Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 4, Article 125, 2006
ON SIMULTANEOUS APPROXIMATION FOR CERTAIN BASKAKOV
DURRMEYER TYPE OPERATORS
VIJAY GUPTA, MUHAMMAD ASLAM NOOR, MAN SINGH BENIWAL, AND M. K. GUPTA
S CHOOL OF A PPLIED S CIENCES
N ETAJI S UBHAS I NSTITUTE OF T ECHNOLOGY
S ECTOR 3 DWARKA
N EW D ELHI 110075, I NDIA
vijay@nsit.ac.in
M ATHEMATICS D EPARTMENT
COMSATS I NSTITUTE OF I NFORMATION T ECHNOLOGY
I SLAMABAD , PAKISTAN
noormaslam@hotmail.com
D EPARTMENT OF A PPLIED S CIENCE
M AHARAJA S URAJMAL I NSTITUTE OF T ECHNOLOGY
C-4, JANAKPURI , N EW D ELHI - 110058, I NDIA
man_s_2005@yahoo.co.in
D EPARTMENT OF M ATHEMATICS
C H C HARAN S INGH U NIVERSITY
M EERUT 250004, I NDIA
mkgupta2002@hotmail.com
Received 06 January, 2006; accepted 16 August, 2006
Communicated by N.E. Cho
A BSTRACT. In the present paper, we study a certain integral modification of the well known
Baskakov operators with the weight function of Beta basis function. We establish pointwise
convergence, an asymptotic formula an error estimation and an inverse result in simultaneous
approximation for these new operators.
Key words and phrases: Baskakov operators, Simultaneous approximation, Asymptotic formula, Pointwise convergence, Error estimation, Inverse theorem.
2000 Mathematics Subject Classification. 41A30, 41A36.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
The work carried out when the second author visited Department of Mathematics and Statistics, Auburn University, USA in fall 2005.
The authors are thankful to the referee for making many valuable suggestions, leading to the better presentation of the paper.
009-06
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V IJAY G UPTA , M UHAMMAD A SLAM N OOR , M AN S INGH B ENIWAL , AND M. K. G UPTA
1. I NTRODUCTION
For
f ∈ Cγ [0, ∞) ≡ {f ∈ C[0, ∞) : |f (t)| ≤ M tγ
for some M > 0, γ > 0} we consider a certain type of Baskakov-Durrmeyer operator as
Z ∞
∞
X
(1.1)
Bn (f (t), x) =
pn,k (x)
bn,k (t)f (t)dt + (1 + x)−n f (0)
0
Zk=1∞
=
Wn (x, t)f (t)dt
0
where
n+k−1
xk
pn,k (x) =
,
k
(1 + x)n+k
tk−1
1
·
bn,k (t) =
B(n + 1, k) (1 + t)n+k+1
and
Wn (x, t) =
∞
X
pn,k (x)bn,k (t) + (1 + x)−n δ(t),
k=1
δ(t) being the Dirac delta function. The norm- || · ||γ on the class Cγ [0, ∞) is defined as
||f ||γ = sup |f (t)|t−γ .
0≤t<∞
The operators defined by (1.1) are the integral modification of the well known Baskakov
operators with weight functions of some Beta basis functions. Very recently Finta [2] also
studied some other approximation properties of these operators. The behavior of these operators
is very similar to the operators recently introduced in [6], [9] and also studied in [8]. These
operators reproduce not only the constant functions but also the linear functions, which is the
interesting property of such operators. The other usual Durrmeyer type integral modification
of the Baskakov operators [5] reproduce only the constant functions, so one can not apply the
iterative combinations easily to improve the order of approximation for the usual Baskakov
Durrmeyer operators. For recent work in this area we refer to [7]. In the present paper we study
some direct results which include pointwise convergence, asymptotic formula, error estimation
and inverse theorem in the simultaneous approximation for the unbounded functions of growth
of order tγ .
2. BASIC R ESULTS
In this section we mention certain lemmas which will be used in the sequel.
Lemma 2.1 ([3]). For m ∈ N ∪ {0}, if the mth order moment be defined as
m
∞
X
k
pn,k (x)
Un,m (x) =
−x
,
n
k=0
then Un,0 (x) = 1, Un,1 (x) = 0 and
(1)
nUn,m+1 (x) = x(1 + x)(Un,m
(x) + mUn,m−1 (x)).
Consequently we have Un,m (x) = O n−[(m+1)/2] .
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
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O N S IMULTANEOUS A PPROXIMATION F OR C ERTAIN BASKAKOV D URRMEYER T YPE O PERATORS
3
Lemma 2.2. Let the function Tn,m (x), m ∈ N ∪ {0}, be defined as
Tn,m (x) = Bn (t − x)m x
Z ∞
∞
X
=
pn,k (x)
bn,k (t)(t − x)m dt + (1 + x)−n (−x)m .
0
k=1
Then Tn,0 (x) = 1, Tn,1 = 0, Tn,2 (x) = 2x(1+x)
and also there holds the recurrence relation
n−1
(1)
(n − m)Tn,m+1 (x) = x(1 + x) Tn,m (x) + 2mTn,m−1 (x) + m(1 + 2x)Tn,m (x).
Proof. By definition, we have
(1)
Tn,m
(x)
=
∞
X
(1)
pn,k (x)
∞
Z
bn,k (t)(t − x)m dt
0
k=1
−m
∞
X
Z
pn,k (x)
∞
bn,k (t)(t − x)m−1 dt
0
k=1
− n(1 + x)−n−1 (−x)m − m(1 + x)−n (−x)m−1 .
Using the identities
(1)
x(1 + x)pn,k (x) = (k − nx)pn,k (x)
and
(1)
t(1 + t)bn,k (t) = [(k − 1) − (n + 2)t]bn,k (t),
we have
(1)
x(1 + x) Tn,m
(x) + mTn,m−1 (x)
Z ∞
∞
X
pn,k (x)
(k − nx)bn,k (t)(t − x)m dt + n(1 + x)−n (−x)m+1
=
=
k=1
∞
X
k=1
0
Z
pn,k (x)
∞
(k − 1) − (n + 2)t + (n + 2)(t − x)
0
+ (1 + 2x) bn,k (t)(t − x)m dt + n(1 + x)−n (−x)m+1
Z ∞
∞
X
(1)
t(1 + t)bn,k (t)(t − x)m dt
=
pn,k (x)
k=1
0
+ (n + 2)[Tn,m+1 (x) − (1 + x)−n (−x)m+1 ]
+ (1 + 2x)[Tn,m (x) − (1 + x)−n (−x)m ] + n(1 + x)−n (−x)m+1
= −(m + 1)(1 + 2x)[Tn,m (x) − (1 + x)−n (−x)m ]
− (m + 2)[Tn,m+1 − (1 + x)−n (−x)m+1 ]
− mx(1 + x)[Tn,m−1 (x) − (1 + x)−n (−x)m−1 ]
+ (n + 2)[Tn,m+1 − (1 + x)−n (−x)m+1 ]
+ (1 + 2x)[Tn,m (x) − (1 + x)−n (−x)m ] + n(1 + x)−n (−x)m+1 .
Thus, we get
(1)
(x) + 2mTn,m−1 (x)] + m(1 + 2x)Tn,m (x).
(n − m)Tn,m+1 (x) = x(1 + x)[Tn,m
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
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4
V IJAY G UPTA , M UHAMMAD A SLAM N OOR , M AN S INGH B ENIWAL , AND M. K. G UPTA
This completes the proof of recurrence relation. From the above recurrence relation, it is easily
verified for all x ∈ [0, ∞) that
Tn,m (x) = O n−[(m+1)/2] .
Remark 2.3. It is easily verified from Lemma 2.1 that for each x ∈ (0, ∞)
(n + i − 1)!(n − i)! i
(n + i − 2)!(n − i)! i−1
Bn (ti , x) =
x + i(i − 1)
x + O(n−2 ).
n!(n − 1)!
n!(n − 1)!
Corollary 2.4. Let δ be a positive number. Then for every γ > 0, x ∈ (0, ∞), there exists a
constant M (s, x) independent of n and depending on s and x such that
Z
γ Wn (x, t)t dt
≤ M (s, x)n−s ,
s = 1, 2, 3, . . .
|t−x|>δ
C[a,b]
Lemma 2.5. There exist the polynomials Qi,j,r (x) independent of n and k such that
X
{x(1 + x)}r Dr pn,k (x) =
ni (k − nx)j Qi,j,r (x)pn,k (x),
2i+j≤r
i,j≥0
where D ≡
d
.
dx
By C0 , we denote the class of continuous functions on the interval (0, ∞) having a compact
support and C0r is the class of r times continuously differentiable functions with C0r ⊂ C0 . The
function f is said to belong to the generalized Zygmund class Liz(α, 1, a, b), if there exists a
constant M such that ω2 (f, δ) ≤ M δ α , δ > 0, where ω2 (f, δ) denotes the modulus of continuity
of 2nd order on the interval [a, b]. The class Liz(α, 1, a, b) is more commonly denoted by
Lip∗ (α, a, b). Suppose G(r) = {g : g ∈ C0r+2 , supp g ⊂ [a0 , b0 ] where [a0 , b0 ] ⊂ (a, b)}. For r
times continuously differentiable functions f with supp f ⊂ [a0 , b0 ] the Peetre’s K-functionals
are defined as
h
n oi
Kr (ξ, f ) = inf f (r) − g (r) 0 0 + ξ g (r) 0 0 + g (r+2) 0 0
,
0 < ξ ≤ 1.
g∈G(r)
C[a ,b ]
C[a ,b ]
C[a ,b ]
For 0 < α < 2, C0r (α, 1, a, b) denotes the set of functions for which
sup ξ −α/2 Kr (ξ, f, a, b) < C.
0<ξ≤1
Lemma 2.6. Let 0 < a0 < a00 < b00 < b0 < b < ∞ and f (r) ∈ C0 with supp f ⊂ [a00 , b00 ]
and if f ∈ C0r (α, 1, a0 , b0 ), we have f (r) ∈ Liz(α, 1, a0 , b0 ) i.e. f (r) ∈ Lip∗ (α, a0 , b0 ) where
Lip∗ (α, a0 , b0 ) denotes the Zygmund class satisfying Kr (δ, f ) ≤ Cδ α/2 .
Proof. Let g ∈ G(r) , then for f ∈ C0r (α, 1, a0 , b0 ), we have
2 (r) 2 (r)
4δ f (x) ≤ 4δ (f − g (r) )(x) + 42δ g (r) (x)
≤ 42δ (f (r) − g (r) )C[a0 ,b0 ] + δ 2 g (r+2) C[a0 ,b0 ]
≤ 4M1 Kr (δ 2 , f ) ≤ M2 δ α .
(r−1)
α
Lemma 2.7. If f is r times differentiable on [0, ∞), such that f
= O(t ), α > 0 as t → ∞,
then for r = 1, 2, 3, . . . and n > α + r we have
Z ∞
∞
(n + r − 1)!(n − r)! X
(r)
pn+r,k (x)
bn−r,k+r (t)f (r) (t)dt.
Bn (f, x) =
n!(n − 1)!
0
k=0
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
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O N S IMULTANEOUS A PPROXIMATION F OR C ERTAIN BASKAKOV D URRMEYER T YPE O PERATORS
5
Proof. First
Bn(1) (f, x)
∞
X
=
(1)
pn,k (x)
Z
∞
bn,k (t)f (t)dt − n(1 + x)−n−1 f (0).
0
k=1
Now using the identities
(1)
(2.1)
pn,k (x) = n[pn+1,k−1 (x) − pn+1,k (x)],
(2.2)
bn,k (t) = (n + 1)[bn+1,k−1 (t) − bn+1,k (t)].
(1)
for k ≥ 1, we have
Bn(1) (f, x)
=
∞
X
Z
∞
n[pn+1,k−1 (x) − pn+1,k (x)]
bn,k (t)f (t)dt − n(1 + x)−n−1 f (0)
0
k=1
Z
∞
bn,1 (t)f (t)dt − n(1 + x)−n−1 f (0)
= npn+1,0 (x)
+n
0
∞
X
Z
[bn,k+1 (t) − bn,k (t)]f (t)dt
0
k=1
−n−1
∞
pn+1,k (x)
∞
Z
(n + 1)(1 + t)−n−2 f (t)dt − n(1 + x)−n−1 f (0)
= n(1 + x)
0
+n
∞
X
Z
∞
pn+1,k (x)
0
k=1
1 (1)
− bn−1,k+1 (t)f (t)dt.
n
Integrating by parts, we get
Bn(1) (f, x)
−n−1
= n(1 + x)
−n−1
Z
∞
f (0) + n(1 + x)
(1 + t)−n−1 f (1) (t)dt
0
−n−1
− n(1 + x)
f (0) +
∞
X
Z
pn+1,k (x)
k=1
=
∞
X
k=0
Z
∞
bn−1,k+1 (t)f (1) (t)dt
0
∞
pn+1,k (x)
bn−1,k+1 (t)f (1) (t)dt.
0
Thus the result is true for r = 1. We prove the result by induction method. Suppose that the
result is true for r = i, then
Z ∞
∞
(n + i − 1)!(n − i)! X
(i)
Bn (f, x) =
pn+i,k (x)
bn−i,k+i (t)f (i) (t)dt.
n!(n − 1)!
0
k=0
Thus using the identities (2.1) and (2.2), we have
Bn(i+1) (f, x)
Z ∞
∞
(n + i − 1)!(n − i)! X
=
(n + i)[pn+i+1,k−1 (x) − pn+i+1,k (x)]
bn−i,k+i (t)f (i) (t)dt
n!(n − 1)!
0
k=1
Z ∞
(n + i − 1)!(n − i)!
+
(−(n + i)(1 + x)−n−i−1 )
bn−i,i (t)f (i) (t)dt
n!(n − 1)!
0
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
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6
V IJAY G UPTA , M UHAMMAD A SLAM N OOR , M AN S INGH B ENIWAL , AND M. K. G UPTA
Z ∞
(n + i)!(n − i)!
=
pn+i+1,0 (x)
bn−i,i+1 (t)f (i) (t)dt
n!(n − 1)!
0
Z ∞
(n + i)!(n − i)!
−
pn+i+1,0 (x)
bn−i,i (t)f (i) (t)dt
n!(n − 1)!
0
Z ∞
∞
X
(n + i)!(n − i)!
pn+i+1,k (x)
+
[bn−i,k+i+1 (t) − bn−i,k+i (t)]f (i) (t)dt
n!(n − 1)! k=1
0
Z ∞
(n + i)!(n − i)!
1
(1)
=
pn+i+1,0 (x)
−
bn−i−1,i+1 (t)f (i) (t)dt
n!(n − 1)!
(n
−
i)
0
Z ∞
∞
X
(n + i)!(n − i)!
1
(1)
+
pn+i+1,k (x)
−
bn−i−1,k+i+1 (t)f (i) (t)dt.
n!(n − 1)! k=1
(n − i)
0
Integrating by parts, we obtain
∞
Bn(i+1) (f, x)
(n + i)!(n − i − 1)! X
=
pn+i+1,k (x)
n!(n − 1)!
k=0
∞
Z
bn−i−1,k+i+1 (t)f (i+1) (t)dt.
0
This completes the proof of the lemma.
3. D IRECT T HEOREMS
In this section we present the following results.
Theorem 3.1. Let f ∈ Cγ [0, ∞) and f (r) exists at a point x ∈ (0, ∞). Then we have
Bn(r) (f, x) = f (r) (x)
as n → ∞.
Proof. By Taylor expansion of f , we have
f (t) =
r
X
f (i) (x)
i=0
i!
(t − x)i + ε(t, x)(t − x)r ,
where ε(t, x) → 0 as t → x. Hence
Z ∞
(r)
Bn (f, x) =
Wn(r) (t, x)f (t)dt
=
0
r
X
i=0
f (i) (x)
i!
Z
∞
Wn(r) (t, x)(t
0
i
Z
− x) dt +
∞
Wn(r) (t, x)ε(t, x)(t − x)r dt
0
=: R1 + R2 .
First to estimate R1 , using the binomial expansion of (t − x)i and Remark 2.3, we have
r
i r Z ∞
X
f (i) (x) X i
i−v ∂
R1 =
(−x)
Wn (t, x)tv dt
r
i!
v
∂x
0
v=0
i=0
(r)
r
f (x) d
(n + r − 1)!(n − r)! r
=
x + terms containing lower powers of x
r! dxr
n!(n − 1)!
(n + r − 1)!(n − r)!
(r)
= f (x)
→ f (r) (x)
n!(n − 1)!
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
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O N S IMULTANEOUS A PPROXIMATION F OR C ERTAIN BASKAKOV D URRMEYER T YPE O PERATORS
7
as n → ∞. Next applying Lemma 2.5, we obtain
Z ∞
R2 =
Wn(r) (t, x)ε(t, x)(t − x)r dt,
0
∞
|Qi,j,r (x)| X
|R2 | ≤
n
|k − nx|j pn,k (x)
r
{x(1 + x)} k=1
2i+j≤r
X
i
i,j≥0
∞
Z
(n + r + 1)!
(1 + x)−n−r |ε(0, x)|xr .
(n
−
1)!
0
The second term in the above expression tends to zero as n → ∞. Since ε(t, x) → 0 as t → x
for a given ε > 0 there exists a δ such that |ε(t, x)| < ε whenever 0 < |t − x| < δ. If
α ≥ max{γ, r}, where α is any integer, then we can find a constant M3 > 0, |ε(t, x)(t − x)r | ≤
M3 |t − x|α , for |t − x| ≥ δ. Therefore
∞
X
X
i
|R2 | ≤ M3
n
pn,k (x)|k − nx|j
bn,k (t)|ε(t, x)||t − x|r dt +
2i+j≤r
i,j≥0
k=0
Z
× ε
bn,k (t)|t − x| dt
Z
r
α
bn,k (t)|t − x| dt +
|t−x|<δ
|t−x|≥δ
=: R3 + R4 .
Applying the Cauchy-Schwarz inequality for integration and summation respectively, we obtain
) 12
(∞
) 12 ( ∞
Z ∞
X
X
X
pn,k (x)
bn,k (t)(t − x)2r dt
.
R3 ≤ εM3
ni
pn,k (x)(k − nx)2j
0
k=1
k=1
2i+j≤r
i,j≥0
Using Lemma 2.1 and Lemma 2.2, we get
R3 = ε · O(nr/2 )O(n−r/2 ) = ε · o(1).
Again using the Cauchy-Schwarz inequality, Lemma 2.1 and Corollary 2.4, we get
Z
∞
X
X
i
j
R4 ≤ M4
n
pn,k (x)|k − nx|
bn,k (t)|t − x|α dt
X
≤ M4
ni
2i+j≤r
i,j≥0
X
≤ M4
2i+j≤r
i,j≥0
=
X
|t−x|≥δ
k=1
2i+j≤r
i,j≥0
ni
∞
X
pn,k (x)|k − nx|j
12 Z
bn,k (t)dt
Z
k=1
|t−x|≥δ
(∞
X
) 12 ( ∞
X
pn,k (x)(k − nx)2j
k=1
12
bn,k (t)(t − x)2α dt
|t−x|≥δ
Z
pn,k (x)
k=1
) 12
∞
bn,k (t)(t − x)2α dt
0
ni O(nj/2 )O(n−α/2 ) = O(n(r−α)/2 ) = o(1).
2i+j≤r
i,j≥0
Collecting the estimates of R1 − R4 , we obtain the required result.
Theorem 3.2. Let f ∈ Cγ [0, ∞). If f (r+2) exists at a point x ∈ (0, ∞). Then
lim n Bn(r) (f, x) − f (r) (x)
n→∞
= r(r − 1)f (r) (x) + r(1 + 2x)f (r+1) (x) + x(1 + x)f (r+2) (x).
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
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8
V IJAY G UPTA , M UHAMMAD A SLAM N OOR , M AN S INGH B ENIWAL , AND M. K. G UPTA
Proof. Using Taylor’s expansion of f, we have
f (t) =
r+2 (i)
X
f (x)
i!
i=0
(t − x)i + ε(t, x)(t − x)r+2 ,
where ε(t, x) → 0 as t → x and ε(t, x) = O((t − x)γ ), t → ∞ for γ > 0. Applying Lemma
2.2, we have
n Bn(r) (f (t), x) − f (r) (x)
" r+2
#
X f (i) (x) Z ∞
=n
Wn(r) (t, x)(t − x)i dt − f (r) (x)
i!
0
i=0
Z ∞
+n
Wn(r) (t, x)ε(t, x)(t − x)r+2 dt
0
=: E1 + E2 .
First, we have
E1 = n
r+2 (i)
i X
f (x) X i
i=0
=
i!
j=0
j
i−j
Z
(−x)
∞
Wn(r) (t, x)tj dt − nf (r) (x)
0
f (r+1) (x) f (r) (x) (r) r
n Bn (t , x) − (r!) +
n (r + 1)(−x)Bn(r) (tr , x) + Bn(r) (tr+1 , x)
r!
(r + 1)!
f (r+2) (x) (r + 2)(r + 1) 2 (r) r
(r) r+1
(r) r+2
n
x Bn (t , x) + (r + 2)(−x)Bn (t , x) + Bn (t , x) .
+
(r + 2)!
2
Therefore, by Remark 2.3, we have
(n + r − 1)!(n − r)!
(r)
E1 = nf (x)
−1
n!(n − 1)!
nf (r+1) (x)
(n + r − 1)!(n − r)!
+
(x − 1)(−x)
(r + 1)!
n!(n − 1)!
(n + r)!(n − r − 1)!
(n + r − 1)!(n − r − 1)!
+
(r + 1)!x + (r + 1)r
r!
n!(n − 1)!
n!(n − 1)!
nf (r+2) (x) (r + 2)(r + 1) 2 (n + r − 1)!(n − r)!
+
x
r!
(r + 2)!
2
n!(n − 1)!
(n + r)!(n − r − 1)!
(n − r − 1)!(n − r − 1)!
+ (r + 2)(−x)
x(r + 1)! + (r + 1)r
r!
2
n!(n − 1)!
(n + r + 1)!(n − r − 2)! (r + 2)! 2
+
x
n!(n − 1)!
2
(n + r)!(n − r − 2)!
+ (r + 2)(r + 1)
(r + 1)!x
+ O(n−2 ).
n!(n − 1)!
In order to complete the proof of the theorem it is sufficient to show that E2 → 0 as n → ∞
which easily follows proceeding along the lines of the proof of Theorem 3.1 and by using
Lemma 2.1, Lemma 2.2 and Lemma 2.5.
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O N S IMULTANEOUS A PPROXIMATION F OR C ERTAIN BASKAKOV D URRMEYER T YPE O PERATORS
9
Lemma 3.3. Let 0 < α < 2, 0 < a < a0 < a00 < b00 < b0 < b < ∞. If f ∈ C0 with
(r)
supp f ⊂ [a00 , b00 ] and Bn (f, ·) − f (r) = O(n−α/2 ), then
C[a,b]
Kr (ξ, f ) = M5 n−α/2 + nξKr (n−1 , f ) .
Consequently Kr (ξ, f ) ≤ M6 ξ α/2 , M6 > 0.
Proof. It is sufficient to prove
Kr (ξ, f ) = M7 {n−α/2 + nξKr (n−1 , f )},
for sufficiently large n. Because supp f ⊂ [a00 , b00 ] therefore by Theorem 3.2 there exists a
function h(i) ∈ G(r) , i = r, r + 2, such that
(r)
Bn (f, •) − h(i) 0 0 ≤ M8 n−1 .
C[a ,b ]
Therefore,
Kr (ξ, f ) ≤ 3M9 n−1 + Bn(r) (f, •) − f (r) C[a0 ,b0 ]
n
o
+ ξ Bn(r) (f, •)C[a0 ,b0 ] + Bn(r+2) (f, •)C[a0 ,b0 ] .
Next, it is sufficient to show that there exists a constant M10 such that for each g ∈ G(r)
(r+2)
Bn (f, •) 0 0 ≤ M10 n{f (r) − g (r) 0 0 + n−1 g (r+2) 0 0 .
(3.1)
C[a ,b ]
C[a ,b ]
C[a ,b ]
Also using the linearity property, we have
(r+2)
Bn (f, •) 0 0 ≤ Bn(r+2) (f − g, •) 0 0 + Bn(r+2) (g, •) 0 0 .
(3.2)
C[a ,b ]
C[a ,b ]
C[a ,b ]
Applying Lemma 2.5, we get
Z
0
∞
r+2
∂
dt ≤
W
(x,
t)
n
∂xr+2
∞
X X
2i+j≤r+2
i,j≥0
ni |k − nx|j
k=1
|Qi,j,r+2 (x)|
{x(1 + x)}r+2
Z
× pn,k (x)
∞
bn,k (t)dt +
0
dr+2
[(1 + x)−n ].
r+2
dx
Therefore by the Cauchy-Schwarz inequality and Lemma 2.1, we obtain
(r)
Bn (f − g, •) 0 0 ≤ M11 n f (r) − g (r) 0 0 ,
(3.3)
C[a ,b ]
C[a ,b ]
where the constant M11 is independent of f and g. Next by Taylor’s expansion, we have
g(t) =
r+1 (i)
X
g (x)
i=0
i!
(t − x)i +
g (r+2) (ξ)
(t − x)r+2 ,
(r + 2)!
R ∞ ∂m
where ξ lies between t and x. Using the above expansion and the fact that 0 ∂x
m Wn (x, t)(t −
x)i dt = 0 for m > i, we get
Z ∞ r+2
(r+2)
(r+2) ∂
r+2 (3.4) Bn (g, •) C[a0 ,b0 ] ≤ M12 g
·
Wn (x, t)(t − x) dt
.
C[a0 ,b0 ] r+2
∂x
0
C[a0 ,b0 ]
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10
V IJAY G UPTA , M UHAMMAD A SLAM N OOR , M AN S INGH B ENIWAL ,
AND
M. K. G UPTA
Also by Lemma 2.5 and the Cauchy-Schwarz inequality, we have
Z ∞ r+2
∂
(t − x)r+2 dt
E≡
W
(x,
t)
n
∂xr+2
0
Z ∞
∞
X X
i
j |Qi,j,r+2 (x)|
≤
n pn,k (x)|k − nx|
bn,k (t)(t − x)r+2 dt
r+2
{x(1
+
x)}
0
2i+j≤r+2 k=1
i,j≥0
+
≤
dr+2
[(−x)r+2 (1 + x)−n ]
dxr+2
X
2i+j≤r+2
i,j≥0
×
|Qi,j,r+2 (x)|
{x(1 + x)}r+2
∞
X
k=1
r+2
Z
∞
pn,k (x)
∞
X
! 12
pn,k (x)(k − nx)2j
k=1
! 12 Z
bn,k (t)(t − x)2r+4 dt
0
∞
12
bn,k (t)dt
0
d
[(−x)r+2 (1 + x)−n ]
dxr+2
X
|Qi,j,r+2 (x)|
−(1+ r2 )
j/2
=
ni
O(n
)O
n
.
r+2
{x(1
+
x)}
2i+j≤r+2
+
i,j≥0
Hence
(3.5)
||Bn(r+2) (g, •)||C[a0 ,b0 ] ≤ M13 ||g (r+2) ||C[a0 ,b0 ] .
Combining the estimates of (3.2)-(3.5), we get (3.1). The other consequence follows form [1].
This completes the proof of the lemma.
Theorem 3.4. Let f ∈ Cγ [0, ∞) and suppose 0 < a < a1 < b1 < b < ∞. Then for all n
sufficiently large, we have
(r)
−1
(r)
− 21
Bn (f, •) − f (r) ,
a,
b
+
M
n
kf
k
,
≤
max
M
ω
f
,
n
15
γ
14
2
C[a1 ,b1 ]
where M14 = M14 (r), M15 = M15 (r, f ).
Proof. For sufficiently small δ > 0, we define a function f2,δ (t) corresponding to f ∈ Cγ [0, ∞)
by
Z δ Z δ
2
2
−2
f (t) − ∆2η f (t) dt1 dt2 ,
f2,δ (t) = δ
− 2δ
− 2δ
2
where η = t1 +t
, t ∈ [a, b] and ∆2η f (t) is the second forward difference of f with step length η.
2
Following [4] it is easily checked that:
(i)
(ii)
(iii)
(iv)
f2,δ has continuous derivatives up to order 2k on [a, b],
(r)
c1 δ −r ω2 (f, δ, a, b),
kf2,δ kC[a1 ,b1 ] ≤ M
c2 ω2 (f, δ, a, b),
kf − f2,δ kC[a1 ,b1 ] ≤ M
c3 kf kγ ,
kf2,δ kC[a1 ,b1 ] ≤ M
ci , i = 1, 2, 3 are certain constants that depend on [a, b] but are independent of f and
where M
n [4].
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O N S IMULTANEOUS A PPROXIMATION F OR C ERTAIN BASKAKOV D URRMEYER T YPE O PERATORS
We can write
(r)
Bn (f, •) − f (r) C[a1 ,b1 ]
(r)
(r)
(r) ≤ Bn (f − f2,δ , •) C[a1 ,b1 ] + Bn (f2,δ , •) − f2,δ C[a1 ,b1 ]
11
(r)
(r) + f − f2,δ C[a1 ,b1 ]
=: H1 + H2 + H3 .
(r)
Since f2,δ = f (r) 2,δ (t), by property (iii) of the function f2,δ , we get
c4 ω2 (f (r) , δ, a, b).
H3 ≤ M
Next on an application of Theorem 3.2, it follows that
r+2 X
(j) −1
c
H2 ≤ M5 n
f2,δ .
C[a,b]
j=r
Using the interpolation property due to Goldberg and Meir [4], for each j = r, r + 1, r + 2, it
follows that
(j) (r+2)
c6 ||f2,δ ||C[a,b] + f
≤M
.
f 2,δ
2,δ
C[a1 ,b1 ]
C[a,b]
Therefore by applying properties (iii) and (iv) of the of the function f2,δ , we obtain
c7 4 · n−1 ||f ||γ + δ −2 ω2 (f (r) , δ) .
H2 ≤ M
Finally we shall estimate H1 , choosing a∗ , b∗ satisfying the conditions 0 < a < a∗ < a1 < b1 <
b∗ < b < ∞. Suppose ψ(t) denotes the characteristic function of the interval [a∗ , b∗ ], then
H1 ≤ Bn(r) (ψ(t)(f (t) − f2,δ (t)), •)C[a1 ,b1 ]
+ Bn(r) ((1 − ψ(t))(f (t) − f2,δ (t)), •)C[a1 ,b1 ]
=: H4 + H5 .
Using Lemma 2.7, it is clear that
Bn(r) ψ(t)(f (t) − f2,δ (t)), x
∞
(n + r − 1)!(n − r)! X
pn+r,k (x)
=
n!(n − 1)!
k=0
Hence
Z
∞
(r)
bn−r,k+r (t)ψ(t)(f (r) (t) − f2,δ (t))dt.
0
(r)
(r) (r)
c8 Bn (ψ(t)(f (t) − f2,δ (t)), •)
≤
M
f
−
f
2,δ C[a1 ,b1 ]
C[a∗ ,b∗ ]
.
Next for x ∈ [a1 , b1 ] and t ∈ [0, ∞) \ [a∗ , b∗ ], we choose a δ1 > 0 satisfying |t − x| ≥ δ1 .
Therefore by Lemma 2.5 and the Cauchy-Schwarz inequality, we have
I ≡ Bn(r) ((1 − ψ(t))(f (t) − f2,δ (t), x)
and
Z ∞
∞
|Qi,j,r (x)| X
j
n
|I| ≤
pn,k (x)|k − nx|
bn,k (t)(1 − ψ(t))|f (t) − f2,δ (t)|dt
{x(1 + x)}r k=1
0
2i+j≤r
X
i
i,j≥0
+
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
(n + r − 1)!
(1 + x)−n−r (1 − ψ(0))|f (0) − f2,δ (0)|.
(n − 1)!
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V IJAY G UPTA , M UHAMMAD A SLAM N OOR , M AN S INGH B ENIWAL ,
AND
M. K. G UPTA
For sufficiently large n, the second term tends to zero. Thus
c9 ||f ||γ
|I| ≤ M
X
n
2i+j≤r
i,j≥0
≤
c9 ||f ||γ δ1−2m
M
i
∞
X
j
pn,k (x)|k − nx|
n
i
2i+j≤r
i,j≥0
c9 ||f ||γ δ −2m
≤M
1
X
2i+j≤r
i,j≥0
bn,k (t)dt
|t−x|≥δ1
k=1
X
Z
ni
∞
X
j
Z
pn,k (x)|k − nx|
∞
12 Z
bn,k (t)dt
k=1
0
(∞
X
) 12 ( ∞
X
pn,k (x)(k − nx)2j
k=1
∞
4m
bn,k (t)(t − x)
12
dt
0
Z
pn,k (x)
k=1
) 12
∞
bn,k (t)(t − x)4m dt
0
Hence by using Lemma 2.1 and Lemma 2.2, we have
(i+ 2j −m)
−2m
c
c11 n−q ||f ||γ ,
I ≤ M10 ||f ||γ δ1 O n
≤M
c11 n−1 kf kγ .
where q = m − 2r . Now choosing m > 0 satisfying q ≥ 1, we obtain I ≤ M
Therefore by property (iii) of the function f2,δ (t), we get
(r) c8 H1 ≤ M
f (r) − f2,δ C[a∗ ,b∗ ]
c11 n−1 ||f ||γ
+M
c12 ω2 (f (r) , δ, a, b) + M
c11 n−1 ||f |kγ .
≤M
1
Choosing δ = n− 2 , the theorem follows.
4. I NVERSE T HEOREM
This section is devoted to the following inverse theorem in simultaneous approximation:
Theorem 4.1. Let 0 < α < 2, 0 < a1 < a2 < b2 < b1 < ∞ and suppose f ∈ Cγ [0, ∞). Then
in the following statements (i) ⇒ (ii)
(r)
(i) ||Bn (f, •)||C[a1 ,b1 ] = O(n−α/2 ),
(ii) f (r) ∈ Lip∗ (α, a2 , b2 ),
where Lip∗ (α, a2 , b2 ) denotes the Zygmund class satisfying ω2 (f, δ, a2 , b2 ) ≤ M δ α .
Proof. Let us choose a0 , a00 , b0 , b00 in such a way that a1 < a0 < a00 < a2 < b2 < b00 < b0 < b1 .
Also suppose g ∈ C0∞ with suppg ∈ [a00 , b00 ] and g(x) = 1 on the interval [a2 , b2 ]. For x ∈ [a0 , b0 ]
d
, we have
with D ≡ dx
Bn(r) (f g, x) − (f g)(r) (x)
= Dr (Bn ((f g)(t) − (f g)(x)), x)
= Dr (Bn (f (t)(g(t) − g(x)), x)) + Dr (Bn (g(x)(f (t) − f (x)), x))
=: J1 + J2 .
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.
O N S IMULTANEOUS A PPROXIMATION F OR C ERTAIN BASKAKOV D URRMEYER T YPE O PERATORS
13
Using the Leibniz formula, we have
Z ∞
∂r
J1 =
Wn (x, t)f (t)[g(t) − g(x)]dt
∂xr 0
r Z ∞
X
r
∂ r−i
=
Wn(i) (x, t) r−i [f (t)(g(t) − g(x))]dt
∂x
i 0
i=0
Z ∞
r−1 X
r (r−i)
(i)
=−
g
(x)Bn (f, x) +
Wn(r) (x, t)f (t)(g(t) − g(x))dt
i
0
i=0
=: J3 + J4 .
Applying Theorem 3.4, we have
J3 = −
r−1 X
r
i
i=0
α
g (r−i) (x)f (i) (x) + O n− 2 ,
uniformly in x ∈ [a0 , b0 ]. Applying Theorem 3.2, the Cauchy-Schwarz inequality, Taylor’s expansions of f and g and Lemma 2.2, we are led to
J4 =
=
r
X
g (i) (x)f (r−i) (x)
i=0
r X
i=0
1
r! + o n− 2
i!(r − i)!
α
r (i)
g (x)f (r−i) (x) + o n− 2 ,
i
uniformly in x ∈ [a0 , b0 ]. Again using the Leibniz formula, we have
J2 =
r Z
X
r
i=0
r X
i
0
∞
Wn(i) (x, t)
∂ r−i
[g(t)(f (t) − f (x))]dt
∂xr−i
r (r−i)
=
g
(x)Bn(i) (f, x) − (f g)(r) (x)
i
i=0
r X
r (r−i)
=
g
(x)f (i) (x) − (f g)(r) (x) + o(n−α/2 )
i
i=0
α
= O n− 2 ,
uniformly in x ∈ [a0 , b0 ]. Combining the above estimates, we get
(r)
Bn (f g, •) − (f g)(r) 0 0 = O n− α2 .
C[a ,b ]
Thus by Lemma 2.5 and Lemma 2.6, we have (f g)(r) ∈ Lip∗ (α, a0 , b0 ) also g(x) = 1 on
the interval [a2 , b2 ], it proves that f (r) ∈ Lip∗ (α, a2 , b2 ). This completes the validity of the
implication (i) ⇒ (ii) for the case 0 < α ≤ 1. To prove the result for 1 < α < 2 for any
interval [a∗ , b∗ ] ⊂ (a1 , b1 ), let a∗2 , b∗2 be such that (a2 , b2 ) ⊂ (a∗2 , b∗2 ) and (a∗2 , b∗2 ) ⊂ (a∗1 , b∗1 ).
Letting δ > 0 we shall prove the assertion α < 2. From the previous case it implies that f (r)
exists and belongs to Lip(1 − δ, a∗1 , b∗1 ). Let g ∈ C0∞ be such that g(x) = 1 on [a2 , b2 ] and
supp g ⊂ (a∗2 , b∗2 ). Then with χ2 (t) denoting the characteristic function of the interval [a∗1 , b∗1 ],
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
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14
V IJAY G UPTA , M UHAMMAD A SLAM N OOR , M AN S INGH B ENIWAL ,
AND
M. K. G UPTA
we have
(r)
Bn (f g, •) − (f g)(r) ∗ ∗
C[a ,b ]
2
2
≤ ||Dr [Bn (g(·)(f (t) − f (·)), •)]||C[a∗2 ,b∗2 ] + ||Dr [Bn (f (t)(g(t) − g(·)), •)]||C[a∗2 ,b∗2 ]
=: P1 + P2 .
To estimate P1 , by Theorem 3.4, we have
P1 = Dr [Bn (g(·)(f (t), •)] − (f g)(r) C[a∗ ,b∗ ]
2 2
r X
r
=
g (r−i) (·)Bn(i) (f, •) − (f g)(r) ∗ ∗
i
i=0
C[a2 ,b2 ]
r
X r g (r−i) (·)f (i) − (f g)(r) + O(n−α/2 )
=
∗ ∗
i
i=0
C[a2 ,b2 ]
−α
=O n 2 .
Also by the Leibniz formula and Theorem 3.2, have
r X
r (r−i)
(r)
P2 ≤ g
(·)Bn (f, •) + Bn (f (t)(g(t) − g(·))χ2 (t), •)
i
i=0
+ O(n−1 )
C[a∗2 ,b∗2 ]
=: ||P3 + P4 ||C[a∗2 ,b∗2 ] + O(n−1 ).
Then by Theorem 3.4, we have
P3 =
r−1 X
r
i
i=0
α
g (r−i) (x)f (i) (x) + O n− 2 ,
uniformly in x ∈ [a∗2 , b∗2 ]. Applying Taylor’s expansion of f, we have
Z
∞
Wn(r) (x, t)[f (t)(g(t) − g(x))χ2 (t)dt
P4 =
i=0
=
Z
r
X
f (i) (x)
i!
i=0
Z
+
∞
Wn(r) (x, t)(t − x)i (g(t) − g(x))dt
0
∞
Wn(r) (x, t)
0
(f (r) (ξ) − f (r) (x))
(t − x)r (g(t) − g(x))χ2 (t)dt,
r!
where ξ lying between t and x. Next by Theorem 3.4, the first term in the above expression is
given by
r X
r
m=0
m
J. Inequal. Pure and Appl. Math., 7(4) Art. 125, 2006
α
g (m) f (r−m) (x) + O n− 2 ,
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O N S IMULTANEOUS A PPROXIMATION F OR C ERTAIN BASKAKOV D URRMEYER T YPE O PERATORS
15
uniformly in x ∈ [a∗2 , b∗2 ]. Also by mean value theorem and using Lemma 2.5, we can obtain the
second term as follows:
Z ∞
(r)
(r)
(f
(ξ)
−
f
(x))
(r)
r
W
(x,
t)
(t
−
x)
(g(t)
−
g(x))χ
(t)dt
2
n
∗ ∗
r!
0
C[a2 ,b2 ]
rZ ∞
(r)
X
(ξ) − f (r) (x)| 0
m+s |Qm,s,r (x)|
δ+r+1 |f
≤
n
W
(x,
t)|t
−
x|
|g
(η)|χ
(t)dt
n
2
x(1 + x)
∗ ∗
r!
0
2m+s≤r
m,s≥0
C[a2 ,b2 ]
δ
= O n− 2 ,
choosing δ such that 0 ≤ δ ≤ 2 − α. Combining the above estimates we get
(r)
Bn (f g, •) − (f g)(r) ∗ ∗ = O n− α2 .
C[a2 ,b2 ]
(a∗2 , b∗2 ),
Since suppf g ⊂
it follows from Lemma 2.5 and Lemma 2.6 that (f g)(r) ∈
Liz(α, 1, a∗2 , b∗2 ). Since g(x) = 1 on [a2 , b2 ], we have f (r) ∈ Liz(α, 1, a∗2 , b∗2 ). This completes
the proof of the theorem.
Remark 4.2. As noted in the first section, these operators also reproduce the linear functions
so we can easily apply the iterative combinations to the operators (1.1) to improve the order of
approximation.
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[3] G. FREUD AND V. POPOV, On approximation by Spline functions, Proceeding Conference on
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[4] S. GOLDBERG AND V. MEIR, Minimum moduli of ordinary differential operators, Proc. London
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[5] V. GUPTA, A note on modified Baskakov type operators, Approximation Theory Appl., 10(3) (1994),
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[6] V. GUPTA, M.K. GUPTA AND V. VASISHTHA, Simultaneous approximation by summationintegral type operators, Nonlinear Funct. Anal. Appl., 8(3), (2003), 399–412.
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