Volume 8 (2007), Issue 4, Article 114, 10 pp.
EQUIVALENT THEOREM ON LUPAŞ OPERATORS
NAOKANT DEO
D EPARTMENT OF A PPLIED M ATHEMATICS
D ELHI C OLLEGE OF E NGINEERING
BAWANA ROAD , D ELHI -110042, I NDIA .
dr_naokant_deo@yahoo.com
Received 10 January, 2007; accepted 22 November, 2007
Communicated by S.S. Dragomir
In memory of Professor Alexandru Lupaş.
A BSTRACT. 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.
Key words and phrases: Lupaş operators, Linear combinations.
2000 Mathematics Subject Classification. 41A36, 41A45.
1. I NTRODUCTION
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
(1.1)
Bn (f, x) =
,
pn,k (x)f
n
k=0
where
pn,k (x) =
n+k−1
k
xk
.
(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
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.
016-07
2
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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) ,
r ∈ N,
i=0
where ni and Ci (n) satisfy:
(i) n = n0 < · · · < nr−1 ≤ An ;
r−1
P
Ci (n) ≤ C;
(ii)
i=0
(iii) Bn,r (1,
x) = 1; (iv) Bn,r (t − x)k , x = 0; k = 1, 2, . . . , r − 1,
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,+∞)
where
41h f (x)
=f
h
x−
2
−f
h
x+
2
,
4k+1 = 41 (4k ),
k∈N
and our K−functional by
(1.4)
(1.5)
n
o
r rλ (r) Kφλ (f, t ) = inf kf − gkC[0,+∞) + t φ g
,
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,+∞)
r
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
ωφr λ (f, t) ∼ Kφλ (f, tr ) ∼ K̃φλ (f, tr ),
(1.6)
(x ∼ y means that there exists c > 0 such that c−1 y ≤ x ≤ cy).
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,+∞)
:= f ∈ C0 : f 0 < +∞ ,
f := sup δnr+α(λ−1) (x)f (r) (x) ,
r
Cλ0
x∈(0,+∞)
Cλr
:= f ∈ C0 : f (r−1) ∈ A · Cloc , f r < +∞ ,
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 114, 10 pp.
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E QUIVALENT T HEOREM ON L UPA Ş O PERATORS
3
o
n
where δn (x) = φ(x) + √1n ∼ max φ(x), √1n , 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
1
|Bn,r (f, x) − f (x)| =O (n− 2 δn1−λ (x))α ,
(1.7)
(1.8)
where δn (x) = φ(x) +
√1
n
ωφr λ (f, t) =O(tα ),
n
o
∼ 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.
2. BASIC R ESULTS
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
+∞
Bn(r) (f, x)
(n + r − 1)! X
pn+r,k (x)4rn−1 f
=
(n − 1)! k=0
k
.
n
Lemma 2.1. Let f (r) ∈ C [0, +∞) and 0 ≤ λ ≤ 1, then
rλ
φ (x)Bn(r) (f, x) ≤ C φrλ f (r) .
(2.1)
∞
Proof. In [3, Sec.9.7], we have
Z r/n (r) k
k
r
−r+1
f
du,
(2.2)
4n−1 f
≤ Cn
+
u
n
n
0
Z r/n
rλ/2 (r)
u
(2.3)
4r −1 f (0) ≤ Cn−r(2−λ)/2
f (u) du.
n
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
≤ C φrλ (x)f (r) ∞ .
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 114, 10 pp.
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4
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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) ≤ √2n . Using
d
pn,k (x) = n pn+1,k−1 (x) − pn+1,k (x)
dx
and
∞
Z
pn,k (x)dx =
0
1
,
n−1
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)
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 ∞ .
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
(2.5)
···
−t/2
t/2
−t/2
δn−β
x+
r
X
!
uj
du1 . . . dur ≤ C(β)tr δn−β (x).
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).
x
Proof. When r = 1, then we have
Z t
−λ
≤ t − x x−λ/2 (1 + t)−λ/2 + (1 + x)−λ/2 t−λ/2 .
(2.7)
φ
(u)du
x
From [3, (9.5.3)],
(2.8)
Bn (t − x)2r , x ≤ Cn−r φ2r (x).
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E QUIVALENT T HEOREM ON L UPA Ş O PERATORS
5
Applying the Hölder inequality, we get
Z t
14 h −λ/2 3
−λ
4
Bn φ (u)du , x ≤ Bn (t − x) , x
x
Bn (1 + t)−2λ/3 , x 4
(2.9)
x
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 .
Applying the Hölder inequality, we get
(2.10)
Bn (t−2λ/3 , x) ≤
−1 ! 3λ2
k
pn,k (x)
≤ Cx−2λ/3 .
n
k=0
+∞
X
Similarly,
Bn ((1 + t)−2λ/3 , x) ≤ C(1 + x)−2λ/3 .
(2.11)
Combining (2.9) to (2.11), we obtain (2.6).
When r > 1, then we have
t − u 2
t − x2
≤ 2
(Trivial for t < u < x),
φ2 (u)
φ (x)
otherwise
t − u x ≤ t − x u
and
u t − u xt − x
≤ 2
(f or t < u < x).
φ2 (u)
φ (x)
Thus
t − ur−2
(2.12)
φ(r−2)λ (u)
≤
t − xr−2
φ(r−2)λ (x)
,
r>2
because
(2.13)
t − u
t − x 1
1
≤
+
φ2 (u)
x
1+x 1+t
otherwise
(u − t)
(x − t)
≤
u
t
and
(2.14)
(Trivial for t < u < x),
1
1
1
≤
+
,
1+u
1+x 1+t
λ
t − x 1
1
≤
+
.
φ2λ (u)
xλ
1+x 1+t
t − u
Because the function tλ (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
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6
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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] .
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
−2λ
−rλ
−λ −(r−2)λ
1/2
. φ (x) + x φ
(x)Bn (1 + t) , x
≤ Cn−r/2 δnr (x)φ−rλ (x).
Lemma 2.5. If r ∈ N and 0 < α < r then
Bn f ≤Cnr/2 f r
0
Bn f ≤C f r
r
(2.17)
(2.18)
Proof. For x ∈ [0, n1 ] and δn (x) ∼
√1 ,
n
(f ∈ Cλ0 ),
(f ∈ Cλr ).
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 ∈
(2.19)
Bn(r) (f, x)
1
, +∞
n
−2r
=φ
r
X
and δn (x) ∼ φ(x), according to [3], we can obtain
+∞
iX
Vi nφ (x) n
pn,k (x)
i=0
2
k=0
k
−x
n
i
4r1/n f
k
,
n
where Vi nφ2 (x) is polynomial in nφ2 (x) of degree (r − i)/2 with constant coefficients and
therefore,
(r+i)/2
−2r
n
1
i
φ Vi (nx)n ≤ C
(2.20)
, for any x ∈
, +∞ .
φ2 (x)
n
Since
(2.21)
k r
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
k φ (x)
r
f φ
(2.22)
pn,k (x)
− x 41/n f
(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).
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 114, 10 pp.
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E QUIVALENT T HEOREM ON L UPA Ş O PERATORS
7
3. D IRECT R ESULTS
Theorem 3.1. Let f ∈ C [0, +∞) , r ∈ N and 0 ≤ λ ≤ 1, then
(3.1)
Bn,r (f, x) − f (x) ≤ Cω r λ f, n−1/2 δn1−λ (x) .
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 ),
φ
r rλ (r) dn φ gn ≤Cωφr λ (f, dn ),
(r) gn ≤Cω r λ (f, dn ).
dr/(1−(λ/2))
n
φ
(3.2)
(3.3)
(3.4)
Now
(3.5)
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:
(t − x)r−1 (r−1)
g
(x) + Rr (gn , t, x),
(r − 1)! n
gn (t) = gn (x) + (t − x)gn0 (x) + · · · +
where
1
Rr (gn , t, x) =
(r − 1)!
Z
t
(t − u)r−1 gn(r) (u)du.
x
Using (i)-(iv) of (1.2) and Lemma 2.4, we obtain
Z t
1
r−1 (r)
Bn,r (gn , x) − gn (x) = 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
(3.7)
Z t
r−1
rλ (r) |t
−
u|
Bn,r (gn , x) − gn (x) ≤ C δn gn Bn,r
du, x rλ
δn (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:
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8
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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 (3.9)
≤ C f − gn + drn φrλ gn(r) + dr/(1−(λ/2))
n
≤ Cωφr λ (f, dn ) .
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
4. I NVERSE R ESULTS
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
ωφr λ (f, t) = O(tα ),
(4.2)
where dn = n−1/2 δn1−λ (x).
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 .
g∈Cλ
Choosing g ∈ Cλr such that
f − g + n−r/2 g ≤ 2Kλα (f, n−r/2 ).
(4.3)
r
0
By (4.1), we can deduce that
kBn,r (f, x) − f (x)k0 ≤ Cn−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
which implies that by [3, 7]
(4.4)
Kλα f, tr ≤ Ctα .
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E QUIVALENT T HEOREM ON L UPA Ş O PERATORS
9
On the other hand, since x + (j − 2r )tφλ (x) ≥ 0, therefore
r λ tφ (x) ≤ x
j−
2
and
r
x+ j−
tφλ (x) ≤ 2x,
2
so that
r λ
(4.5)
δn x + j −
tφ (x) ≤ 2δn (x).
2
Thus, for f ∈ Cλ0 , we get
(4.6)
r
4
tφλ (x)
f (x) ≤ f r X
r
δnα(1−λ)
j
j=0
2r α(1−λ)
≤ 2 δn
(x)f 0 .
0
!
r λ
x+ j−
tφ (x)
2
From Lemma 2.3, for g ∈ Cλr , 0 < tφλ (x) < 1/8r, x ± rtφλ (x)/2 ∈ [0, +∞) , we have
!
Z (t/2)φλ (x)
r
Z (t/2)φλ (x)
X
r
(r)
···
g
x+
uj du1 . . . dur 4tφλ (x) g(x) ≤ −(t/2)φλ (x)
−(t/2)φλ (x)
j=1
!
Z (t/2)φλ (x)
Z (t/2)φλ (x)
r
X
(4.7)
uj du1 . . . dur
≤ g ···
δn−r+α(1−λ) x +
r
−(t/2)φλ (x)
−(t/2)φλ (x)
j=1
≤ Ctr δn−r+α(1−λ) (x)g r .
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
α(1−λ)
r r(λ−1)
≤ Cδn
(x) f − g 0 + t δn
(x)g r
!
r
t
≤ Cδnα(1−λ) (x)Kλα f, r(1−λ)
δn
(x)
r
≤ Ct .
Remark 4.2. 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).
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J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 114, 10 pp.
http://jipam.vu.edu.au/