q ! + f (x) :
It is easy to see that ~Cq
n(t x; x) = 0: Let g 2 CB2[0; 1). By using Taylor expansion
of g, we obtain
g (t) g (x) = (t x) g0(x) + Z t
x
(t u) g00(u) du Applying the operator ~Cq
n to the above equality, we get
~ Cnq(g; x) g (x) = g0(x) ~Cnq(t x; x) C~nq Z t x (t u) g00(u) du; x = C~nq Z t x (t u) g00(u) du; x = Cnq Z t x (t u) g00(u) du; x Z x+ 1 1+q[n]qbn x x + 1 1 + q bn [n]q u ! g00(u) du: Thus, we have ~ Cnq(g; x) + g (x) 6 Cnq Z t x (t u) g00(u) du ; x + Z x+ 1 1+q[n]qbn x x + 1 1 + q bn [n]q u ! g00(u) du : Since Z t x (t u) g00(u) du 6 (t x)2 g00 we get Z x+ 1 1+q[n]qbn x x + 1 1 + q bn [n]q u ! g00(u) du 6 1 1 + q bn [n]q !2 g00 : We can write ~ Cnq(g; x) g (x) 6 8 < :C q n (t x) 2 ; x + 1 1 + q bn [n]q !29= ; g 00 : Then, by using Lemma 2.3, we may write
~
Cnq(g; x) g (x) 6 n(x) kg00k :
jCnq(f ; x) f (x)j 6 C!2 f; 1 2 p n(x) + ! f ; bn [n]q ! ; where C > 0 is a constant.
Proof. Assume that f 2 CB[0; 1) by using the de…nition of ~Cnq(f ; x), we get
jCnq(f ; x) f (x)j 6 C~nq(f g; x) +j(f g) (x)j + ~Cnq(g; x) g (x) + f x + 1 1 + q bn [n]q ! f (x) and ~ Cnq(f ; x) 6 kfk Cnq(1; x) + 2 kfk = 3 kfk : Thus, we obtain jCnq(f ; x) f (x)j 6 4 kf gk + ~Cnq(g; x) g (x) + ! f ; bn [n]q !
and using Lemma 5.1,
jCnq(f ; x) f (x)j 6 4 kf gk + 1 4 n(x) kg 00k + ! f ; bn [n]q ! : Taking the in…mum over all g 2 C2
B[0; 1) on the right hand side of above inequality
and using (5.1), the proof is …nished.
6. Rete of convergence in weighted space
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 R+ (see [18]). Let
(f; ) = sup
jhj< ; x2R+
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 6.1. Let f 2 Ck
x2[0; 1) : Then,
i) (f; ) is a monotonically increasing function of ; > 0: ii) For every f 2 Cx2[0; 1) ; lim
!0 (f; ) = 0:
iii) For each > 0;
(f; )6 2 (1 + ) 1 + 2 (f; ) : (6.1)
From the inequality (6.1) and de…nition of (f; ) we get
for every f 2 Cx2[0; 1) and x; t 2 R+.
Theorem 6.2. Let 0 < q6 1 and f 2 Cx2[0; 1). Then, we have sup 06x6bn jCnq(f; x) f (x)j (1 + x2)3 6 C f; s bn [n]qn !
where C is an absolute constant. Proof. Using (6.2) , we get
jCnq(f; x) f (x)j = Cnq(jf (t) f (x)j ; x)
6 2 1 + x2 1 + 2
Cnq 1 + (t x)2 1 +jt xj ; x (f; ) also we can write that
Cnq 1 + (t x)2 1 +jt xj ; x = 1 + Cnq (t x)2; x +1Cnq(jt xj ; x) +1Cnq jt xj (t x)2; x 6 1 + Cq n (t x) 2 ; x +1 r Cnq (t x)2; x +1 r Cnq (t x)2; x r Cnq (t x)4; x :
From (4.1) and (??), we know that
Cnq (t x)2; x = O bn [n]qn x2+ x + 1 and Cnq (t x)4; x = O b 2 n [n]2 qn x4+ x3+ x2+ x + 1 : Choosing = q bn [n]qn we have jCnq(f; x) f (x)j 6 2 1 + x2 1 + 2 Cnq 1 + (t x) 2 1 +jt xj ; x (f; ) 6 4 1 + x2 f; s bn [n]qn ! 1 + O [n]bn qn x2+ x + 1 +p(x2+ x + 1) s O b 2 n [n]2 qn (x4+ x3+ x2+ x + 1) ! ; which proves the theorem.
7. Statistical Convergence of Cq n(f ; x)
In 1951, Fast [14] and Steinhaus [22] de…ned the notion of statistical convergence for sequences of real numbers as:
Let M be a subset of the set of natural numbers N. Then, Mn = fk 6 n : k 2 Mg.
The natural density of M is de…ned by (M ) = limnn1jMnj provided that the limit
exists, where jMnj denotes the cardinality of the set Mn. A sequence x = (xk) is
called statistically convergent to the number ` 2 R, denoted by st lim x = `. For each > 0; the set M"= fk 2 N : jxk `j > g has a natural density zero, that is
lim
n!1
1
njfk 6 n : jxk `j > gj = 0:
This concept was used in approximation theory by Gadjiev and Orhan [16] in 2002. They proved the Bohman–Korovkin type approximation theorem [12] for statistical convergence. Currently, researchers studying statistical convergence have devoted their e¤ort to statistical approximation.
In this section, we examine the statistical approximation properties of the Cq n(f ; x).
Theorem 7.1. q := (qn) ; 0 < qn6 1 be a sequence satisfying the following
condi-tions: st lim n!1qn= 1; st nlim!1q n n = a; st nlim !1 bn [n]qn = 0: (7.1) Let f be a monotone increasing function on [0; 1) then,
st lim
n!1kC qn
n (f ) f kx2= 0:
Proof. Since Cnq(f ; x) is a linear-positive operator, if we show that
st limn!1kCnqn(ei) eikx2 = 0, where ei= xi; i = 0; 1; 2, then we are done. It
is clear that st lim n!1kC qn n (e0) e0kx2 = 0: (7.2) Using (2.2), we get (Cqn n (e1) e1) = 1 1 + qn bn [n]q n : Let " > 0, then we de…ne the following set:
K := fk : kCqk n (e1) e1kx2 > "g = ( k : 1 1 + qk bk [k]qk > " ) : One can obtain that
fk 6 n : kCqk n (e1) e1kx2 > "g 6 ( k6 n : 1 1 + qk bk [k]qk ) : Thus, we get st lim n!1kC qn n (e1) e1kx2 = 0: (7.3)
Using (2.3), we can write (Cqn n (e2) e2) = x2 [n]q n + xq 2 n+ 3qn+ 2 q2 n+ qn+ 1 bn [n]q n + 1 q2 n+ qn+ 1 b2 n [n]2qn: Thus, we get kCqn n (e2) e2kx26 1 [n]qnke2kx2+ q2n+ 3qn+ 2 q2 n+ qn+ 1 bn [n]q ke1kx2+ 1 q2 n+ qn+ 1 b2n [n]2q: Let " > 0; we de…ne the following sets:
U := fk := kCqk n (e2) e2kx2 > "g ; U1:= ( k := a 2 [k]qk > " 3 ) ; U2:= ( k := aq 2 k+ 3qk+ 2 (q2 k+ qk+ 1) bk [k]qk > " 3 ) ; U3:= ( k := b 2 k (q2 k+ qk+ 1) [k]2qk >"3 ) : It is clear that U U1[ U2[ U3: Thus,
fk 6 n : kCqk n (e2) e2kx2 > "g 6 ( k6 n : a 2 [k]qk > " 3 ) + ( k6 n : aq 2 k+ 3qk+ 2 (q2 k+ qk+ 1) bk [k]qk > " 3 ) + ( k6 n : b 2 k (q2 k+ qk+ 1) [k] 2 qk > " 3 )
and since (qn) satis…es (7.1), we obtain
st lim
n!1kC qn
n (e2) e2kx2 = 0: (7.4)
Hence, by using statistical Korovkin theorem, the desired result follows from (7.2)-(7.4).
Note: It is obvious that f (x) > 0 does not guarantee the positivity of the operators Cq
n(f ; x). Thus, we assumed that f is a monotone increasing function
on [0; bn]. By using this assumption, we showed the statistical convergence of the
operators via Korovkin theorem. However, this assumption is not su¢ cient to investigate the rate of convergence and order of approximation because of the usual de…nition of q integral. In order to solve this problem, there are two ways proposed by Gauchman[17] and Marinkovic [20]. They de…ned di¤erent types of q integrals namely, restricted q integral and Riemann type q integral respectively.
In this study, we rede…ne q Chlodowsky-Kantorovich operators by using Rie-mann type q integral.
De…nition 1. (Riemann type q integral) Let 0 < q < 1 and 0 < a < b. The Riemann type q integral is de…ned as follows:
Z b a f (x) dRqx = (1 q) (b a) 1 X j=0 f a + (b a) qj qj:
The modi…ed version of Cnq(f ; x) via Riemann type q integral is as follows: ^ Cnq(f ; x) = [n]q n X k=0 q k n k q x bn k 1 x bn n k q Z [k+1]q/[n]q [k]q/[n]q f (bnt) dRqt:
Lemma 7.2. Let 0 < q < 1 and 0 < a < b, m1 + n1 = 1, for Rq(jfgj ; a; b)
(Rq(jfjm; a; b))1=m(Rq(jgjm; a; b))1=n, where Rq(f ; a; b) =
Rb
a f (x) dRqx.
Proof. : Given in [13].
Remark 7.3. From ([13]), we can obtain the following integrals via making neces-sary computations . Z [k+1]q/[n]q [k]q/[n]q dRqt = q k [n]q; Z [k+1]q/[n]q [k]q/[n]q bnt dRqt = qk[k]q bn [n]2q + q2k [2]q bn [n]2q; Z [k+1]q/[n]q [k]q/[n]q b2nt2dRqt = qk[k]2q b 2 n [n]3q + 2q2k[k] q [2]q b2 n [n]3q + q3k [3]q b2 n [n]3q:
Lemma 7.4. By using the above q Riemann type integrals, we can obtain the following formulas for the moments of ^Cq
n(f ; x): ^ Cnq(1; x) = 1; ^ Cnq(t; x) = 2q 1 + qx + bn [n]q; ^ Cnq t2; x = q2 4q2+ q + 1 q3+ 2q2+ 2q + 1 [n 1]q [n]q x 2+ q 4q2+ 5q + 3 q3+ 2q2+ 2q + 1 bn [n]qx + 1 q2+ q + 1 b2 n [n]2q:
Now, we can give the statistical approximation of ^Cq
n(f ; x) in the following
Theorem 7.5. Let q := (qn) be sequence satisfying (7.1) and let f be a function
de…ned on [0; 1) by f 2 Cx2[0; 1), then we have st lim n!1 ^ Cqn n (f ) f x2= 0: Proof. It is clear that
st lim n!1 ^ Cqn n (e0) e0 x2 = 0: (7.5) Secondly, k ^Cqn n (e1) e1kx2 = qn 1 qn+ 1ke 1kx2+ bn [n]qn (7.6)
Now, for a given " > 0, we de…ne the following sets: L := n k : C^qk n (e1) e1 x2> " o L1:= k : qk 1 qk+ 1 > " 2 ; L2:= ( k : bk [k]q k >2" ) : From (7.6), we see that L L1[ L2: So, we get,
n k6 n : ^Cqk n (e1) e1 x2 > " o 6 k6 n : qk 1 qk+ 1 > " 2 + ( k6 n : bk [n]qk > " 2 ) : Since st lim n!1 qn 1 qn+1 = 0 and st nlim!1 bn [n]qn = 0, we have st lim n!1 ^ Cqn n (e1) e1 x2 = 0: (7.7) Finally, we have ^ Cqn n (e2; x) e2(x) = 4q4 n+ qn3+ qn2 q3 n+ 2qn2+ 2qn+ 1 [n 1]qn [n]qn 1 ! x2 + 4q 3 n+ 5q2n+ 3qn q3 n+ 2q2n+ 2q + 1 bn [n]qnx + 1 q2 n+ qn+ 1 b2 n [n]2qn: Using [n 1]qn < [n]qn, ^ Cqn n (e2) e2 x26 4q4 n qn2 2qn 1 q3 n+ 2qn2+ 2qn+ 1 k e2kx 2 + 4q 3 n+ 5q2n+ 3qn q3 n+ 2qn2+ 2qn+ 1 bn [n]q n k e1kx2 + 1 q2 n+ qn+ 1 b2 n [n]2qn: Let n := 4q4 n q2n 2qn 1 q3 n+ 2q2n+ 2qn+ 1 and n:= 4q 3 n+ 5qn2+ 3qn q3 n+ 2qn2+ 2qn+ 1 bn [n]q n ;
it is easy to see that the followings hold: st lim
n!1 n= 0; st nlim!1 n= 0 and st nlim!1
b2 n
[n]2qn = 0 Similarly, for a given " > 0, we de…ne the following sets:
N := n k : C^qk n (e2) e2 x2 > " o ; N1:= n k : k> " 3 o ; N2:= n k : k >" 3 o ; N3:= ( k : 1 q2 k+ qk+ 1 b2 k [k]2qk > " 3 ) : It is obtained that N N1[ N2[ N3: So, we may write,
n k6 n : ^Cqk n (e2; :) e2 x2> " o 6 nk6 n : k > " 3 o + n k6 n : k> " 3 o + ( k6 n : 1 q2 k+ qk+ 1 b2 k [k]2qk > " 3 ) : Thus, we obtain st lim n!1 ^ Cqn n (e2; :) e2 x2= 0: (7.8)
The proof is …nished using (7.5), (7.7) and (7.8) via statistical Korovkin’s theorem.
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Current address : A. Karaisa, Department of Mathematics–Computer Sciences, Necmettin Erbakan University, 42090 Meram, Konya, Turkey
E-mail address : akaraisa@konya.edu.tr, alikaraisa@hotmail.com
Current address : A. Aral, Department of Mathematics, K¬r¬kkale University,71450 Yah¸sihan, K¬r¬kkale, Turkey
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