Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 1, Article 21, 2005
DIRECT RESULTS FOR CERTAIN FAMILY OF INTEGRAL TYPE OPERATORS
VIJAY GUPTA AND OGÜN DOĞRU
S CHOOL OF A PPLIED S CIENCES
N ETAJI S UBHAS I NSTITUTE OF T ECHNOLOGY
S ECTOR 3 DWARKA , N EW D ELHI -110045, INDIA
vijay@nsit.ac.in
A NKARA U NIVERSITY
FACULTY OF S CIENCE
D EPARTMENT OF M ATHEMATICS
TANDO ĞAN , A NKARA -06100, TURKEY
dogru@science.ankara.edu.tr
Received 08 September, 2004; accepted 22 January, 2005
Communicated by D. Hinton
A BSTRACT. In the present paper we introduce a certain family of linear positive operators and
study some direct results which include a pointwise convergence, asymptotic formula and an
estimation of error in simultaneous approximation.
Key words and phrases: Linear positive operators, Simultaneous approximation, Steklov mean, Modulus of continuity.
2000 Mathematics Subject Classification. 41A25, 41A30.
1. I NTRODUCTION
We consider a certain family of integral type operators, which are defined as
Z ∞
∞
1X
(1.1)
Bn (f, x) =
pn,ν (x)
pn,ν−1 (t)f (t)dt + (1 + x)−n−1 f (0), x ∈ [0, ∞),
n ν=1
0
where
1
tν (1 + t)−n−ν−1
B(n, ν + 1)
with B(n, ν + 1) = ν!(n − 1)!/(n + ν)! the Beta function.
Alternatively the operators (1.1) may be written as
Z ∞
Bn (f, x) =
Wn (x, t)f (t)dt,
pn,ν (t) =
0
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
The authors are thankful to the referee for his/her useful recommendations and valuable remarks.
165-04
2
V IJAY G UPTA AND O GÜN D O ĞRU
where
∞
1X
Wn (x, t) =
pn,ν (x)pn,ν−1 (t) + (1 + x)−n−1 δ(t),
n ν=1
δ(t) being the Dirac delta function. The operators Bn are discretely defined linear positive
operators. It is easily verified that these operators reproduce only the constant functions. As far
as the degree of approximation is concerned these operators are very similar to the operators
considered by Srivastava and Gupta [5], but the approximation properties of these operators are
different. In this paper, we study some direct theorems in simultaneous approximation for the
operators (1.1).
2. AUXILIARY R ESULTS
In this section we mention some lemmas which are necessary to prove the main theorems.
Lemma 2.1 ([3]). For m ∈ N0 , if the m-th order moment is defined as
m
∞
ν
1X
Un,m (x) =
pn,ν (x)
−x
,
n ν=0
n+1
then
0
(n + 1)Un,m+1 (x) = x(1 + x) Un,m
(x) + mUn,m−1 (x) .
Consequently
Um,n (x) = O n−[(m+1)/2] .
Lemma 2.2. Let the function µn,m (x), n > m and m ∈ N0 , be defined as
Z ∞
∞
1X
µn,m (x) =
pn,ν (x)
pn,ν−1 (t)(t − x)m dt + (−x)m (1 + x)−n−1 .
n ν=1
0
Then
2x
2nx(1 + x) + 2x(1 + 4x)
, µn,2 (x) =
(n − 1)
(n − 1)(n − 2)
and there holds the recurrence relation
µn,0 (x) = 1, µn,1 (x) =
(n − m − 1)µn,m+1 (x)
= x(1 + x) µ0n,m (x) + 2mµn,m−1 (x) + [m(1 + 2x) + 2x] µn,m (x).
Consequently for each x ∈ [0, ∞), we have from this recurrence relation that
µn,m (x) = O n−[(m+1)/2] .
Proof. The values of µn,0 (x), µn,1 (x) and µn,2 (x) easily follow from the definition. We prove
the recurrence relation as follows
Z ∞
∞
1X
0
0
x(1 + x)µn,m (x) =
x(1 + x)pn,ν (x)
pn,ν−1 (t)(t − x)m dt
n ν=1
0
Z ∞
∞
1X
−m
x(1 + x)pn,ν (x)
pn,ν−1 (t)(t − x)m−1 dt
n ν=1
0
m
−n−2
− (n + 1)(−x) (1 + x)
+ m(−x)m−1 (1 + x)−n−1 x(1 + x).
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Now using the identities x(1 + x)p0n,ν (x) = [ν − (n + 1)x] pn,ν (x), we obtain
x(1 + x) µ0n,m (x) + mµn,m−1 (x)
∞
1X
=
[ν − (n + 1)x] pn,ν (x)
n ν=1
Z ∞
×
pn,ν−1 (t)(t − x)m dt + (n + 1)(−x)m+1 (1 + x)−n−1
1
=
n
0
∞
X
Z
pn,ν (x)
ν=1
∞
[{(ν − 1) − (n + 1)t} + (n + 1)(t − x) + 1] pn,ν−1 (t)(t − x)m dt
0
+ (n + 1)(−x)m+1 (1 + x)−n−1
Z ∞
∞
1X
pn,ν (x)
t(1 + t)p0n,ν−1 (t)(t − x)m dt
=
n ν=1
0
Z ∞
∞
1X
pn,ν (x)
+ (n + 1)
pn,ν−1 (t)(t − x)m+1 dt
n ν=1
0
Z ∞
∞
1X
+
pn,ν (x)
pn,ν−1 (t)(t − x)m dt + (n + 1)(−x)m+1 (1 + x)−n−1
n ν=1
0
Z
∞
∞
1X
=
pn,ν (x)
(1 + 2x)(t − x) + (t − x)2 + x(1 + x) p0n,ν−1 (t)(t − x)m dt
n ν=1
0
Z ∞
∞
1X
pn,ν (x)
+ (n + 1)
pn,ν−1 (t)(t − x)m+1 dt
n ν=1
0
Z
∞
∞
1X
+
pn,ν (x)
pn,ν−1 (t)(t − x)m dt + (n + 1)(−x)m+1 (1 + x)−n−1
n ν=1
0
= −(m + 1)(1 + 2x) µn,m (x) − (−x)m (1 + x)−n−1
− (m + 2) µn,m+1 (x) − (−x)m+1 (1 + x)−n−1
− mx(1 + x) µn,m−1 (x) − (−x)m−1 (1 + x)−n−1
+ (n + 1) µn,m+1 (x) − (−x)m+1 (1 + x)−n−1
+ µn,m (x) − (−x)m (1 + x)−n−1 + (n + 1)(−x)m+1 (1 + x)−n−1
= − [m(1 + 2x) + 2x] µn,m (x) + (n − m − 1)µn,m+1 (x) − mx(1 + x)µn,m−1 (x).
This completes the proof of recurrence relation.
Remark 2.3. It is easily verified from Lemma 2.2 and by the principle of mathematical induction, that for n > i and each x ∈ (0, ∞)
Bn (ti , x) =
(n + i)!(n − i − 1)! i
x
n!(n − 1)!
(n + i − 1)!(n − i − 1)! i−1
+ i(i − 1)
x + i(i − 1)(i − 2)O n−2 .
n!(n − 1)!
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V IJAY G UPTA AND O GÜN D O ĞRU
Corollary 2.4. Let δ be a positive number. Then for every n > γ > 0, x ∈ (0, ∞), there exists
a constant M (s, x) independent of n and depending on s and x such that
∞
1X
pn,ν (x)
n ν=1
Z
pn,ν−1 (t)tγ dt ≤ M (s, x)n−s , s = 1, 2, 3, . . . .
|t−x|>δ
Lemma 2.5 ([3]). There exist the polynomials Qi,j,r (x) independent of n and ν such that
X
{x(1 + x)}r Dr [pn,ν (x)] =
(n + 1)i [ν − (n + 1)x]j Qi,j,r (x)pn,ν (x),
2i+j≤r
i,j≥0
where D ≡
d
.
dx
Lemma 2.6. Let f be r times differentiable on [0, ∞) such that f (r−1) is absolutely continuous
with f (r−1) (t) = O(tγ ) for some γ > 0 as t → ∞. Then for r = 1, 2, 3, . . . and n > γ + r we
have
∞
Bn(r) (f, x)
(n + r − 1)!(n − r − 1)! X
=
pn+r,ν (x)
n!(n − 1)!
ν=0
Z
∞
pn−r,ν+r−1 (t)f (r) (t)dt.
0
Proof. It follows by simple computation the following relations:
p0n,ν (t) = n [pn+1,ν−1 (t) − pn+1,ν (t)] ,
(2.1)
where t ∈ [0, ∞).
Furthermore, we prove our lemma by mathematical induction. Using the above identity (2.1),
we have
∞
Bn0 (f, x)
1X 0
=
p (x)
n ν=1 n,ν
=
∞
X
Z
∞
pn,ν−1 (t)f (t)dt − (n + 1)(1 + x)−n−2 f (0)
0
Z
[pn+1,ν−1 (x) − pn+1,ν (x)]
∞
pn,ν−1 (t)f (t)dt
0
ν=1
− (n + 1)(1 + x)−n−2 f (0)
Z ∞
= npn+1,0 (x)
pn,0 (t)f (t)dt − (n + 1)(1 + x)−n−2 f (0)
+
0
∞
X
∞
Z
[pn,ν (t) − pn,ν−1 (t)] f (t)dt
pn+1,ν (x)
0
ν=1
−n−2
Z
∞
n(1 + t)−n−1 f (t)dt
= (n + 1)(1 + x)
0
+
∞
X
Z
∞
pn+1,ν (x)
0
ν=1
−n−2
− (n + 1)(1 + x)
J. Inequal. Pure and Appl. Math., 6(1) Art. 21, 2005
−1
n−1
p0n−1,ν (t)f (t)dt
f (0) .
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Applying the integration by parts, we get
Bn0 (f, x)
−n−2
= (n + 1)(1 + x)
Z
−n−2
∞
f (0) + (n + 1)(1 + x)
(1 + t)−n f 0 (t)dt
0
+
=
1
n−1
1
n−1
∞
X
∞
X
∞
Z
pn−1,ν (t)f 0 (t)dt − (n + 1)(1 + x)−n−2 f (0)
pn+1,ν (x)
0
ν=1
Z
∞
pn+1,ν (x)
pn−1,ν (t)f 0 (t)dt ,
0
ν=0
which was to be proved.
If we suppose that
∞
Bn(i) (f, x)
(n + i − 1)!(n − i − 1)! X
=
pn+i,ν (x)
n!(n − 1)!
ν=0
Z
∞
pn−i,ν+i−1 (t)f (i) (t)dt
0
then by (2.1), and using a similar method to the one above it is easily verified that the result is
true for r = i + 1. Therefore by the principle of mathematical induction the result follows. 3. S IMULTANEOUS A PPROXIMATION
In this section we study the rate of pointwise convergence of an asymptotic formula and
an error estimation in terms of a higher order modulus of continuity in simultaneous approximation for the operators defined by (1.1). Throughout the section, we have Cγ [0, ∞) :=
{f ∈ C[0, ∞) : |f (t)| ≤ M tγ for some M > 0, γ > 0} .
Theorem 3.1. Let f ∈ Cγ [0, ∞), γ > 0 and f (r) exists at a point x ∈ (0, ∞), then
Bn(r) (f, x) = f (r) (x) + o(1) as n → ∞.
Proof. By Taylor’s expansion of f , we have
f (t) =
r
X
f (i) (x)
i!
i=0
(t − x)i + ε(t, x)(t − x)r ,
where ε(t, x) → 0 as t → x.
Hence
Bn(r) (f, x)
Z
∞
=
Wn(r) (t, x)f (t)dt
0
=
Z
r
X
f (i) (x)
i=0
i!
∞
Wn(r) (t, x)(t − x)i dt
0
Z
+
∞
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)m , Lemma 2.2 and Remark 2.3, we
have
r
i r Z ∞
X
f (i) (x) X i
i−ν ∂
R1 =
(−x)
Wn (t, x)tν dt
r
i!
ν
∂x
0
ν=0
i=0
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V IJAY G UPTA AND O GÜN D O ĞRU
Z ∞
f (r) (x) ∂ r
=
Wn (t, x)tr dt
r! ∂xr 0
f (r) (x) (n + r)!(n − r − 1)!
r! + terms containing lower powers of x
=
r!
n!(n − 1)!
= f (r) (x) + o(1), n → ∞.
Using Lemma 2.5, we obtain
Z ∞
R2 =
Wn(r) (t, x)ε(t, x)(t − x)r dt
0
∞
Qi,j,r (x) X
pn,ν (x)
=
n
[ν − (n + 1)x]j
r
{x(1 + x)} ν=1
n
2i+j≤r
X
i
i,j≥0
∞
Z
pn,ν−1 (t)ε(t, x)(t − x)r dt + (−1)r
×
0
(n + r)!
(1 + x)−n−r−1 ε(0, x)(−x)r
(n + 1)!
=: R3 + R4 .
Since ε(t, x) → 0 as t → x for a given ε > 0 there exists a δ > 0 such that |ε(t, x)| < ε
whenever 0 < |t − x| < δ. Thus for some M1 > 0, we can write
Z
∞
X
X
j
i−1
|R3 | ≤ M1
n
pn,ν (x) |ν − (n + 1)x| ε
pn,ν−1 (t) |t − x|r dt
|t−x|<δ
ν=1
2i+j≤r
i,j≥0
pn,ν−1 (t)M2 t dt
Z
γ
+
|t−x|≥δ
=: R5 + R6 ,
where
|Qi,j,r (x)|
r
2i+j≤r {x(1 + x)}
M1 = sup
i,j≥0
and M2 is independent of t.
Applying Schwarz’s inequality for integration and summation respectively, we obtain
Z ∞
12
∞
X
X
j
R5 ≤ εM1
ni−1
pn,ν (x) |ν − (n + 1)x|
pn,ν−1 (t)dt
0
ν=1
2i+j≤r
i,j≥0
Z
×
∞
12
pn,ν−1 (t)(t − x) dt
2r
0
≤ εM1
X
2i+j≤r
ni
∞
X
∞
1X
pn,ν (x)[ν − (n + 1)x]2j
n ν=1
pn,ν (x)
ν=1
! 12
i,j≥0
∞
×
1X
pn,ν (x)
n ν=1
Z
! 12
∞
pn,ν−1 (t)(t − x)2r dt
.
0
Using Lemma 2.1 and Lemma 2.2, we get
R5 ≤ εM1 O nj/2 O n−r/2 = εO (1) .
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Again using the Schwarz inequality, Lemma 2.1 and Corollary 2.4, we obtain
X
R6 ≤ M2
n
i−1
∞
X
j
pn,ν−1 (t)tγ dt
pn,ν (x) |ν − (n + 1)x|
|t−x|≥δ
ν=1
2i+j≤r
Z
i,j≥0
X
≤ M2
n
i−1
∞
X
j
Z
pn,ν (x) |ν − (n + 1)x|
|t−x|≥δ
ν=1
2i+j≤r
12
pn,ν−1 (t)dt
i,j≥0
21
pn,ν−1 (t)t2γ dt
Z
×
|t−x|≥δ
∞
X
≤ M2
ni
2i+j≤r
1X
pn,ν (x)[ν − (n + 1)x]2j
n ν=1
! 12
i,j≥0
! 12
Z ∞
∞
1X
×
pn,ν (x)
pn,ν−1 (t)t2γ dt
n ν=1
0
X
=
ni O nj/2 O n−s/2
2i+j≤r
i,j≥0
for any s > 0.
Choosing s > r we get R6 = o(1). Thus, due to arbitrariness of ε > 0, it follows that
R3 = o(1). Also R4 → 0 as n → ∞ and hence R2 = o(1). Collecting the estimates of R1 and
R2 , we get the required result.
The following result holds.
Theorem 3.2. Let f ∈ Cγ [0, ∞), γ > 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) + [2x(1 + r) + r]f (r+1) (x) + x(1 + x)f (r+2) (x).
Proof. Using Taylor’s expansion of f , we have
f (t) =
r+2 (i)
X
f (x)
i=0
i!
(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)β , t → ∞ for some β > 0. Applying
Lemma 2.2, we have
" r+2
#
X f (i) (x) Z ∞
n[Bn(r) (f, x) − f (r) (x)] = n
Wn(r) (x, t)(t − x)i dt − f (r) (x)
i!
0
i=0
Z ∞
(r)
r+2
+ n
Wn (x, t)ε(t, x)(t − x) dt
0
=: E1 + E2 .
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V IJAY G UPTA AND O GÜN D O ĞRU
E1 = n
r+2 (i)
i X
f (x) X i
i=0
=
i!
j=0
j
i−j
Z
(−x)
∞
Wn(r) (x, t)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)
r!
(r + 1)!
f (r+2) (x) (r + 2)(r + 1) 2 (r) r
(r) r+1
+Bn (t , x) +
n
x Bn (t , x)
(r + 2)!
2
+ (r + 2)(−x)Bn(r) (tr+1 , x) + Bn(r) (tr+2 , x) .
Therefore by applying Remark 2.3, we get
(n + r)!(n − r − 1)!
(r)
E1 = nf (x)
−1
n!(n − 1)!
(n + r)!(n − r − 1)!
f (r+1) (x)
+n
(r + 1)(−x)r!
(r + 1)!
n!(n − 1)!
(n + r + 1)!(n − r − 2)!
(n + r)!(n − r − 2)!
+
(r + 1)!x + r(r + 1)
r!
n!(n − 1)!
n!(n − 1)!
f (r+2) (x) (r + 2)(r + 1)x2
(n + r)!(n − r − 1)!
+n
· r!
(r + 2)!
2
n!(n − 1)!
(n + r + 1)!(n − r − 2)!
+ (r + 2)(−x)
(r + 1)!x
n!(n − 1)!
(n + r)!(n − r − 2)!
+ r(r + 1)
r!
n!(n − 1)!
(n + r + 2)!(n − r − 3)! (r + 2)! 2
+
x
n!(n − 1)!
2
(n + r + 1)!(n − r − 3)!
(r + 1)!x
+ O n−2 .
+(r + 1)(r + 2)
n!(n − 1)!
In order to complete the proof of the theorem it is sufficient to show that E2 → 0 as n → ∞,
which can be easily proved along the lines of the proof of Theorem 3.1 and by using Lemma
2.1, Lemma 2.2 and Lemma 2.5.
Let us assume that 0 < a < a1 < b1 < b < ∞, for sufficiently small δ > 0, the m-th order
Steklov mean fm,δ (t) corresponding to f ∈ Cγ [0, ∞) is defined by
Z δ Z δ Z δ
m
2 2
2
Y
m
−m
f (t) − ∆η f (t)
dti
...
fm,δ (t) = δ
− 2δ
1
m
Pm
− 2δ
− 2δ
i=1
∆m
η f (t)
is the m−th forward difference with step length η.
where η =
i=1 ti , t ∈ [a, b] and
It is easily checked (see e. g. [1], [4]) that
(i) fm,δ has
continuous derivatives up to order m on [a, b];
(r) (ii) fm,δ ≤ M1 δ −r ωr (f, δ, a1 , b1 ), r = 1, 2, 3, . . . , m;
C[a1 ,b1 ]
(iii) kf − fm,δ kC[a1 ,b1 ] ≤ M2 ωm (f, δ, a, b);
(iv) kfm,δ kC[a1 ,b1 ] ≤ M3 kf kγ ,
where Mi , for i = 1, 2, 3 are certain unrelated constants independent of f and δ. The r−th order
modulus of continuity ωr (f, δ, a, b) for a function f continuous on the interval [a, b] is defined
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by:
ωr (f, δ, a, b) = sup {|∆rh f (x)| : |h| ≤ δ; x, x + h ∈ [a, b]} .
For r = 1, ω1 (f, δ) is written simply ωf (δ) or ω(f, δ).
The following error estimation is in terms of higher order modulus of continuity:
Theorem 3.3. Let f ∈ Cγ [0, ∞), γ > 0 and 0 < a < a1 < b1 < b < ∞. Then for all n
sufficiently large
n
o
(r)
(r)
−1/2
−1
Bn (f, ∗) − f (r) (x)
≤
max
M
ω
(f
,
n
,
a,
b),
M
n
kf
k
3 2
4
γ
C[a1 ,b1 ]
where M3 = M3 (r), M4 = M4 (r, f ).
Proof. First by the linearity property, we have
(r)
(r)
(r)
(r) Bn (f, ∗) − f (r) ≤ Bn ((f − f2,δ ), ∗) C[a1 ,b1 ] + Bn (f2,δ , ∗) − f2,δ C[a1 ,b1 ]
C[a1 ,b1 ]
(r)
(r) + f − f2,δ C[a1 ,b1 ]
=: A1 + A2 + A3 .
By property (iii) of the Steklov mean, we have
A3 ≤ C1 ω2 (f (r) , δ, a, b).
Next using Theorem 3.2, we have
A2 ≤ C2 n
−(k+1)
r+2 X
(j) f2,δ j=r
.
C[a,b]
By applying the interpolation property due to Goldberg and Meir [2] for each j = r, r + 1,
r + 2, we have
(j) (r+2) ≤ C3 kf2,δ kC[a,b] + f2,δ .
f2,δ C[a,b]
C[a,b]
Therefore by applying properties (ii) and (iv) of the Steklov mean, we obtain
n
o
A2 ≤ C4 n−1 kf kγ + δ −2 ω2 (f (r) , δ) .
Finally we estimate A1 , choosing a∗ , b∗ satisfying the condition 0 < a < a∗ < a1 < b1 <
b < b < ∞. Also let ψ(t) denote the characteristic function of the interval [a∗ , b∗ ], then
A1 ≤ 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 ]
∗
=: A4 + A5 .
We may note here that to estimate A4 and A5 , it is enough to consider their expressions without
the linear combinations. By Lemma 2.6, we have
Bn(r) (ψ(t)(f (t) − f2,δ (t)), x)
∞
(n − r − 1)!(n + r − 1)! X
=
pn+r,ν (x)
n!(n − 1)!
ν=0
Hence
Z
∞
pn−1,ν+r−1 (t)f (r) (t)dt.
0
(r)
(r)
(r) Bn (ψ(t)(f (t) − f2,δ (t)), ∗)
≤
C
f
−
f
5
2,δ C[a,b]
C[a∗ ,b∗ ]
J. Inequal. Pure and Appl. Math., 6(1) Art. 21, 2005
.
http://jipam.vu.edu.au/
10
V IJAY G UPTA
AND
O GÜN D O ĞRU
Now 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 Schwarz inequality, we have
I = Bn(r) ((1 − ψ(t))(f (t) − f2,δ (t)), x)
X
|Qi,j,r (x)|
≤
ni
xr
2i+j≤r
i,j≥0
∞
1X
×
pn,ν (x) |ν − (n + 1)x|j
n ν=1
Z
∞
pn,ν−1 (t)(1 − ψ(t)) |f (t) − f2,δ (t)| dt
0
+ (1 + x)−n−1 |(−n − 1)(−n) · · · (−n − r)|(1 − ψ(0)) |f (0) − f2,δ (0)|


∞
 X
X
≤ C6 kf kγ
ni−1
pn,ν (x) |ν − (n + 1)x|j

2i+j≤r
ν=1
i,j≥0
Z
−n−1
|(−n − 1)(−n) · · · (−n − r)|
×
pn,ν−1 (t)dt + (1 + x)
|t−x|≥δ1


Z ∞
21
∞

X
X
j
−2s
i−1
≤ C6 kf kγ δ1
n
pn,ν (x) |ν − (n + 1)x|
pn,ν−1 (t)dt

0

ν=1
2i+j≤r
i,j≥0
Z
∞
×
)
12
pn,ν−1 (t)(t − x)4s dt
+ (1 + x)−n−1 |(−n − 1)(−n) · · · (−n − r)|
0
≤
C6 kf kγ δ1−2s
(
×
X
ni
2i+j≤r
∞
1X
pn,ν (x) [ν − (n + 1)x]2j − (1 + x)−n−1 − {−(n + 1)x}2j
n ν=0
) 12
i,j≥0
(
×
∞
1X
pn,ν (x)
n ν=0
−n−1
− (1 + x)
∞
Z
pn,ν−1 (t)(t − x)4s dt − (1 + x)−n−1 (−x)4s
0
4s
(−x)
12
+ C6 kf kγ (1 + x)−n−1 |(−n − 1)(−n) · · · (−n − r)|.
Hence by Lemma 2.1 and Lemma 2.2, we have
j
r
I ≤ C7 kf kγ δ1−2s O n(i+ 2 −s) ≤ C7 n−q kf kγ , q = s − ,
2
where the last term vanishes as n → ∞. Now choosing q satisfying q ≥ 1, we obtain
I ≤ C7 n−1 kf kγ .
Therefore by property (iii) of the Steklov mean, we get
(r)
(r) A1 ≤ C8 f − f2,δ + C7 n−1 kf kγ
C[a∗ ,b∗ ]
≤ C9 ω2 (f (r) , δ, a, b) + C7 n−1 kf kγ .
Choosing δ = n−1/2 , the theorem follows.
J. Inequal. Pure and Appl. Math., 6(1) Art. 21, 2005
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D IRECT R ESULTS FOR C ERTAIN FAMILY OF I NTEGRAL T YPE O PERATORS
11
R EFERENCES
[1] G. FREUD AND V. POPOV, On approximation by spline functions, Proc. Conf. on Constructive
Theory Functions, Budapest (1969), 163–172.
[2] S. GOLDBERG AND V. MEIR, Minimum moduli of ordinary differential operators, Proc. London
Math. Soc., 23 (1971), 1–15.
[3] V. GUPTA AND G.S. SRIVASTAVA, Convergence of derivatives by summation-integral type operators, Revista Colombiana de Matematicas, 29 (1995), 1–11.
[4] E. HEWITT AND K. STROMBERG, Real and Abstract Analysis, McGraw Hill, New York, 1956.
[5] H.M. SRIVASTAVA AND V. GUPTA, A certain family of summation integral type operators, Mathematical and Computer Modelling, 37 (2003), 1307–1315.
J. Inequal. Pure and Appl. Math., 6(1) Art. 21, 2005
http://jipam.vu.edu.au/