MATEMATIQKI VESNIK
UDK 517.984
originalniresearch
nauqnipaper
rad
57 (2005), 87–94
THE VORONOVSKAYA THEOREM FOR GENERALIZED
BASKAKOV-KANTOROVICH OPERATORS IN
POLYNOMIAL WEIGHT SPACES
Abdul Wafi and Salma Khatoon
Abstract. Recently, K. Bogalska has applied a new technique to establish a Voronovskayatype theorem. In this paper we will also prove the Voronovskaya Theorem for geenralized
Baskakov-Kantorovich-type operators in polynomial weight spaces using the same approach.
1. Introduction
For two times differentiable functions, Voronovskaya [7] was the first to prove
a theorem for Bernstein polynomials known as Voronovskaya Theorem. Later on,
it was studied by Butzer and Nessel [5], Rempulska and Skorupka [6] for some other
linear positive operators.
Recently, K. Bogalska [4] has given a different proof of the Voronovskaya Theorem for Baskakov-Kantorovich operators. In this paper we will also prove a similar
theorem for generalized Baskakov-Kantorovich operators defined in [1].
Consider the notation used in M. Becker [3], i.e., N = {1, 2, . . . }, W = N∪{0},
and let for a fixed p ∈ W , x ∈ [0, ∞), weights wp (x) be given by
w0 (x) = 1,
wp (x) = (1 + xp )−1
if p > 1.
(1.1)
Let Cp = { f ∈ C[0, ∞) : wp f is uniformly continuous and bounded on [0, ∞) }
and the norm in Cp be defined by the formula
kf kp = sup wp (x)|f (x)|,
x>0
and so wp (x)|f (x)| 6 kf kp ; let m ∈ N, p ∈ W be fixed numbers, and denote
Cpm = { f ∈ Cp : f (k) ∈ Cp , k = 0, 1, 2, . . . , m }.
AMS Subject Classification: 41A36
Keywords and phrases: Voronovskaya Theorem, polynomial weighted spaces.
(1.2)
88
A. Wafi, S. Khatoon
The Generalized Baskakov-Kantorovich operator Vna (f ; x)[1] is connected with
the Taylor’s series of functions
½∞
¾½ ∞ µ
¶ ¾
P (at)k
P n−1+k k
1
eat
=
t
(1 − t)n
k
k=0 k!
k=0
x
and x ∈ [0, ∞) yields
1+x
µ ¶
µ
¶k
ax
∞ P
k
P
k (n − 1 + i)! ak−i
x
e 1+x (1 + x)n =
.
(n − 1)! k!
1+x
k=0 i=0 i
for t ∈ [0, 1) and n = 1, 2, . . . , which for t =
We define the operator as follows
Vna (f ; x) = n
n
P
k=0
where
pn,k (x, a) = e
µZ
(k+1)/n
¶
f (t) dt pn,k (x, a),
(1.3)
k/n
ax
− 1+x
with
pk (n, a) =
·
pk (n, a)
xk
·
,
k!
(1 + x)n+k
k
P
k
i=0
Ci (n)i ak−i ,
(1.4)
(1.5)
and (n)0 = 1, (n)i = n(n + 1)(n + 2) · · · (n + i − 1), for i > 1; a > 0 is a fixed
absolute constant.
2. Some properties and lemmas
We shall give here some properties of the operator (1.3).
Lemma 2.1. For a, x > 0, n ∈ N, the following identities hold:
Vna (1; x) = 1,
(2.1)
ax
1
Vna (t; x) = x +
+
,
(2.2)
n(1 + x) 2n
1+n 2
1
2x
a2 x2
2ax2
2ax
Vna (t2 ; x) =
x + 2+
+ 2
+
+ 2
,
(2.3)
2
n
3n
n
n (1 + x)
n(1 + x) n (1 + x)
(n + 1)(n + 2) 3 9(n + 1) 2
7x
1
n + 1 3ax3
Vna (t3 ; x) =
x
+
x
+
+
+
n2
2n2
2n2
4n3
n2 n(1 + x)
2 3
3 3
2
2 2
3a x
a x
9ax
9a x
7ax
+ 2
+ 3
+ 2
+ 3
+ 3
,
2
3
2
n (1 + x)
n (1 + x)
n (1 + x) 2n (1 + x)
2n (1 + x)
(2.4)
Vna (t4 ; x) =
(n + 1)(n + 2)(n + 3) 4 8(n + 1)(n + 2) 3 15(n + 1) 2 6x
x +
x +
x + 3
n3
n3
n3
n
(n + 1)(n + 2) 4ax4
n + 1 6a2 x4
4a3 x4
1
+
+ 3
+ 4+
5n
n2
n(1 + x)
n2 n(1 + x)2
n (1 + x)3
The Voronovskaya theorem in polynomial weight spaces
a4 x 4
n + 1 24ax3
24a2 x3
8a3 x3
+
+
+
n4 (1 + x)4
n2 n(1 + x) n3 (1 + x)2
n4 (1 + x)3
2
2 2
30ax
15a x
6ax
+ 3
+ 4
+ 4
.
2
n (1 + x) n (1 + x)
n (1 + x)
89
+
(2.5)
Proof. Identities (2.1), (2.2) and (2.3) are proved in [1]. Identities (2.4) and
(2.5) can be calculated similarly using (1.3).
Lemma 2.2. For a, x > 0, n ∈ N, we have:
·
¸
1
1
ax
+
Vna ((t − x); x) =
,
(2.6)
n 1+x 2
1
2ax
a2 x2
x(1 + x)
+ 2+ 2
+ 2
,
(2.7)
Vna ((t − x)2 ; x) =
n
3n
n (1 + x) n (1 + x)2
3
6
6
16
3
15
5
1
Vna ((t − x)4 ; x) = 2 x4 + 3 x4 + 2 x3 + 3 x3 + 2 x2 + 3 x2 + 3 x + 4
n
n
n
n ¸ n
n
5n
· n
1
6a2 x3
10ax3
2
3
+ 3 6ax + 8ax +
+
n
1+x
1+x
·
¸
2 2
3 3
1
6ax
15a x
8a x
a4 x4
+ 4
+
+
+
.
(2.8)
n 1 + x (1 + x)2
(1 + x)3
(1 + x)4
Proof. (2.6) and (2.7) are proved in [1]. For (2.8), lineraity of the operator
implies that
Vna ((t − x)4 ; x) = Vna (t4 ; x) − 4xVna (t3 ; x) + 6x2 Vna (t2 ; x) − 4x3 Vna (t; x) + Vna (1; x).
Using Lemma 2.1 we arrive at (2.8).
Using the mathematical induction we can prove a formula for Vna (tq ; x) similar
to the one given by Bogalska [4] for Baskakov-Kantorovich operators.
Lemma 2.3. For q ∈ W , n ∈ N, x ∈ [0, ∞) there exist positive coefficients
ξq,n,j and δj,n,i , 0 6 j 6 q, 1 6 i 6 j depending only on q, j, n, i and bounded for n
and x such that
¶i ¾
½ j
µ
q
q
P
P
P
a
a q
j j−q
j−q−1
Vn (t ; x) =
ξq,n,j x n
+
n
δj,n,i
xj ,
(2.9)
1+x
j=0
j=1
i=1
where 1 6 ξq,n,j 6 q!.
Lemma 2.4. For every fixed p ∈ W , there exists a positive constant M1 (p, a)
depending only on p and a such that for all n ∈ N, we have
° µ
¶°
° a
°
1
°Vn
° 6 M1 (p, a).
;
x
(2.10)
°
°
wp (t)
p
90
A. Wafi, S. Khatoon
Proof. For p = 0, (2.10) is obvious. For p > 1, we have
µ
¶
1
; x = wp (x)Vna (1 + tp ; x) = wp (x)[Vna (1; x) + Vna (tp ; x)]
wp (t)
½ j
µ
¶i ¾ ¸
·
p
p
P
P
P
a
j j−p
j−p−i
δj,n,i
= wp (x) 1 +
ξp,n,j x n
+
n
xj
1+x
i=1
j=0
j=1
½ j
µ
¶i ¾
p
P
P
P
xj
a
xj
1 + p! xp p−1
j−p
j−p−1
+
ξ
n
+
n
δ
.
6
p,n,j
j,n,i
1 + xp
1 + xp j=1
1+x
1 + xp
j=o
i=1
wp (x)Vna
xj
1 + p! xp
But 0 6
6 1 for x ∈ [0, ∞) and 1 6 j 6 p − 1. Also
6 p! and
p
1 + xp
µ
¶i1 + x
a
6 ai . Therefore,
1+x
µ
wp (x)Vna
¾
¶
½ j
p−1
p
P
P
P
1
δj,n,i ai
; x 6 p! +
ξp,n,j nj−p +
nj−p−1
wp (t)
i=1
j=0
j=1
6 M1 (p, a).
Using Lemma 2.4, we shall prove the following lemma.
Lemma 2.5. For every fixed p ∈ W , there exists a positive constant M1 (p, a)
depending only on p and a such that for any f ∈ Cp ,
kVna (f ; x)kp 6 kf kp M1 (p, a),
n ∈ N.
(2.11)
Proof. From (1.2) we have kVna (f ; x)kp = supx>0 wp (x)|Vna (f ; x)|. But
wp (x)|Vna (f ; x)| 6 wp (x)n
∞
P
µZ
k=0
6 kf kp wp (x)Vna
(k+1)/n
f (t)wp (t)
k/n
µ
¶
1
;x ;
wp (t)
¶
1
dt pn,k (x, a)
wp (t)
hence, from Lemma 2.4 we obtain kVna (f ; x)kp 6 kf kp M1 (p, a).
Lemma 2.6 For some x0 ∈ [0, ∞) there exists a positive constant M2 (x0 , a)
depending only on x0 and a such that for all n ∈ N we have
Vna ((t − x)4 ; x) 6 M2 (x0 , a)n−2 .
(2.12)
The Voronovskaya theorem in polynomial weight spaces
91
Proof. From (2.8) we get
Vn ((t − x0 )4 ; x0 ) =
3 4
6
6
16
3
15
5
1
x0 + 3 x40 + 2 x30 + 3 x30 + 2 x20 + 3 x20 + 3 x0 + 4
2
n
n
n
n
n
5n
· n
¸ n
2 3
3
1
6a
x
10ax
0
0
+ 3 6ax20 + 8ax30 +
+
n
1 + x0
1 + x0
·
¸
1
6ax0
15a2 x20
8a3 x30
a4 x40
+ 4
+
+
+
,
n 1 + x0
(1 + x0 )2
(1 + x0 )3
(1 + x0 )4
Vn ((t − x0 )4 ; x0 ) 6 (9x40 + 22x30 + 18x20 + 5x0 + 0.2)n−2 + (6ax0 + 24ax20
+ 8ax30 + 6a2 x0 + 6a2 x20 + 6a + 15a2 + 8a3 + a4 )n−2
6 M2 (x0 , a)n−2 .
Lemma 2.7. For every x0 ∈ [0, ∞) there hold
Lt nVna ((t − x0 ); x0 ) =
n→∞
and
1
ax0
+
2 1 + x0
Lt nVna ((t − x0 )2 ; x0 ) = x0 (1 + x0 ).
n→∞
(2.13)
(2.14)
Proof. From Lemma 2.2 we know that for a, x0 > 0,
1
ax0
Vna ((t − x0 ); x0 ) =
+
,
2n n(1 + x0 )
x0 (1 + x0 )
1
2ax0
a2 x20
Vna ((t − x0 )2 ; x0 ) =
+ 2+ 2
+ 2
,
n
3n
n (1 + x0 ) n (1 + x0 )2
which yield (2.13) and (2.14).
The following lemma [2] is needed for further estimations.
Lemma 2.8. For each p ∈ W there exists a constant M3 (p, a) such that for all
n ∈ N,
µ
¶
2
x(1 + x)
a (t − x)
wp (x)Vn
; x 6 M3 (p, a)
.
wp (t)
n
Lemma 2.9. Let x0 ∈ [0, ∞) be a fixed point and let g(t; x0 ) ∈ Cp with p ∈ W
such that
Lt g(t; x0 ) = 0.
(2.15)
t→x0
Then
Lt Vna (g(t; x0 ); x0 ) = 0.
t→x0
(2.16)
Proof. Let ε > 0 and the constant M1 (p, a) be as in Lemma 2.4. Then from
(2.15) and properties of g(t; x0 ) there exist positive constants δ = δ(ε, M ) and M4
such that
ε
wp (t)|g(t; x0 )| <
, for |t − x0 | < δ
(2.17)
2M1 (p, a)
92
A. Wafi, S. Khatoon
and
wp (t)|g(t; x0 )| < M4 ,
Then for all n ∈ N we have
n
P
wp (x0 )|Vna (g(t; x0 ); x0 )| 6 wp (x0 )n
= wp (x0 )n ¯
µZ
P
¯
¯
¯k
¯ −x0 ¯<δ
¯
¯n
for all t ∈ [0, ∞).
µZ
k=0
(k+1)/n
¶
|g(t; x0 )| dt pn,k (x0 , a)
k/n
¶
|g(t; x0 )| dt pn,k (x0 , a)
(k+1)/n
k/n
µZ
P
+ wp (x0 )n ¯
¯
¯
¯k
¯ −x0 ¯>δ
¯
¯n
(k+1)/n
¶
|g(t; x0 )| dt pn,k (x0 , a).
k/n
From 2.17 and Lemma 2.4, we obtain
µZ
P
ε
S1 6
wp (x0 )n ¯
¯
2M1 (p, a)
¯k
¯
µ
(k+1)/n
k/n
¯ −x0 ¯<δ
¯n
¯
ε
wp (x0 )Vna
6
2M1 (p, a)
1
; x0
wp (t)
¶
6
¶
1
dt pn,k (x0 , a)
wp (t)
ε
ε
M1 (p, a) = .
2M1 (p, a)
2
Moreover, by (2.18) we get
S2 6 M4 wp (x0 )n ¯
µZ
P
¯
¯
¯k
¯ −x0 ¯>δ
¯
¯n
(k+1)/n
k/n
But from [4] we know that
Z (k+1)/n
k/n
1
dt 6
wp (t)
µ
nwp
¶
1
dt pn,k (x0 , a).
wp (t)
1
2p+1
1
¶6
µ ¶,
k+1
k
n
wp
n
n
therefore,
P
1
µ ¶ pn,k (x0 , a)
k
wp
n
µ
¶2
P
1 k
1
p+1
µ ¶ pn,k (x0 , a)
6 2 M4 wp (x0 ) ¯
− x0
2
¯
k
δ
n
¯k
¯
¯ −x0 ¯>δ
wp
¯n
¯
n
µ
¶2
∞
P
2p+1
k
1
µ ¶ pn,k (x0 , a)
6 M4 2 wp (x0 )
− x0
k
δ
n
k=0
wp
n
µ
¶
2
2p+1
(t
−
x
)
0
6 M4 2 wp (x0 )Vna
; x0 .
δ
wp (t)
S2 6 2p+1 M4 wp (x0 ) ¯
(2.18)
¯
¯k
¯
¯ −x0 ¯>δ
¯n
¯
93
The Voronovskaya theorem in polynomial weight spaces
From [2], we have
2p+1
x0 (1 + x0 )
x0 (1 + x0 )
M3 (p, a)
.
≡ M5 (p, a)
2
δ
n
nδ 2
For the fixed positive numbers ε, δ, M5 and x0 > 0 there exists a natural number
x0 (1 + x0 )
ε
n1 , depending only on ε, δ and M5 , such that M5 (p, a)
<
for all
nδ 2
2
ε
n > n1 . Thus S2 < and hence
2
S2 6 M4
wp (x0 )|Vna (g(t; x0 ); x0 )| < ε for all n > n1 ,
i.e. Ltn→∞ Vna (g(t; x0 ); x0 ) = 0.
(2.16).
Above estimation and Definition 1.1 imply
3. The Voronovskaya Theorem
The main theroem is as follows.
Theorem 3.1. Let f ∈ Cp2 with some p ∈ W . Then for every x ∈ [0, ∞) we
have
½
¾
ª
© a
1
ax
x(1 + x) 00
Lt Vn (f ; x) − f (x) =
+
f 0 (x) +
f (x).
(3.1)
n→∞
2 1+x
2
Proof. Take a fixed point x0 ∈ [0, ∞). Then for f ∈ Cp2 , t ∈ [0, ∞) we have by
Taylor’s formula
1
f (t) = f (x0 ) + f 0 (x0 )(t − x0 ) + f 00 (x0 )(t − x0 )2 + ϕ(t; x0 )(t − x0 )2 ,
2
where ϕ(t; x0 ) ∈ Cp and Ltt→x0 ϕ(t; x0 ) = 0. Multiplying both sides by npn,k (x, a),
k
k+1
integrating with respect to t between the limits
and
, then summing over
n
n
k, we get
Vna (f ; x0 ) = f (x0 )Vna (1; x0 ) + f 0 (x0 )Vna ((t − x0 ); x0 )
1
+ f 00 (x0 )Vna ((t − x0 )2 ; x0 ) + Vna (ϕ(t; x0 )(t − x0 )2 ; x0 ).
2
Using Lemma 2.7 we obtain
½
¾
©
ª
1
ax0
Lt n Vna (f ; x0 ) − f (x0 ) =
+
f 0 (x0 )
n→∞
2 1 + x0
x0 (1 + x0 ) 00
f (x0 ) + Lt nVna (ϕ(t; x0 )(t − x0 )2 ; x0 ). (3.2)
+
n→∞
2
From Hölder’s inequality and Lemma 2.6, we have
1
1
Vna (ϕ(t; x0 )(t − x0 )2 ; x0 ) 6 {Vna (ϕ2 (t; x0 ); x0 )} 2 {Vna ((t − x0 )4 ; x0 )} 2
1
1
6 {Vna (ϕ2 (t; x0 ); x0 )} 2 {M2 (x0 , a)n−2 } 2 .
(3.3)
94
A. Wafi, S. Khatoon
The properties of the function ϕ(t; x0 ) imply that for ψ(t; x0 ) ≡ ϕ2 (t; x0 ) we have
Ltt→x0 ψ(t; x0 ) = 0. By Lemma 2.9 we get
Lt Vna (ψ(t; x0 ); x0 ) = 0.
n→∞
(3.4)
Combining (3.2), (3.3) and (3.4), we arrive at (3.1).
Remark. When a = 0, (3.1) reduces to Bogalska estimate [4].
Acknowledgement. The authors are thankful to the referee for his valuable
suggestions and comments.
REFERENCES
[1] A. Wafi and S. Khatoon, Uniform approximation with generalized Baskakaov-Kantorovich
operators, Alig. Bull. Math. 22 (2) (2003), 119–133.
[2] A. Wafi and S. Khatoon, Direct and inverse theorems for generalized Baskakaov operators in
polynomial weight spaces, Anal. Stiint. Ale Univ. “Al. I. Cuza”, Iaşi, Vol. L, S.I.a, Matematica
(2004), 159–173.
[3] M. Becker, Global approximation theorems for Szász-Mirakjan and Baskakov operators in
polynomial weight spaces, Indiana Univ. Math J. 27 (1) (1978), 127–142.
[4] K. Bogalska, The Voronovskaya type theorem for the Baskakov-Kantorovich operators, Fasc.
Math. 30 (1999), 5–13.
[5] P. L. Butzer and R. J. Nessel, Fourier Analysis and Approximation, Vol. I, Birkhäuser, Bessel
and Academic Press, New York, 1971.
[6] L. Rempulska and M. Skorupka, The Voronovskaya theorem for some operators of the SzázMirakjan type, Le Mathematiche 50 (2) (1995), 251–261.
[7] E. Voronovskaja, Dètermination de la forme asymptotique d’approximation des functions
par ploynômes de M. Bernstein, C. R. Acad. Sci. URSS 79 (1932), 79–85.
(received 24.09.2003, in revised form 27.07.2005)
Department of Mathematics, Jamia Millia Islamia, Jamia Nagar, New Delhi 110025, INDIA