EQUIVALENT THEOREM ON LUPAŞ OPERATORS
NAOKANT DEO
Department of Applied Mathematics
Delhi College of Engineering
Bawana Road, Delhi-110042, India.
EMail: dr_naokant_deo@yahoo.com
Received:
10 January, 2007
Accepted:
22 November, 2007
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
41A36, 41A45.
Key words:
Abstract:
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
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Contents
Acknowledgements:
Dedicatory:
JJ
II
Lupaş operators, Linear combinations.
J
I
The purpose of this present paper is to give an equivalent theorem on Lupaş operators with ωφr λ (f, t), where ωφr λ (f, t) is Ditzian-Totik modulus of smoothness
for linear combination of Lupaş operators.
Page 1 of 18
This research is supported by UNESCO (CAS-TWAS postdoctoral fellowship).
The author is extremely thankful to the referee for his valuable comments, leading to a better presentation of the paper.
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In memory of Professor Alexandru Lupaş.
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Contents
1
Introduction
3
2
Basic Results
6
3
Direct Results
12
4
Inverse Results
15
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
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1.
Introduction
Let C [0, +∞) be the set of continuous and bounded functions defined on [0, +∞) .
For f ∈ C [0, +∞) and n ∈ N, the Lupaş operators are defined as
+∞
X
k
pn,k (x)f
Bn (f, x) =
,
n
k=0
(1.1)
where
Naokant Deo
pn,k (x) =
n+k−1
k
k
x
.
(1 + x)n+k
Since the Lupaş operators cannot be used for the investigation of higher orders of
smoothness, we consider combinations of these operators, which have higher orders
of approximation. The linear combinations of Lupaş operators on C[0, +∞) are
defined as (see [3, p.116])
Bn,r (f, x) =
(1.2)
r−1
X
Ci (n)Bni (f, x) ,
i=0
where ni and Ci (n) satisfy:
(i) n = n0 < · · · < nr−1 ≤ An ;
(ii)
Equivalent Theorem on Lupaş
Operators
r−1
P
Ci (n) ≤ C;
i=0
(iii) Bn,r (1, x) = 1;
k
(iv) Bn,r (t − x) , x = 0; k = 1, 2, . . . , r − 1,
r ∈ N,
vol. 8, iss. 4, art. 114, 2007
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where constants A and C are independent of n.
Ditzian [1] used ωφ2 λ (f, t) (0 ≤ λ ≤ 1) and studied an interesting direct result
for Bernstein polynomials and unified the results with ω 2 (f, t) and ωφ2 (f, t). In [2],
ωφr λ (f, t) was also used for polynomial approximation.
To state our results, we give some notations (c.f. [4]). Let C [0, +∞) be the set of
continuous and bounded functions on [0, +∞) . Our modulus of smoothness is given
by
r
(1.3)
ωφr λ (f, t) = sup
sup
4hφλ (x) f (x) ,
0<h≤t x±(rhφλ (x)/2)∈[0,+∞)
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
where
41h f (x)
=f
h
x−
2
−f
h
x+
2
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,
k+1
4
1
k
= 4 (4 ),
k∈N
and our K−functional by
(1.4)
(1.5)
n
o
Kφλ (f, tr ) = inf kf − gkC[0,+∞) + tr φrλ g (r) C[0,+∞) ,
n
K̃φλ (f, tr ) = inf kf − gkC[0,+∞) + tr φrλ g (r) C[0,+∞)
o
r/(1−λ/2) (r) +t
g
,
C[0,+∞)
where
the infimum is taken on the functions satisfying g (r−1) ∈ A · Cloc , φ(x) =
p
x(1 + x) and 0 ≤ λ ≤ 1. It is well known (see [1]) that
(1.6)
ωφr λ (f, t) ∼ Kφλ (f, tr ) ∼ K̃φλ (f, tr ),
(x ∼ y means that there exists c > 0 such that c−1 y ≤ x ≤ cy).
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To prove the inverse, we need the following notations. Let us denote
C0 := {f ∈ C [0, +∞) , f (0) = 0} ,
f := sup δnα(λ−1) (x)f (x) ,
0
x∈(0,+∞)
0
Cλ := f ∈ C0 : f 0 < +∞ ,
f := sup δnr+α(λ−1) (x)f (r) (x) ,
r
x∈(0,+∞)
:= f ∈ C0 : f (r−1) ∈ A · Cloc , f r < +∞ ,
n
o
1
1
√
√
where δn (x) = φ(x) + n ∼ max φ(x), n , 0 ≤ λ ≤ 1, r ∈ N and 0 < α < r.
Now, we state the main results.
If f ∈ C [0, +∞) , r ∈ N, 0 < α < r, 0 ≤ λ ≤ 1, then the following statements
are equivalent
− 12 1−λ
α
(1.7)
|Bn,r (f, x) − f (x)| =O (n δn (x)) ,
Cλr
ωφr λ (f, t) =O(tα ),
n
o
where δn (x) = φ(x) + √1n ∼ max φ(x), √1n .
In this paper, we will consider the operators (1.2) and obtain an equivalent approximation theorem for these operators.
Throughout this paper C denotes a constant independent of n and x. It is not
necessarily the same at each occurrence.
(1.8)
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
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2.
Basic Results
In this section, we mention some basis results, which will be used to prove the main
results.
If f ∈ C [0, +∞) , r ∈ N, then by [3] we know that
+∞
(n + r − 1)! X
k
r
(r)
Bn (f, x) =
pn+r,k (x)4n−1 f
.
(n − 1)! k=0
n
Lemma 2.1. Let f
(2.1)
(r)
∈ C [0, +∞) and 0 ≤ λ ≤ 1, then
rλ
φ (x)Bn(r) (f, x) ≤ C φrλ f (r) .
∞
Proof. In [3, Sec.9.7], we have
Z r/n (r) k
k
−r+1
r
(2.2)
4n−1 f
≤ Cn
f
+ u du,
n
n
0
Z r/n
rλ/2 (r)
u
(2.3)
4rn−1 f (0) ≤ Cn−r(2−λ)/2
f (u) du.
0
Using (2.2), (2.3) and the Hölder inequality, we get
+∞
X
rλ
(n
+
r
−
1)!
k φ (x)Bn(r) (f, x) ≤ φrλ (x)
pn+r,k (x)4rn−1 f
(n − 1)! k=0
n (n + r − 1)! rλ
≤
φ (x)pn+r,0 (x)4rn−1 f (0)
(n − 1)! +∞
X
k +
φrλ (x)pn+r,k (x)4rn−1 f
n k=1
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
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≤ C φrλ (x)f (r) ∞ .
Lemma 2.2. Let f ∈ C [0, +∞) , r ∈ N and 0 ≤ λ ≤ 1, then for n > r, we get
rλ
φ (x)Bn(r) (f, x) ≤ Cnr/2 δn−r(1−λ) (x)f .
∞
Proof. We consider x ∈ [0, 1/n]. Then we have δn (x) ∼ √1n , φ(x) ≤
d
pn,k (x) = n pn+1,k−1 (x) − pn+1,k (x)
dx
and
Z
0
∞
√2 .
n
Using
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
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1
pn,k (x)dx =
,
n−1
Contents
we have
rλ
φ (x)Bn(r) (f, x) ≤ Cnr/2 δn−r(1−λ) (x)f .
∞
Now we consider the interval x ∈ (1/n, +∞), then φ(x) ∼ δn (x). Using Lemma
4.5 in [6], we get
r
φ (x)Bn(r) (f, x) ≤ Cnr/2 f .
(2.4)
∞
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Therefore
rλ
φ (x)Bn(r) (f, x) = φr(λ−1) (x) φr (x)Bn(r) (f, x)
≤ Cφr(λ−1) (x)nr/2 f ∞
≤ Cδn−r(λ−1) (x)nr/2 f .
∞
Close
Lemma 2.3. Let r ∈ N, 0 ≤ β ≤ r, x ± rt/2 ∈ I and 0 ≤ t ≤ 1/8r, then we have
the following inequality:
!
Z t/2
Z t/2
r
X
(2.5)
···
δn−β x +
uj du1 . . . dur ≤ C(β)tr δn−β (x).
−t/2
−t/2
j=1
Proof. The result follows from [7, (4.11)].
Lemma 2.4. For x, t, u ∈ (0, +∞) , x < u < t, t, r ∈ N and λ ∈ [0, 1], then
Z t
r−1 −rλ
(2.6)
Bn t−u
φ (u)du , x ≤ Cn−r/2 δnr (x)φ−rλ (x).
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
x
Proof. When r = 1, then we have
Z t
−λ/2
−λ
(2.7)
(1 + t)−λ/2 + (1 + x)−λ/2 t−λ/2 .
φ (u)du ≤ t − x x
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x
From [3, (9.5.3)],
(2.8)
2r
−r 2r
Bn (t − x) , x ≤ Cn φ (x).
Applying the Hölder inequality, we get
Z t
−λ
(2.9)
Bn φ (u)du , x
x
3
1 h
≤ Bn (t − x)4 , x 4 x−λ/2 Bn (1 + t)−2λ/3 , x 4
3 i
+(1 + x)−λ/2 Bn (t−2λ/3 , x) 4
h
3
≤ Cn−1/2 δn (x) x−λ/2 Bn (1 + t)−2λ/3 , x 4
3 i
+ (1 + x)−λ/2 Bn (t−2λ/3 , x) 4 .
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Applying the Hölder inequality, we get
(2.10)
−1 ! 3λ2
k
pn,k (x)
≤ Cx−2λ/3 .
n
k=0
+∞
X
Bn (t−2λ/3 , x) ≤
Similarly,
−2λ/3
Bn ((1 + t)
(2.11)
−2λ/3
, x) ≤ C(1 + x)
.
Equivalent Theorem on Lupaş
Operators
Naokant Deo
Combining (2.9) to (2.11), we obtain (2.6).
When r > 1, then we have
t − x2
t − u2
≤ 2
(Trivial for t < u < x),
φ2 (u)
φ (x)
otherwise
t − ux ≤ t − xu
and
u t − u
φ2 (u)
≤
xt − x
φ2 (x)
(f or t < u < x).
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Thus
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t − ur−2
(2.12)
φ(r−2)λ (u)
≤
t − xr−2
φ(r−2)λ (x)
,
r>2
because
(2.13)
vol. 8, iss. 4, art. 114, 2007
t − u
t − x 1
1
≤
+
φ2 (u)
x
1+x 1+t
(Trivial for t < u < x),
Close
otherwise
(u − t)
(x − t)
≤
u
t
and
(2.14)
1
1
1
≤
+
,
1+u
1+x 1+t
λ
t − x 1
1
≤
+
.
φ2λ (u)
xλ
1+x 1+t
t − u
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
λ
Because the function t (0 ≤ λ ≤ 1) is subadditive, using (2.12) and (2.14), we
obtain
(
λ λ )
t − ur−1
t − xr−1
1
1
.
(2.15)
≤ (r−2)λ
+
φrλ (u)
φ
(x)xλ
1+x
1+t
Since Bn (1 + t)−r , x) ≤ C(1 + x)−r (r ∈ N, x ∈ [0, +∞)), using the Hölder
inequality, we have
(2.16)
Bn (1 + t)−2λ , x) ≤ C(1 + x)−2λ
x ∈ [0, +∞) , λ ∈ [0, 1] .
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Using (2.8), (2.15), (2.16) and the Hölder inequality, we get
Z t
r−1 −rλ
2r Bn t − u φ (u)du , x ≤ Bn1/2 t − u , x
x
. φ−rλ (x) + x−λ φ−(r−2)λ (x)Bn1/2 (1 + t)−2λ , x
≤ Cn−r/2 δnr (x)φ−rλ (x).
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Lemma 2.5. If r ∈ N and 0 < α < r then
Bn f ≤Cnr/2 f (2.17)
0
r
Bn f ≤C f (2.18)
r
r
(f ∈ Cλ0 ),
(f ∈ Cλr ).
Proof. For x ∈ [0, n1 ] and δn (x) ∼ √1n , according to Lemma 2.2, we get
r+α(λ−1)
δn
(x)Bn(r) (f, x) ≤ Cnr/2 f 0 .
On the other hand, for x ∈ n1 , +∞ and δn (x) ∼ φ(x), according to [3], we can
obtain
i
r
+∞
X
iX
k
k
(r)
−2r
2
r
(2.19) Bn (f, x) = φ
Vi nφ (x) n
pn,k (x)
− x 41/n f
,
n
n
i=0
k=0
2
2
where Vi nφ (x) is polynomial in nφ (x) of degree (r − i)/2 with constant coefficients and therefore,
(r+i)/2
−2r
n
1
i
(2.20)
φ Vi (nx)n ≤ C
, for any x ∈
, +∞ .
φ2 (x)
n
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
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Since
k r
(2.21)
41/n f
≤ C f 0 φα(λ−1) (x),
n
using the Hölder inequality, we get
+∞
i
2 i/2
X
α(λ−1)
k φ (x)
k
r
f φ
− x 41/n f
(x).
(2.22) pn,k (x)
≤C
0
n
n n
k=0
From (2.19) to (2.22) we can deduce (2.17) easily. Similarly, like Lemma 2.1 we can
obtain (2.18).
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3.
Direct Results
Theorem 3.1. Let f ∈ C [0, +∞) , r ∈ N and 0 ≤ λ ≤ 1, then
Bn,r (f, x) − f (x) ≤ Cω r λ f, n−1/2 δn1−λ (x) .
(3.1)
φ
Proof. Using (1.5) and (1.6) and taking dn = dn (x, λ) = n−1/2 δn1−λ (x), we can
choose gn = gn,x,λ for fixed x and λ satisfying:
f − gn ≤Cω r λ (f, dn ),
(3.2)
φ
r rλ (r) (3.3)
dn φ gn ≤Cωφr λ (f, dn ),
(r) gn ≤Cω r λ (f, dn ).
(3.4)
dr/(1−(λ/2))
n
φ
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
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Now
(3.5)
Contents
Bn,r (f, x) − f (x) ≤ Bn,r (f, x) − Bn,r (gn , x) + f − gn (x)
+ Bn,r (gn , x) − gn (x)
≤ C f − gn + Bn,r (gn , x) − gn (x).
∞
By using Taylor’s formula:
1
Rr (gn , t, x) =
(r − 1)!
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(t − x)r−1 (r−1)
g
(x) + Rr (gn , t, x),
(r − 1)! n
gn (t) = gn (x) + (t − x)gn0 (x) + · · · +
where
JJ
Z
x
t
(t − u)r−1 gn(r) (u)du.
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Using (i)-(iv) of (1.2) and Lemma 2.4, we obtain
Bn,r (gn , x) − gn (x)
Z t
1
r−1 (r)
= Bn,r
(t − u) gn (u)du, x (r − 1)! x
Z t
r−1
rλ (r) |t
−
u|
,x
(3.6)
≤ C φ gn Bn,r du
φrλ (u)
x
≤ Cφ−rλ (x)n−r/2 δnr (x) φrλ gn(r) .
Again using (i)-(iv) of (1.2) and (2.12), we get
Bn,r (gn , x) − gn (x)
Z
rλ (r) ≤ C δn gn Bn,r
(3.7)
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
t
|t − u|r−1
du,
x
δnrλ (u)
x
1/2
≤ C δnrλ gn(r) nrλ/2 Bn,r (t − x)2r , x ≤ Cn−r/2 δnr (x)nrλ/2 δnrλ gn(r) .
We will take the following two cases:
Case-I: For x ∈ [0, 1/n], δn (x) ∼ √1n . Then by (3.2) – (3.5) and (3.7), we have
Bn,r (f, x) − f (x)
≤ C f − gn + drn δnrλ gn(r) (3.8)
≤ C f − gn + drn φrλ gn(r) + drn n−rλ/2 gn(r) (r) gn ≤ C f − gn + drn φrλ gn(r) + dr/(1−(λ/2))
(3.9)
n
≤ Cωφr λ (f, dn ) .
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Case-II: For x ∈ (1/n, +∞), δn (x) ∼ φ(x). Then by (3.2) – (3.3) and (3.5) –
(3.6), we get
Bn,r (f, x) − f (x) ≤ C f − gn + drn φrλ gn(r) ≤ Cω r λ (f, dn ) .
(3.10)
φ
Equivalent Theorem on Lupaş
Operators
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vol. 8, iss. 4, art. 114, 2007
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4.
Inverse Results
Theorem 4.1. Let f ∈ C [0, +∞) , r ∈ N, 0 < α < r, 0 ≤ λ ≤ 1. Then
Bn,r (f, x) − f (x) = O(dαn ).
(4.1)
implies
Equivalent Theorem on Lupaş
Operators
ωφr λ (f, t) = O(tα ),
(4.2)
Naokant Deo
where dn = n−1/2 δn1−λ (x).
vol. 8, iss. 4, art. 114, 2007
Proof. Since Bn (f, x) preserves the constant, hence we may assume f ∈ C0 . Suppose that (4.1) holds. Now we introduce a new K-functional as
Kλα (f, tr ) = infr f − g 0 + tr g r .
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g∈Cλ
Choosing g ∈
(4.3)
Cλr
such that
f − g + n−r/2 g ≤ 2Kλα (f, n−r/2 ).
0
r
By (4.1), we can deduce that
kBn,r (f, x) − f (x)k0 ≤ Cn
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−α/2
.
Thus, by using Lemma 2.5 and (4.3), we obtain
Kλα (f, tr ) ≤ f − Bn,r (f )0 + tr Bn,r (f )r
≤ Cn−α/2 + tr Bn,r (f − g)r + Bn,r (g)r
≤ C n−α/2 + tr nr/2 f − g 0 + g r
tr
≤ C n−α/2 + −r/2 Kλα f, n−r/2
n
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which implies that by [3, 7]
Kλα f, tr ≤ Ctα .
(4.4)
On the other hand, since x + (j − 2r )tφλ (x) ≥ 0, therefore
r λ tφ (x) ≤ x
j−
2
and
Equivalent Theorem on Lupaş
Operators
Naokant Deo
r
x+ j−
tφλ (x) ≤ 2x,
2
vol. 8, iss. 4, art. 114, 2007
so that
δn
(4.5)
r λ
x+ j−
tφ (x) ≤ 2δn (x).
2
Thus, for f ∈ Cλ0 , we get
(4.6)
r
4
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≤ f f
(x)
λ
tφ (x)
0
r X
r
δnα(1−λ)
j
j=0
2r α(1−λ)
≤ 2 δn
(x)f 0 .
!
r λ
x+ j−
tφ (x)
2
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From Lemma 2.3, for g ∈ Cλr , 0 < tφλ (x) < 1/8r, x ± rtφλ (x)/2 ∈ [0, +∞) ,
we have
r
4tφλ (x) g(x)
Z
!
Z (t/2)φλ (x)
r
(t/2)φλ (x)
X
≤
···
g (r) x +
uj du1 . . . dur −(t/2)φλ (x)
−(t/2)φλ (x)
j=1
Close
(4.7)
(4.8)
≤ g r
≤
Z
(t/2)φλ (x)
Z
(t/2)φλ (x)
···
−(t/2)φλ (x)
−(t/2)φλ (x)
δn−r+α(1−λ)
x+
r
X
!
uj
du1 . . . dur
j=1
Ctr δn−r+α(1−λ) (x)g r .v
Using (4.4), (4.6) and (4.7), for 0 < tφλ (x) < 1/8r, x ± rtφλ (x)/2 ∈ [0, +∞)
and choosing appropriate g, we get
r
4 λ f (x) ≤ 4r λ (f − g)(x) + 4r λ g(x)
tφ (x)
tφ (x)
tφ (x)
n
o
≤ Cδnα(1−λ) (x) f − g 0 + tr δnr(λ−1) (x)g r
!
r
t
≤ Cδnα(1−λ) (x)Kλα f, r(1−λ)
δn
(x)
r
≤ Ct .
Remark 1. Very recently Gupta and Deo [5] have studied two dimensional modified
Lupaş operators. In the same manner we can obtain an equivalent theorem with
ωφr λ (f, t).
Equivalent Theorem on Lupaş
Operators
Naokant Deo
vol. 8, iss. 4, art. 114, 2007
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References
[1] Z. DITZIAN, Direct estimate for Bernstein polynomials, J. Approx. Theory, 79
(1994), 165–166.
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