Ann Univ Ferrara DOI 10.1007/s11565-015-0227-6
Approximation by complex q-modified Bernstein–Schurer operators on compact disks P. N. Agrawal1 · A. Sathish Kumar1
Received: 21 June 2014 / Accepted: 20 April 2015 © Università degli Studi di Ferrara 2015
Abstract The purpose of this paper is to study the exact order of approximation and Voronovskaja type results with quantitative estimates for the complex q-modified Bernstein–Schurer operators (0 < q < 1) attached to analytic functions on compact disks. In this way, we show the overconvergence phenomenon for these operators, namely the extensions of the approximation properties with quantitative estimates from the real intervals to compact disks in the complex plane. Keywords q-modified Bernstein–Schurer operators · q-integers · Voronovskaja-type result with quantitative upper estimate · Exact degrees of approximation in compact disks Mathematics Subject Classification
30E10 · 41A25
1 Introduction In the recent years, the application of q-calculus has played an important role in the fields of approximation theory, number theory and theoretical physics. In this paper, by applying q-calculus we study the convergence properties of modified Bernstein– Schurer operators based on q-integers on compact disks. In the complex case, Bernstein [20] initiated the study by proving that if f : G → C is an analytic function in the open set G ⊂ C, with D1 ⊂ G (where D1 = {z ∈ C : |z| < 1}), then the complex Bernstein
B
A. Sathish Kumar
[email protected] P. N. Agrawal
[email protected]
1
Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee 247667, India
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polynomials Bn ( f )(z) = nk=0 nk z k (1 − z)n−k f nk , converges uniformly to f in D1 (see e.g., Lorentz [20], p. 88). Gal [10] pioneered the study of the upper quantitative estimates for the uniform convergence of Bn ( f ) to f. On the other hand, quantitative estimates for certain other variants of Durrmeyer type operators were given in [1,2,5,12,13] and [16], . Also, the convergence properties for several integral operators were studied in [11] and [17]. Very recently, the complex approximation properties of q-Durrmeyer operators and their modifications had been discussed in [15] and [1] respectively. A q-analogue of genuine Bernstein–Durrmeyer operators on compact disks was given in [21]. To make the convergence faster, Ren and Zeng used the King type approach for the Kantorovich-type q-Bernstein–Stancu operators on compact disks in [23]. Similar results for the complex case of the Stancu generalizations of Bernstein, Kantorovich, Szász and Baskakov operators were obtained in [6,7,9,18] and [14]. Further, the geometric properties for the complex Favard–Szász–Mirakjan operators were studied by Gal in [8]. In the present paper, we study some approximation properties of the modified Bernstein–Schurer operators based on q-integers defined in (1.2). First, we obtain an upper bound and then we give Voronovskaja-type theorem for these operators on compact disks. Finally, we give the exact order of approximation of the analytic functions by (1.2) on compact disks. In what follows, we follow the notations used in the book [3] and [19] on q-operators. From [3], for s = 0, 1, 2 . . . , by the definition of q-beta function of first kind, we have
1
t s pn,k (q; qt)dq t =
0
1 q k [n]q ![k + s]q ! n qk t k+s (1 − qt)qn−k dq t = [n + s + 1]q ![k]q ! k q 0 (1.1)
Definition 1 Let f ∈ C[0, 1 + p] and 0 < q < 1, the q-modified Bernstein–Schurer operators are defined as
Sn, p ( f, q, z) =
n+ p [n + p + 1]q bn+ p,k (q; z)q −k (1 + p)2n+2 p+1 k=0 1+ p × f (t)bn+ p,k (q; qt)dq (t), ∀ z ∈ C
(1.2)
0
n+ p−k where bn+ p,k (q; z) = n+k p q z k (1 + p − z)q . If q → 1− and p = 0, then the operators (1.2) reduce to the classical complex Bernstein–Durrmeyer operators [2] and only for p = 0, these operators reduce to the operators discussed in [22]. Let D R be the disk D R := {z ∈ C : |z| < R} in the complex plane C. Let us denote by H (D R ), the space of all analytic functions on D R . For f ∈ H (D R ), we may write m f (z) = ∞ m=0 cm z .
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2 Preliminaries In the sequel, we require the following results. Lemma 1 Let Sn, p ( f, q, z) be as defined in (1.2) and 0 < q < 1. Then, Sn, p (t m , q, z) is a polynomial in z of degree ≤ min{m, n} and Sn, p (t m , q, z) =
m [n + p + 1]q ! 1 cs (m, q)[n]qs Bn, p (es , q, z), [n + p + m + 1]q ! (1 + p)n+ p−m s=0
(2.1) where cs (m, q) ≥ 0, are certain constants depending on m and q and Bn, p (f, q, z) n+ p [k] is the q-Bernstein–Schurer polynomial defined by k=0 bn+ p,k (q, z) f [n]qq . Proof We may write Sn, p (t m , q, z) =
1+ p n+ p [n + p + 1]q −k b (q; z)q bn+ p,k (q; qt)t m dq t. n+ p,k (1 + p)2n+2 p+1 0 k=0
(2.2) Let us consider
1+ p
I1 = 0
n+p n+ p−k m (qt)k (1 + p − qt)q t dq t. k q
Putting t = (1 + p)u, we have I1 = (1 + p)
n+ p+m+1
1 n+p n+ p−k k q u k+m (1 − qu)q dq u. k 0 q
Applying the Eq. (1.1), we obtain I1 =
[n + p]q ! [k + m]q ! (1 + p)n+ p+m+1 . [k]q ! [n + p + m + 1]q !
(2.3)
Using (2.3) in (2.2), we get Sn, p (t m , q, z) =
n+ p [n + p + 1]q ! [k + m]q ! 1 . bn+ p,k (q; z) n+ p−m [n + p + m + 1]q ! (1 + p) [k]q ! k=0
(2.4) For m = 0 in (2.4), we have
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Sn, p (1, q, z) =
n+ p 1 1 bn+ p,k (q; z) = , n+ p (1 + p) (1 + p)n+ p k=0
thus the result is true for m = 0 with c0 (0, q) = 1. Putting m = 1 in (2.4), we obtain n+ p 1 1 Sn, p (t, q, z) = bn+ p,k (q; z)[k + 1]q . [n + p + 2]q (1 + p)n+ p−1 k=0
Using [k + 1]q = 1 + q[k]q , we have 1 1 = [n + p + 2]q (1 + p)n+ p−1 =
1 1 [n + p + 2]q (1 + p)n+ p−1
n+ p
n+ p
[k]q bn+ p,k (q; z) + q[n]q bn+ p,k (q; z) [n]q k=0 k=0
1 cs (1, q)[n]qs Bn, p (es , q, z) .
s=0
Hence, the result is true for m = 1 with c0 (1, q) = 1 and c1 (1, q) = q > 0. Putting m = 2 in (2.4), we get Sn, p (t 2 , q, z) =
1 1 [n + p + 2]q [n + p + 3]q (1 + p)n+ p−2 ×
n+ p
bn+ p,k (q; z)[k + 1]q [k + 2]q .
k=0
Since [k + 2]q = [2]q + q 2 [k]q and [k + 1]q = 1 + q[k]q , we obtain n+ p 1 1 bn+ p,k (q; z) [2]q + q[2]q [k]q n+ p−2 [n + p + 2]q [n + p + 3]q (1 + p) k=0 +q 2 [k]q + q 3 [k]q2 n+ p 1 1 = bn+ p,k (q; z) [2] q [n + p + 2]q [n + p + 3]q (1 + p)n+ p−2 k=0
n+ n+ p p [k]q2 [k]q 2 2 3 bn+ p,k (q; z) + [n]q q bn+ p,k (q; z) 2 +[n]q (q[2]q + q ) [n]q [n]q k=0 k=0
2 1 1 s = cs (2, q)[n]q Bn, p (es , q, z) . [n + p + 2]q [n + p + 3]q (1 + p)n+ p−2
=
s=0
Hence, the result is true for m = 2 with c0 (2, q) = [2]q > 0, c1 (2, q) = q[2]q +q 2 > 0 and c2 (2, q) = q 3 > 0.
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Now, we shall show that the result is true for m = r. Using [k +s]q = [s]q +q s [k]q , we get r
[k]q [k + 1]q [k + 2]q . . . [k + r ]q =
([s]q + q [k]q ) = s
s=0
r
cs (r )[k]qs ,
s=0
where cs (r ) > 0, s = 0, 1, 2, . . . r are certain constants independent of k. Therefore, we have Sn, p (t r , q, z) =
n+ r p [n + p + 1]q ! 1 c (r ) bn+ p,k (q, z)[k]qs s [n + p + r + 1]q ! (1 + p)n+ p−r s=0
=
[n + p + 1]q ! 1 [n + p + r + 1]q ! (1 + p)n+ p−r
k=0
r
cs (r )[n]qs Bn, p (es , q, z).
s=0
Hence, the result is true for m = r. This completes the proof of the Lemma.
Lemma 2 Let 0 < q < 1. Then, for all m, n ∈ N and p ∈ N ∪ {0} such that m ≤ n + p, we have [n + p + 1]q ! cs (m, q)[n + p]qs ≤ 1. [n + p + m + 1]q ! m
s=0
Proof Using Lemma 1, with em (t) = t m we have Sn, p (t m , q, 1 + p) =
[n + p + 1]q ! 1 [n + p + m + 1]q ! (1 + p)n+ p−m m × cs (m, q)[n]qs Bn, p (es , q, 1 + p).
(2.5)
s=0
Now, consider Bn, p (es , q, z) =
n+ p k=0
n+p k
z (1 + p − k
n+ p−k z)q
q
[k]q [n]q
s .
Let us put z = (1 + p)w, then n+ p k=0
n+p k
n+ p−k
z k (1 + p − z)q q
= (1 + p)n+ p
n+ p k=0
n+p k
[k]q [n]q
s
n+ p−k
w k (1 − w)q q
[k]q [n]q
s
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n+ p (1 + p)n+ p n + p n+ p−k w k (1 − w)q [k]qs . k [n]qs q k=0 s n n k [k] = 1 at w = 1 for From ([4], p.61, Theorem 1.5.6), k=0 k q w (1 − w)qn−k [n]qq =
all s = 0, 1, 2 . . . therefore, n+ p k=0
n+p k
z (1 + p − k
n+ p−k z)q
q
[k]q [n]q
s =
(1 + p)n+ p [n + p]qs [n]qs
, at
z = 1 + p for all s = 0, 1, 2 . . . In view of (2.5), we have Sn, p (t m , q, 1 + p) =
[n + p + 1]q ! 1 [n + p + m + 1]q ! (1 + p)n+ p−m m [n + p]qs × cs (m, q)[n]qs (1 + p)n+ p [n]qs s=0
[n + p + 1]q ! (1 + p)m cs (m, q)[n + p]qs . (2.6) [n + p + m + 1]q ! m
=
s=0
From, bn+ p,k (q; z) =
n+p z k (1 + p − z)(1 + p − qz) 1 + p − q 2 z . . . k q
1 + p − q n+ p−k−1 z ,
it is immediate that bn+ p,k (q; 1 + p) = 0, for k = 0, 1, 2 . . . n + p − 1 and bn+ p,k (q; 1 + p) = (1 + p)n+ p for k = n + p, hence (1.2) leads us to Sn, p (t m , q, 1 + p) =
[n + p + 1]q −(n+ p) q bn+ p,n+ p (q; 1 + p) (1 + p)2n+2 p+1 1+ p bn+ p,n+ p (q; qt)t m dq t × 0
1+ p [n + p + 1]q n+ p = (1 + p) t n+ p+m dq t (1 + p)2n+2 p+1 0 [n + p + 1]q (1 + p)m = ≤ (1 + p)m . [n + p + m + 1]q Combining (2.6)–(2.7), we get
(2.7)
[n + p + 1]q ! cs (m, q)[n + p]qs ≤ 1. [n + p + m + 1]q ! m
s=0
This completes the proof.
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Lemma 3 Let r ≥ 1 + p and 0 < q < 1. Then, for all m, n ∈ N ∪ {0} and |z| ≤ r, we have |Sn, p (em , q, z)| ≤ r m . Proof From Gal ([4] p. 61, proof of Theorem 1.5.6), we have |Bn,q (es , z)| ≤ r s , whenever |z| ≤ r and r ≥ 1. Thus, in view of Lemmas 1 and 2, for all m, n ∈ N ∪ {0} and |z| ≤ r, r ≥ 1 + p we obtain |Sn, p (em , q, z)| ≤
m [n + p + 1]q ! 1 cs (m)[n]qs |Bn, p (es , q, z)| [n + p + m + 1]q ! (1 + p)n+ p−m s=0
[n + p + 1]q ! 1 ≤ [n + p + m + 1]q ! (1 + p)n+ p−m s m n+ p r s (1 + p) s × cs (m)[n]q [n + p]q [n]qs 1+ p s=0
≤
m [n + p + 1]q ! rm (1 + p)m cs (m)[n + p]qs [n + p + m + 1]q ! (1 + p)m s=0
≤
[n + p + 1]q ! rm [n + p + m + 1]q !
m
cs (m)[n + p]qs ≤ r m .
s=0
Hence, the proof is completed. Remark 1 By simple computations, we have
z(1 + p − z)Dq (bn+ p,k (q; z)) = bn+ p,k (q; z) [k]q (1 + p) − [n + p]q z and t (1 + p − qt)Dq (bn+ p,k (q; qt)) = bn+ p,k (q; qt) [k]q (1 + p) − [n + p]q qt . Lemma 4 Let 0 < q < 1. For all em (t) = t m , m ∈ N ∪ {0} and z ∈ C we have Sn, p (em+1 , q, z) =
q m+1 z(1 + p − z) Dq (Sn, p (em , q, z)) [n + p + m + 2]q [m + 1]q (1 + p) + q m+1 [n + p]q z Sn, p (em , q, z). + [n + p + m + 2]q
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Proof Using Remark 1, we have z(1 + p − z)Dq (Sn, p (em , q, z)) [n + p + 1]q z(1 + p − z)Dq (bn+ p,k (q; z))q −k (1 + p)2n+2 p+1 k=0 1+ p × bn+ p,k (q; qt)t m dq t n+ p
=
0
n+ p [n + p + 1]q b (q; z) [k] (1 + p) − [n + p] z q −k n+ p,k q q (1 + p)2n+2 p+1 k=0 1+ p × bn+ p,k (q; qt)t m dq t
=
0
n+ p [n + p + 1]q bn+ p,k (q; z) [k]q (1 + p) − [n + p]q qt = (1 + p)2n+2 p+1 k=0 + [n + p]q qt − [n + p]q z q −k
1+ p
×
bn+ p,k (q; qt)t m dq t
0
n+ p [n + p + 1]q bn+ p,k (q; z) [k]q (1 + p) − [n + p]q qt q −k = (1 + p)2n+2 p+1 k=0 1+ p × bn+ p,k (q; qt)t m dq t 0
+ q[n + p]q Sn, p (em+1 , q, z) − [n + p]q zSn, p (em , q, z) n+ p [n + p + 1]q bn+ p,k (q; z)q −k (1 + p)2n+2 p+1 k=0 1+ p × Dq (bn+ p,k (q; qt))(1 + p − qt)t m+1 dq t
=
0
+ q[n + p]q Sn, p (em+1 , q, z) − [n + p]q zSn, p (em , q, z). m+1 t Let δ(t) = (1 + p − t) , then δ(qt) = (1 + p − qt)t m+1 . q Applying the formula for q-integration by parts, we have
1+ p
Dq (bn+ p,k (q; qt))δ(qt)dq (t)
0
=
123
1+ p [δ(t)bn+ p,k (q; qt)]0
− 0
1+ p
bn+ p,k (q; qt)Dq (δ(t))dq (t)
(2.8)
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=−
1 q m+1
1+ p
bn+ p,k (q; qt)Dq t m+1 (1 + p) − t m+2 dq (t)
0
= −[m + 1]q (1 + p)q −(m+1) + [m + 2]q q −(m+1)
1+ p
bn+ p,k (q; qt)t m dq (t)
0
1+ p
bn+ p,k (q; qt)t m+1 dq (t).
(2.9)
0
Combining (2.8)–(2.9), we obtain the required result.
3 Main results 3.1 Upper bound
Let Pn (z) be a polynomial of degree n of complex variable z with derivative Pn (z). Then, by the Bernstein inequality and the complex mean value theorem, we have
|Dq (Pn (z))| ≤ Pn r ≤
n
Pn r , for all |z| ≤ r, r
(3.1)
where . r denotes the sup-norm on |z| ≤ r. Our first main result is the following upper estimate. m Theorem 1 Let 0 < q < 1, f (z) = ∞ m=0 cm z , for all |z| < R and let 1 + p ≤ r < R. Then, for all |z| ≤ r and n ∈ N, we have | Sn, p ( f, q, z) − f (z) |≤ where Cr, p ( f ) = (1 + r )(1 + p)
∞
Cr, p ( f ) , [n + p + 1]q
m=1 |cm |m(m
+ 1)r m−1 < ∞.
cm Sn, p (em , q, z), where em (z) = Proof First we show that Sn, p ( f, q, z) = ∞ m=0 m j z m , m = 0, 1, 2, . . . Indeed, denoting f m (z) = j=0 c j z , |z| ≤ r, m ∈ N, by applying the linearity of Sn, p , we get Sn, p ( f m , q, z) =
m
c j Sn, p (e j , q, z).
j=0
For any fixed n ∈ N and |z| ≤ r with r ≥ 1 + p, it is enough to show that lim Sn, p ( f m , q, z) = Sn, p ( f, q, z). But this is immediate from limm→∞ m→∞ f m − f r = 0 and from the inequality | Sn, p ( f m , q, z) − Sn, p ( f, q, z) | ≤
n+ p [n + p + 1]q |bn+ p,k (q; z)|q −k (1 + p)2n+2 p+1 k=0
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1+ p
×
bn+ p,k (q; qt) | f m (t) − f (t) | dq t
0
≤ K r,n, p f m − f r , valid for all |z| ≤ r, where K r,n, p =
n+ p [n + p + 1]q n + p n+ p−k −k r k (1 + p + r )q q k (1 + p)2n+2 p+1 q k=0
1+ p
×
bn+ p,k (q; qt)dq t
0
=
n+ p n + p 1 n+ p−k −k r k (1 + p + r )q q . k (1 + p)n+ p q k=0
Therefore, we get |Sn, p ( f, q, z) − f (z)| ≤
∞
|cm ||Sn, p (em , q, z) − em (z)|
m=0
=
∞
|cm ||Sn, p (em , q, z) − em (z)|,
m=1
since Sn, p (e0 , q, z) = e0 (z) = 1. Now, from Lemma 4 for all m ∈ N, we have Sn, p (em , q, z) − em (z) =
[m]q (1 + p) + q m [n + p]q z q m z(1 + p − z) Dq (Sn, p (em−1 , q, z)) + [n + p + m + 1]q [n + p + m + 1]q
(Sn, p (em−1 , q, z) − em−1 (z)) [m]q (1 + p) + q m [n + p]q z m−1 z − zm [n + p + m + 1]q [m]q (1 + p) + q m [n + p]q z q m z(1 + p − z) = Dq (Sn, p (em−1 , q, z)) + [n + p + m + 1]q [n + p + m + 1]q +
(Sn, p (em−1 , q, z) − em−1 (z)) +
[m]q (1 + p) q m [n + p]q − [n + p + m + 1]q m z m−1 + z . [n + p + m + 1]q [n + p + m + 1]q
Since q m [n + p]q − [n + p + m + 1]q = −[m]q − q n+ p+m , we have | Sn, p (em , q, z) − em (z) |≤
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q m r (1 + p + r ) |Dq (Sn, p (em−1 , q, z))| [n + p + m + 1]q
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+
q m [n + p]q r + [m]q (1 + p) [n + p + m + 1]q
|Sn, p (em−1 , q, z) − em−1 (z)| + +
[m]q (1 + p) r m−1 [n + p + m + 1]q
([m]q + q n+ p+m ) m r . [n + p + m + 1]q
(3.2)
Now, by using (3.1) and Lemma 3 we obtain q m r (1 + p + r ) |Dq (Sn, p (em−1 , q, z))| [n + p + m + 1]q ≤
q m r (1 + p)(1 + r ) (m − 1)
Sn, p (em−1 , q, .) r [n + p + m + 1]q r
≤
q m (1 + p)(1 + r )(m − 1)
Sn, p (em−1 , q, .) r [n + p + 1]q
≤
(1 + p)(1 + r )(m − 1) m−1 r . [n + p + 1]q
Also, we have [m + 1]q [m]q + q n+ p+m m r ≤ r m and [n + p + m + 1]q [n + p + 1]q [m + 1]q (1 + p) m [m]q (1 + p) rm ≤ r . [n + p + m + 1]q [n + p + 1]q Since
q m [n + p]q r + [m]q (1 + p) ≤ r, in view of (3.2) we get [n + p + m + 1]q (1 + p)(1 + r )(m − 1) m−1 r [n + p + 1]q +r |Sn, p (em−1 , q, z) − em−1 (z)|
| Sn, p (em , q, z) − em (z) |≤
+ ≤
[m + 1]q (1 + p) m−1 [m + 1]q r + rm [n + p + 1]q [n + p + 1]q
(1 + p)(1 + r )(m − 1) m−1 r + r |Sn, p (em−1 , q, z) − em−1 (z)| [n + p + 1]q +
[m + 1]q (1 + p) m−1 r (1 + r ) [n + p + 1]q
≤ 2m
(1 + p)(1 + r ) m−1 r + r |Sn, p (em−1 , q, z) − em−1 (z)|. [n + p + 1]q
By writing last inequality, for m = 2, 3 . . . we can easily obtain, step by step the following
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| Sn, p (em , q, z) − em (z) | 2m(1 + p)(1 + r ) m−2 ≤ r r |Sn, p (em−2 , q, z) − em−2 (z)| + r [n + p + 1]q +
2(m + 1)(1 + p)(1 + r ) m−1 r [n + p + 1]q
= r 2 |Sn, p (em−2 , q, z) − em−2 (z)| + +
2m(1 + p)(1 + r ) m−1 r [n + p + 1]q
2(m + 1)(1 + p)(1 + r ) m−1 r [n + p + 1]q
= r 2 |Sn, p (em−2 , q, z) − em−2 (z)| + ≤ ··· ≤
2(1 + p)(1 + r ) m−1 r (1 + m + m) [n + p + 1]q
(1 + p)(1 + r ) m(m + 1)r m−1 . [n + p + 1]q
Therefore, we have | Sn, p ( f, q, z) − f (z) | ≤
∞
|cm | | Sn, p (em , q, z) − em (z) |
m=1
≤
∞ m=1
≤
|cm |
(1 + p)(1 + r ) m(m + 1)r m−1 [n + p + 1]q
∞ (1 + p)(1 + r ) |cm |m(m + 1)r m−1 [n + p + 1]q m=1
≤
Cr, p ( f ) , [n + p + 1]q
m−1 . where Cr, p ( f ) = (1 + p)(1 + r ) ∞ m=1 |cm |m(m + 1)r This completes the proof of the theorem.
Remark 2 Let q ∈ (0, 1) be fixed. Since [n]1 q → 1 − q, when n → ∞, by applying limit n → ∞ in the Theorem 1, Sn, p ( f, q, z) does not converge to f (z) but this can be achieved by taking a sequence q = qn satisfying 0 < qn < 1 with qn → 1 and qnn → 0 as n → ∞. In this case [n]1q → 0, as n → ∞. Therefore from Theorem 1, we n have Sn, p ( f, qn , z) → f (z) as n → ∞ uniformly for |z| ≤ r, when 1 + p ≤ r < R. 3.2 Voronovskaja type result Our next main result is the following Vornovskaja type theorem with a quantitative estimate.
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Theorem 2 Let q ∈ (0, 1), R > 1 + p and suppose that f : D R → C is analytic in k D R , i.e. we can write f (z) = ∞ k=0 ck z , for all z ∈ D R . For any fixed r ∈ [1 + p, R) and for all n ∈ N, |z| ≤ r, we have Sn, p ( f, q, z) − f (z) − z(1 + p − z) f (z) + (1 + p − 2z) f (z) [n] q
≤
Mr, p ( f ) + 3(1 − q) [n]q2
where Mr, p ( f ) =
∞
k=1 |ck |r
∞
|ck |r k k 4 ,
k=1
kk F k,r, p
< ∞ and
Fk,r, p = (1 + p)2 2(k − 1)3 + 8k 2 (k + 1) + 7k(k + 1)2 +4(1 + p)k(k − 1)2 (1 + r ). Proof Let ek (z) = z k , k = 0, 1, 2, . . . Since f is an analytic function, we can write Sn, p ( f, q, z) = ∞ k=0 ck Sn, p (ek , q, z). Also, ∞
z(1 + p − z) f (z) + (1 + p − 2z) f (z) z(1 + p − z) = ck k(k − 1)z k−2 [n]q [n]q + =
k=2 ∞
(1 + p − 2z) [n]q
1 [n]q
∞
ck kz k−1
k=1
ck (k 2 (1+ p)−k(k + 1)z)z k−1 .
k=1
For all z ∈ D R and n ∈ N, we have Sn, p ( f, q, z) − f (z) − z(1 + p − z) f (z) + (1 + p − 2z) f (z) [n]q ∞ (k 2 (1 + p) − k(k + 1)z) k−1 ≤ |ck | Sn, p (ek , q, z) − ek (z) − z . [n]q k=1
If we consider E k,n, p (q, z) = Sn, p (ek , q, z) − ek (z) −
(k 2 (1 + p) − k(k + 1)z) k−1 z , [n]q
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then it is clear that E k,n, p (q, z) is a polynomial in z of degree ≤ k. Using Lemma 4, we have E k,n, p (q, z) =
q k z(1 + p − z) E (z) [n + p + k + 1]q k−1,n, p +
([k]q (1 + p) + q k [n + p]z) E k−1,n, p (z) + X k,n, p (q, z), [n + p + k + 2]q
where X k,n, p (q, z) z k−2 2 k 2 2 (1 + p) q (k − 1) [k − 2]q + [k]q (k − 1) = [n]q [n + p + k + 1]q + z(1 + p) q k [k − 1]q [n]q − q k k(k − 1)[k − 1]q − q k (k − 1)2 [k − 2]q k 2 2 + q [n + p]q (k − 1) + [k]q [n]q − [k]q k(k − 1) − k [n + p + k + 1]q 2 + z − q k [n]q [k − 1]q + q k k(k − 1)[k − 1]q + q k [n + p]q [n]q − q [n + p]q k(k − 1) − [n]q [n + p + k + 1]q + k(k + 1)[n + p + k + 1]q k
=
z k−2 (1 + p)2 Ak,n (q) + z(1 + p)Bk,n (q) + z 2 Ck,n (q) . [n]q [n + p + k + 1]q
It is clear that |Ak,n (q)| ≤ 2(k − 1)3 . Next, we estimate Bk,n (q). Using [n + p]q = [n]q +q n [ p]q and [n + p + k + 1]q = [n] + q n [ p + k + 1]q , we have k k 2 2 Bk,n (q) = [n]q q [k − 1]q + [k]q + q (k − 1) − k + q n+k [ p]q (k − 1)2 − q n k 2 [ p + k + 1]q − q k k(k − 1)[k − 1]q − q k (k − 1)2 [k − 2]q − k(k − 1)[k]q . (3.3) Now, we consider k k 2 2 [n]q q [k − 1]q + [k]q + q (k − 1) − k k−2 k−1 = [n]q q k (q − 1) [ j]q + (k − 1) + (q − 1) [ j]q + k j=0
+ q k (k − 1)2 − k 2
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k−1 k−2 = (1 − q n ) − q k [ j]q − [ j]q − k(k − 1)[k]q . j=0
Thus,
j=0
[n]q q k [k − 1]q + [k]q + q k (k − 1)2 − k 2 ≤
(k − 1)(k − 2) k(k − 1) + + k 2 (k − 1). 2 2
Hence, in view of (3.3), we have |Bk,n (q)| ≤ (1 + p) (k − 1)2 (k − 2) + k 2 (k − 1) + k 2 (k − 1) + k(k − 1)2 2 2 2 2 + k (k + 1) + k(k − 1) + (k − 1) (k − 2) + k (k − 1) ≤ 8(1 + p)k 2 (k + 1). Now, we estimate Ck,n (q). Using [n + p]q = [n]q + q n [ p]q and [n + p + k + 1]q = [n] + q n [ p + k + 1]q , we obtain 2 k Ck,n (q) = [n]q (q − 1) + [n]q − q k [k − 1]q + q k+n [ p]q − q k k(k − 1) n −q [ p + k + 1]q + k(k + 1) −q k+n k(k − 1)[ p]q + q n k(k + 1)[ p + k + 1]q + q k k(k − 1)[k − 1]q . (3.4) First, we consider
[n]q2 (q k − 1) + [n]q
− q k [k − 1]q + q k+n [ p]q − q k k(k − 1) − q n [ p + k + 1]q + k(k + 1)
k−2 [ j]q (q − 1) + (k − 1) = [n]q2 (q k − 1) + [n]q − q k j=0
+ q k+n [ p]q − q k k(k − 1) − q n ([k + 1]q + q k+1 [ p]q ) + k(k + 1) k−2 [ j]q (q − 1) − q k (k − 1) = [n]q2 (q k − 1) + [n]q − q k j=0
+q
k+n
[ p]q (1 − q) − q k(k − 1) k
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− qn
k
[ j]q (q − 1) + (k + 1) + k(k + 1)
j=0
= [n]q2 (q k − 1) + [n]q
⎧ ⎨ ⎩
−q k
k−2 k [ j]q (q − 1) − q n [ j]q (q − 1) j=0
j=0
+ q k+n [ p]q (1−q) + k 2 [k]q (1 − q) + k[n]q (1 − q) + (q k − 1) + (1 − q n )
⎫ ⎬ ⎭
= [n]q2 [k]q (q − 1) + [n]q2 k(1 − q) + [n]q2 (1 − q) + [n]q k−2 k − qk [ j]q (q − 1) − q n [ j]q (q − 1) j=0
j=0
+ q k+n [ p]q (1 − q) + k 2 [k]q (1 − q) + [k]q (q − 1) = [n]q (1 − q n )(1 + k − [k]q ) + (1 − q n ) k−2 k k n k+n 2 q [ j]q + q [ j]q + q [ p]q + k [k]q − [k]q . j=0
j=0
Therefore, we get |Ck,n (q)| ≤ (1 − q n )[n]q 3k + (k − 1)2 (k − 2) + k(k + 1)2 + k(k + 1)2 (1 + p) + k 3 + k(k + 1)2 + (1 + p)k(k + 1)2 + k(k + 1)2 + k(k − 1)2 ≤ 8(1 + p)k(k + 1)2 + 3k 3 [n]q (1 − q n ). Therefore, we have r k−2 2 3 2 2 2 |X k,n, p (q; z)| ≤ (1 + p) 2(k − 1) + 8r k (k + 1) + 8r k(k + 1) [n]q2 3r k 3 k (1 − q n ) [n]q r k−2 2 3 2 2 2 2(k − 1) (1 + p) + 8r k (k + 1) + 8r k(k + 1) ≤ [n]q2 +
+ 3r k k 3 (1 − q). From Theorem 1, we obtain (1 + p)(1 + r ) k(k + 1)r k−1 [n + p + 1]q (1 + p)(1 + r ) ≤ k(k + 1)r k−1 . [n]q
|Sn, p (ek , q, z) − ek (z)| ≤
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For all k, n ∈ N, and |z| ≤ r, we have |E k,n, p (q, z)| ≤
r (1 + r )(1 + p) |E k−1,n, p (q, z)| + r |E k−1,n, p (q, z)| [n]q + |X k,n, p (q, z)|.
(3.5)
Now, we shall find an estimate of |E k−1,n, p (q, z)|, for k ≥ 2. k−1
E k−1,n, p r r k−1 ≤
Sn, p (ek−1 , q, .) − ek−1 r r ((k − 1)2 (1 + p) − k(k − 1)e1 )ek−2 + [n]q r k − 1 (1 + p)(1 + r ) k−2 ≤ k(k − 1)r r [n]q ((1 + p)(1 + r ) + k(k − 1)r k−2 [n]q
|E k−1,n, p (q, z)| ≤
≤
2k(k − 1)2 (1 + r ) k−2 r . [n]q
Thus, we have r (1 + r )(1 + p) 4k(k − 1)2 (1 + r )(1 + p) k |E k−1,n, p (z)| ≤ r . [n]q [n]q2 In view of (3.5), we get |E k,n, p (q, z)| ≤
4k(k − 1)2 (1 + r )(1 + p) k r + r |E k−1,n, p (z)| [n]q2 +|X k,n, p (q, z)|,
where r k−2 2 3 2 2 2 |X k,n, p (q; z)| ≤ (1 + p) 2(k − 1) + 8r k (k + 1) + 8r k(k + 1) [n]q2 +3r k k 3 (1 − q) rk ≤ Dk, p + 3r k k 3 (1 − q), for all |z| ≤ r, k ≥ 1 and n ∈ N, [n]q2 where Dk, p = (1 + p)2 2(k − 1)3 + 8k 2 (k + 1) + 8k(k + 1)2 .
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Thus, for all |z| ≤ r, k ≥ 1 and n ∈ N, we have |E k,n, p (q, z)| ≤
rk Fk,r, p + r |E k−1,n, p (z)| + 3r k k 3 (1 − q), [n]q2
where Fk,r, p is a polynomial of degree 3 in k defined as Fk,r, p = Dk, p + 4(1 + p)k(k − 1)2 (1 + r ). But E 0,n, p (q, z) = 0, for any z ∈ C and therefore by writing the last inequality for k = 1, 2, . . . we easily obtain step by step the following k k rk k |E k,n, p (q, z)| ≤ F j,r, p + 3r (1 − q) j3 [n]q2 j=1
j=1
rk ≤ k Fk,r, p + 3r k k 4 (1 − q) [n]q2 Therefore, we conclude that Sn, p ( f, q, z) − f (z) − z(1 + p − z) f (z) + (1 + p − 2z) f (z) [n] q
≤
∞
|ck ||E k,n, p (q, z)|
k=1
≤
∞ ∞ 1 k |c |r k F + 3(1 − q) |ck |r k k 4 . k k,r, p [n]q2 k=1
k=1
∞ ∞ k−4 and the series k As f (4) (z) = k=4 ck k(k − 1)(k − 2)(k − 3)z k=0 ck z is ∞ k(k − 1)(k − 2)(k − absolutely convergent in |z| ≤ r, it easily follows that k=4 ck ∞ k < ∞ and 4 k |c |k F r 3)r k−4 < ∞, which implies that ∞ k k,r, p k=1 k=1 |ck |k r < ∞. Hence, the proof is completed.
Remark 3 Let 0 < q < 1 be fixed. Since [n]1 q → 1 − q, when n → ∞, by applying limit n → ∞ in the Theorem 2, we don’t get the convergence. But this can be achieved by taking 0 < qn < 1 with qn → 1 as n → ∞ and qnn → a(a < 1) as n → ∞. In this case [n]1q → 0 as n → ∞ and 1 − qn ≤ n12 ≤ [n]12 . Therefore, from Theorem 2 n
qn
we have
Sn, p ( f, qn , z) − f (z) − z(1 + p − z) f (z) + (1 + p − 2z) f (z) [n] qn
∞ Mr, p ( f ) 3 ≤ + |ck |r k k 4 , [n]q2n [n]q2n k=1
that is, the order of approximation is O
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1 [n]q2n
.
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3.3 Exact order approximation In this section, we will find the exact order of approximation for complex q-modified Bernstein–Schurer operators. Throughout the section, we assume that {qn } is a sequence such that 0 < qn < 1 with qn → 1 as n → ∞ and qnn → a(a < 1) as n → ∞. Theorem 3 Let R > 1+ p, D R k= {z ∈ C; |z| < R} and f : D R → C be analytic, i.e. we can write f (z) = ∞ k=0 ck z , for all z ∈ D R . If f is a non-constant polynomial, then for any r ∈ [1 + p, R) we have
Sn, p ( f, qn , .) − f r ≥
Cr, p ( f ) [n]qn
, n ∈ N,
where the constant Cr, p ( f ) > 0 depends on f, r and p on the sequence {qn }n∈N but it is independent of n. Proof For all z ∈ D R and n ∈ N, we have 1 z(1 + p − z) f (z) + (1 + p − 2z) f (z) Sn, p ( f, qn , z) − f (z) = [n]qn 1 2 [n]qn Sn, p ( f, qn , z) − f (z) + [n]qn z(1 + p − z) f (z) + (1 + p − 2z) f (z) − [n]qn Now, using the following identity
F + G r ≥ | F r − G r | ≥ F r − G r , we get
Sn, p ( f, qn , .) − f r 1
e1 (1 + p − e1 ) f + (1 + p − 2e1 ) f r ≥ [n]qn 1 e1 (1 + p − e1 ) f + (1 + p − 2e1 ) f 2 [n]qn Sn, p ( f, qn , .)− f − . − [n]qn [n]qn r (3.6) Since f is a non-constant polynomial in D R , we get e1 (1 + p − e1 ) f + (1 + p − 2e1 ) f r > 0. Indeed, supposing the contrary it follows that z(1 + p − z) f (z) + (1 + p − 2z) f (z) = 0, for all |z| ≤ r, that is (z(1 + p − z) f (z)) = 0, for all |z| ≤ r with z = 0. The last inequality is equivalent to z(1 + p − z) f (z) = C, where C is a constant, for all |z| ≤ r with z = 0.
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Therefore, we have f (z) =
C , for all |z| ≤ r with z = 0. z(1 + p − z)
But since f is analytic in Dr and r ≥ 1 + p, we necessarily have C = 0 (otherwise, we would get that f (z) is not differentiable at z = 1 + p, this is not possible because f (z) is also analytic on Dr and r ≥ 1 + p), which implies that f (z) = 0 and hence f (z) = C for all z ∈ Dr , a contradiction to the hypothesis. But from Remark 3, we get e1 (1 + p − e1 ) f + (1 + p − 2e1 ) f 2 [n]qn Sn, p ( f, qn , .) − f − [n]qn r ∞ ≤ Mr, p ( f ) + |ck |r k k 4 , k=1
with [n]1q → 0 as n → ∞. Therefore, there exists a positive integer n 0 depending n only on f and r, such that for all n ≥ n 0 , for the expression in the curly brackets on the right hand side of (3.6) we have
e1 (1 + p − e1 ) f + (1 + p − 2e1 ) f r 1 e1 (1 + p − e1 ) f + (1 + p − 2e1 ) f 2 [n]qn Sn, p ( f, qn , .) − f − − [n]qn [n]qn r 1 ≥ e1 (1 + p − e1 ) f + (1 + p − 2e1 ) f r , 2 which implies that
Sn, p ( f, qn , .) − f r ≥
1
e1 (1 + p − e1 ) f + (1 + p − 2e1 ) f r , 2[n]qn
∀ n ≥ n0. For n ∈ {1, 2 . . . , n 0 − 1}, we obviously have Sn, p ( f, qn , .) − f r ≥ with Cr,n, p ( f ) = [n]qn Sn, p ( f, qn , .) − f r > 0. Therefore, finally we get Sn, p ( f, qn , .) − f r ≥
Cr,n, p ( f ) [n]qn
Cr, p ( f ) [n]qn
, for all n
1 where Cr, p ( f ) = min Cr,1, p , Cr,2, p , . . . , Cr,n 0 −1, p , e1 (1 + p − e1 ) f 2 +(1 + p − 2e1 ) f r .
Hence, the proof is completed.
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Ann Univ Ferrara Acknowledgments The authors are extremely grateful to the learned referees for a very careful reading of the manuscript and making valuable comments and suggestions which greatly improved our paper. The second author is thankful to the “Ministry of Human Resource and Development” India for financial support to carry out his research work.
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