Báo cáo hóa học: "Research Article On the Generalized Favard-Kantorovich and Favard-Durrmeyer Operators in Exponential Function Spaces" ppt

Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 75142, 12 pages
doi:10.1155/2007/75142
Research Article
On the Generalized Favard-Kantorovich and Favard-Durrmeyer
Operators in Exponential Function Spaces
Grzegorz Nowak and Aneta Sikorska-Nowak
Received 18 January 2007; Revised 12 June 2007; Accepted 14 November 2007
Recommended by Ulrich Abel
We consider the Kantorovich- and the Durrmeyer-type modifications of the generalized
Favard operators and we prove an inverse approximation theorem for functions f such
that w
σ
f ∈ L
p
(R), where 1 ≤ p ≤∞and w
σ
(x) =ex p(−σx
2
), σ>0.
Copyright © 2007 G. Nowak and A. Sikorska-Nowak. This is an open access article dis-
tributed under the Creative Commons Attribution License, which per mits unrestricted
use, distribution, and reproduction in any medium, provided the original work is prop-
erly cited.
1. Preliminaries
Let
L
p,σ
(R) =


f :


w
σ
f


p
< ∞

for 1 ≤ p ≤∞ (1.1)
be the weighted function space, where w
σ
(x) =ex p(−σx
2
), σ>0,
g
p
=


∞
−∞


g(x)


p

dx

1/p
if 1 ≤ p<∞,
g
∞
= essup
x∈R


g(x)


.
(1.2)
We define the generalized Favard operators F
n
for functions f : R→R by
F
n
f (x) =
∞

k=−∞
f (k/n)p
n,k
(x; γ)(x ∈R, n ∈ N), (1.3)
2 Journal of Inequalities and Applications
where N
={1,2, },

p
n,k
(x; γ) =
1
nγ
n
√
2π
exp

−
1
2γ
2
n

k
n
−x

2

(1.4)
and γ
= (γ
n
)
∞
n=1
is a positive sequence convergent to zero (see [1]). In the case where

γ
2
n
= ϑ/(2n) with a positive constant ϑ, F
n
become the known Favard operators intro-
duced by Favard [2]. Some approximation properties of the classical Favard operators for
continuous functions f on R are presented in [3, 4]. Some approximation properties of
their generalization can be found, for example, in [1, 5]. Denote by F
∗
n
the Kantorovich-
type modification of operators F
n
,definedby
F
∗
n
f (x) = n
∞

k=−∞
p
n,k
(x; γ)

(k+1)/n
k/n
f (t)dt (x ∈R, n ∈ N), (1.5)
and by


F
n
the Durrmeyer-type modification of operators F
n

F
n
f (x) = n
∞

k=−∞
p
n,k
(x; γ)

∞
−∞
p
n,k
(t;γ) f (t)dt (x ∈R, n ∈ N), (1.6)
where f
∈ L
p,σ
(R). Some estimates concerning the rates of pointwise convergence of the
operators F
∗
n
f and


F
n
f can be found in [6, 7].
Recently, several autors investigated the conditions under which global smoothness of
a function f , as measured by its modulus of continuity ω( f ;
◦), is retained by the elements
of approximating sequences (L
n
f ) (see, e.g., [8, 9]). For example, Kratz and Stadtm
¨
uller
considered in [10]awideclassofdiscreteoperatorsL
n
and derived estimates of the form
ω

L
n
f ;t

≤
Kω( f ;t)(t>0), (1.7)
with a positive constant K independent of f ,n,andt. For bounded functions f
∈ C(R)
and operators F
n
satisfying
γ
2
1

≥
1
2
π
−2
log2, n
2
γ
2
n
≥
1
2
π
−2
logn if n ≥2, (1.8)
they obtained the inequality
ω(F
n
f ;t) ≤ 140ω( f ;t)+16π·tf  (t>0), (1.9)
where
f =sup{|f (x)| : x ∈ R}.
Forbounded functions f
∈ C
m
(R) ={f : w
m
f 
∞
< ∞}, w

m
(x) = (1 + x
2m
)
−1
, m ∈ N
and for operators F
n
satisfying nγ
2
n
≥ c>0foralln ∈ N, Pych-Taberska [5]obtainedthe
inequality
ω
2

F
n
f ;t

m
≤ K

(1 +t
2
0
)ω
2
( f ;t)
m

+ t
2
f 
m

(0 <t≤ t
0
) (1.10)
for all n
∈ N, n ≥n
c
where n
c
∈ N and K is a constant.
G. Nowak and A. Sikorska-Nowak 3
In this paper, we obtained an analogous inequality for the rth weighted modulus of
smoothness of the function
f
∈ L
p,σ
(R),σ>0,1 ≤ p ≤∞,
ω
r
( f ;t)
σ,p
= sup
0<h≤t


w

σ
Δ
r
h
f 
p
(r ∈ N),
(1.11)
where
Δ
r
h
f (x) =
r

i=0

r
i

(−1)
i
f

x + h

r/2 −i

. (1.12)
Namely, suppose that (γ

n
) is a positive null sequence satisfying nγ
r/2+1
n
≥
c max
n∈N
{γ
r/2−1
n
} > 0foralln ∈ N and σ
1
>σ>0. Then there exist positive constants,
K,K
1
, such that for all n ≥K
1
and for arbitrary positive number t
0
ω
r

L
n
f ,t

σ
1
,p
≤ K


(1 +t
2
0
)ω
r
( f ,t)
σ,p
+ t
r


w
σ
f


p


0 <t≤ t
0

, (1.13)
where L
n
denotes the Favard-Kantorovich operator or the Favard-Durrmeyer operator.
Throughout the paper, the sy mbols K(σ,σ
1
, ), K

j
(σ,σ
1
, )(j = 1,2, )willmean
some positive constants, not necessarily the same at each occurrence, depending only on
the parameters indicated in parentheses.
2. Preliminary results
Let γ
= (γ
n
)
∞
n=1
be a positive sequence and let nγ
2
n
≥ c for all n ∈N, with a positive abso-
lute constant c.Asisknown[5], for v
∈ N
0
={0}∪N, n ∈ N, x ∈ R,
∞

k=−∞




k
n

−x




v
p
n,k
(x; γ) ≤ 15A
c

2
e

v/2

(2v)!γ
v
n
, (2.1)
where A
c
= max {1,(2cπ
2
)
−1
}. A simple calculation and the known Schwarz inequality
lead to

∞

−∞




k
n
−t




v
p
n,k
(t;γ)dt ≤

(2v)!!
γ
v
n
n

k ∈ Z =

0,±1,±2,

. (2.2)
Let us choose n
∈ N, j ∈ N

0
and let us write
G
∗
n, j
f (x) = n
∞

k=−∞
p
n,k
(x; γ)

k
n
−x

j

(k+1)/n
k/n
f (t)dt,
(2.3)

G
n, j
f (x) = n
∞

k=−∞

p
n,k
(x; γ)

k
n
−x

j

∞
−∞
p
n,k
(t;γ) f (t)dt, (2.4)
where f
∈ L
p,σ
(R), 1 ≤ p ≤∞, σ>0. Obviously, G
∗
n,0
f (x) = F
∗
n
f (x)and

G
n,0
f (x) =


F
n
f (x).
4 Journal of Inequalities and Applications
Lemma 2.1. Let γ
= (γ
n
)
∞
n=1
beapositivesequenceconvergentto0andletnγ
2
n
≥ c for all
n
∈ N, with a positive absolute constant c.Thenforj ∈N
0
, f ∈ L
p,σ
(R), σ>0, 1 ≤ p ≤∞,
and σ
1
>σ,


w
σ
1
G
∗

n, j
f


p
≤ 15A
c
exp

σ
1

σ + σ
1

σ
1
−σ



2 j

!2
j/2
γ
j
n



w
σ
f


p
(2.5)
for all n
∈ N such that γ
2
n
≤ (σ
1
−σ)/(4σ(σ + σ
1
)),


w
σ
1

G
n, j
f


p
≤ 30A
c


(2 j)!2
j/2
γ
j
n


w
σ
f


p
(2.6)
for all n
∈ N such that γ
n
≤ max{(σ
1
−σ)/(2
√
σ(σ + σ
1
));(
√
σ
1
−σ)/(
√

2(σ + σ
1
))}.
Proof. In view of definition (2.3),
exp

−
σ
1
x
2

|
G
∗
n, j
f (x)|≤n
∞

k=−∞
exp

−
σ
1
x
2

p
n,k

(x; γ)




k
n
−x




j
× exp

σ(|k|+1

2
/n
2


(k+1)/n
k/n
exp

−
σt
2


|
f (t)|dt.
(2.7)
Using the inequality
(u +v)
2
≤
σ + σ
1
2σ
u
2
+
σ + σ
1
σ
1
−σ
v
2
(u ∈ R, v ∈ R), (2.8)
we can easily observe, that
p
n,k
(x; γ)exp

−
σ
1
x

2

exp

σ

k +1
n

2

≤
√
2exp

σ
1
(σ + σ
1
)
σ
1
−σ

p
n,k

x;
√
2γ


,
p
n,k
(x; γ)exp

−
σ
1
x
2

exp

σ

k
n

2

≤
√
2p
n,k

x;
√
2γ


,
(2.9)
for n
∈ N such that γ
2
n
≤ (σ
1
−σ)/(4σ(σ + σ
1
)) (see [9]), where the symbol
√
2γ means
the sequence (
√
2γ
n
)
∞
n=1
. Therefore,
exp(
−σ
1
x
2
)


G

∗
n, j
f (x)


≤
exp

σ
1

σ + σ
1

σ
1
−σ

√
2n
∞

k=−∞
p
n,k

x;
√
2γ


×




k
n
−x




j

(k+1)/n
k/n
exp

−
σt
2

|
f (t)|dt.
(2.10)
From (2.2), we have


w
σ

1
G
∗
n, j
f


1
≤ exp

σ
1

σ + σ
1

σ
1
−σ



2 j

!!(
√
2)
j+1
γ
j

n


w
σ
f


1
. (2.11)
G. Nowak and A. Sikorska-Nowak 5
Instead, for p
=∞,from(2.1) it follows that


w
σ
1
G
∗
n, j
f


∞
≤
√
2exp

σ

1

σ + σ
1

σ
1
−σ



w
σ
f


∞
essup
x∈R

∞

k=−∞
p
n,k

x;
√
2γ






k
n
−x




j

≤
15
√
2A
c
exp

σ
1

σ + σ
1

σ
1
−σ



2
j

2 j

!γ
j
n
w
σ
f 
∞
.
(2.12)
Finally, by Riesz-Thorin theorem, we have (2.5).
In view of definition (2.4) and the inequality
p
n,k
(x; γ)p
n,k
(t;γ)exp(−σ
1
x
2
)exp(σt
2
) ≤ 2p
n,k
(x;

√
2γ)p
n,k
(t;
√
2γ), (2.13)
for n
∈ N such that γ
n
≤ max{(σ
1
−σ)/(2
√
σ(σ + σ
1
));
√
σ
1
−σ/(
√
2(σ + σ
1
))} (see [6]),
we have
exp

−
σ
1

x
2




G
n, j
f (x)


≤
2n
∞

k=−∞
p
n,k

x;
√
2γ





k
n
−x





j

∞
−∞
exp

−
σt
2

p
n,k

√
2t;γ



f (t)


dt.
(2.14)
Applying (2.1)and(2.2), we get
w
σ

1

G
n, j
f 
1
≤ 30A
c

(2 j)!!γ
j
n
2
j/2
w
σ
f 
1
,
w
σ
1

G
n, j
f 
∞
≤ 30A
c


(2 j)!γ
j
n
2
j/2
w
σ
f 
∞
.
(2.15)
Finally, by Riesz-Thorin theorem, we have (2.6).
Further , for δ>0, x ∈ R,andr ∈N we define Stieklov function of f
f
(δ,2r)
(x) =
1
δ
2r
2

2r
r


δ/2
−δ/2
···

δ/2

−δ/2
r

i=1

2r
r
−i

(−1)
i−1
f

x + i

t
1
+ ···+ t
2r

dt
1
···dt
2r
.
(2.16)

Lemma 2.2. For all r = 1,2, , 0 <δ≤1, σ
1
>σ>0, 1 ≤ p ≤∞,andx ∈ R,



w
σ
1
f
(r)
(δ,2r)


p
≤ K

r,σ,σ
1

1
δ
r
ω
r
( f ;δ)
σ,p
, (2.17)


w
σ
1


f
(δ,2r)
− f



p
≤ K

r,σ,σ
1

ω
r
( f ;δ)
σ,p
. (2.18)
Proof. It is easy to see by induction that
f
(r)
(δ,2r)
(x) =
2

2r
r

r

i=1

(−1)
i−1

2r
r
−i

1
(iδ)
2r
×

iδ/2
−iδ/2
···

iδ/2
−iδ/2
Δ
r
iδ
f

x + u
1
+ ···+ u
r

du
1

···du
r
.
(2.19)
6 Journal of Inequalities and Applications
Let σ
2
= (2σ
1
+ σ)/3. In view of the inequality
exp

−
σ
1
x
2
+ σ
2
(x + u)
2

≤
exp

σ
2
σ
1
σ

1
−σ
2
u
2

, (2.20)
where 0 <δ
≤ 1andu = u
1
+ ···+ u
r
,(u ≤r
2
/2), we have


w
σ
1
f
(r)
(δ,2r)


p
≤
2

2r

r

exp

σ
1
σ
2
σ
1
−σ
2
r
4
4

r

i=1

2r
r
−i

1
(iδ)
r
w
σ
2

Δ
r
iδ
f 
p
. (2.21)
Applying the Minkowski inequality and the fact that for 0
≤ l
i
≤ i −1(0≤i ≤ r), 0 <h≤
1,
exp

−
σ
2
x
2
+ σ

x + h

l
1
+ ···+ l
r
−
r(i −1)
2


2

≤
exp

σσ
2
σ
2
−σ

r(i −1)
2

2

,
(2.22)
we obtain


w
σ
2
Δ
r
iδ
f



p
=sup
|h|≤δ


∞
−∞




exp

−
σ
2
x
2

i−1

l
1
=0
···
i−1

l
r
=0

Δ
r
h
f

x+h

l
1
+ ···+ l
r
−
r(i −1)
2





p
dx

1/p
≤ exp

σσ
2
σ
2
−σ

r
2
(i −1)
2
4

i
r
ω
r
( f ;δ)
σ,p
.
(2.23)
So (2.17) is evident. It is easy to see that
f
(δ,2r)
(x) − f (x) =
(−1)
r−1
δ
2r
1

2r
r


δ/2
−δ/2

···

δ/2
−δ/2
Δ
2r
t
1
+···+t
2r
f (x)dt
1
···dt
2r
. (2.24)
By Minkowski inequality, for 1
≤ p ≤∞,wehave(2.18). 
Lemma 2.3. Suppose that γ = (γ
n
)
∞
n=1
is a positive sequence convergent to 0 and that
nγ
r/2+1
n
≥ cK(r),wherer ∈ N, r ≥ 2, K(r) = max
n∈N
{γ
r/2−1

n
}, c is a positive absolute con-
stant and let a
r
= 1 for even r and a
r
= 2 for odd r.Thenfor f ∈L
p,σ
(R), σ>0, 1 ≤ p ≤∞
and σ
1
>σ,wehave


w
σ
1

F
∗
n
f

(r)
−(n/a
r
)
r
F
∗

n
Δ
r
a
r
/n
f



p
≤ K(σ,σ
1
,c,r)


w
σ
f


p
(2.25)
for all n
∈ N such that γ
2
n
≤ (σ
1
−σ)/(4σ(σ + σ

1
)) and nγ
n
> 4a
2
r
r
2
,and


w
σ
1


F
n
f

(r)
−(n/a
r
)
r

F
n
Δ
r

a
r
/n
f



p
≤ K(σ,σ
1
,c,r)


w
σ
f


p
(2.26)
for all n
∈ N such that γ
n
≤ max{(σ
1
− σ)/(2
√
σ(σ + σ
1
));

√
σ
1
−σ/(
√
2(σ + σ
1
))} and
nγ
n
>r
2
/4.
G. Nowak and A. Sikorska-Nowak 7
Proof. We consider an even r.Letr
= 2r
1
, r
1
∈ N, x ∈ R.Then
n
r
F
∗
n

Δ
r
1/n
f (x)


=
n
2r
1
+1
∞

k=−∞
p
n,k
(x; γ)
2r
1

i=0

2r
1
i

(−1)
i

(k+r
1
−i+1)/n
(k+r
1
−i)/n

f (t)dt
= n
2r
1
+1
r
1
−1

i=0

2r
1
i

(−1)
i
×
∞

k=−∞

p
n,k−(r
1
−i)
(x; γ)+p
n,k+(r
1
−i)

(x; γ)


(k+1)/n
k/n
f (t)dt
+ n
2r
1
+1
∞

k=−∞
⎛
⎝
2r
1
r
1
⎞
⎠
(−1)
r
1
p
n,k
(x; γ)

(k+1)/n
k/n

f (t)dt.
(2.27)
It is easy to see that
p
n,k−(r
1
−i)
(x; γ)+p
n,k+(r
1
−i)
(x; γ)
=p
n,k
(x; γ)

exp

r
1
−i
nγ
2
n

k
n
−x

−

(r
1
−i)
2
2n
2
γ
2
n

+exp

−
r
1
−i
nγ
2
n

k
n
−x

−
(r
1
−i)
2
2n

2
γ
2
n

=
p
n,k
(x; γ)
∞

l=1
(−1)
l
l!
[l/2]

j=0

l
2 j

2
2j+1−l

k
n
−x

2j

n
2j−2l
γ
−2l
n
(r
1
−i)
2l−2 j
+2p
n,k
(x; γ).
(2.28)
Consequently, using definition (2.3), we get
n
r
F
∗
n

Δ
r
1/n
f (x)

=
2r
1

l=1

[l/2]

j=0
n
2(r
1
+j−l)
γ
−2l
n
(−1)
l
2
2j+1−l

2 j

!

l −2 j

!
×
r
1
−1

i=0

2r

1
i

(−1)
i

r
1
−i

2l−2 j
G
∗
n,2j
f (x)
+
∞

l=2r
1
+1
[l/2]

j=0
n
2(r
1
+j−l)
γ
−2l

n
(−1)
l
2
2j+1−l

2 j

!

l −2 j

!
×
r
1
−1

i=0

2r
1
i

(−1)
i

r
1
−i


2l−2 j
G
∗
n,2j
f (x)
= S
n,1
f (x)+S
n,2
f (x).
(2.29)
8 Journal of Inequalities and Applications
In view of (2.5) and using Stirling formula, we obtain


w
σ
1
S
n,2
f


p
≤ K
1

σ,σ
1

,c



w
σ
f


p
4
r
1
n
2r
1
∞

l=2r
1
+1
r
2l
1
n
2l
γ
2l
n
2

l
[l/2]

j=0

(4 j

!2
j

2 j

!

l −2 j

!
n
2j
γ
2j
n
4
j
r
−2j
1
≤ K
2


σ,σ
1
,c,r



w
σ
f


p
n
2r
1
∞

l=2r
1
+1

r
2
1
/2)
l
(n
2
γ
2

n
)
l
[l/2]

j=0

n
2
γ
2
n

j
64
j
≤ K
3

σ,σ
1
,c,r



w
σ
f



p


16r
2
1

2r
1
+1
n
2
γ
2r
1
+2
n
+ n
2r
1
∞

l=2r
1
+2

16r
2
1
nγ

n

l

.
(2.30)
Assuming (16r
2
1
)/(nγ
n
) < 1 and using the condition nγ
r
1
+1
n
≥ cK(r), we get
w
σ
1
S
n,2
f 
p
≤ K
4
(σ,σ
1
,c,r)w
σ

f 
p
. (2.31)
Now observe that
r
1
−1

i=0

2r
1
i

(−1)
i

r
1
−i

2s
=
⎧
⎨
⎩
0if0<s<r
1
,
(2r

1
)!/2ifs = r
1
.
(2.32)
The equality follows simply from properties of finite differences since the left-hand side
of the equation is a half of the finite difference of the polynomial (r
1
−x)
2s
. Therefore,
S
n,1
f (x) =
2r
1

l=r
1
(−1)
l
2
2j+1−l
l!n
2l−2 j−2r
1
γ
2l
n


l
2 j

r
1
−1

i=0

2r
1
i

(−1)
i

r
1
−i

2l−2 j
G
∗
n,2j
f (x)
=
r
1

l=0

l
−1

j=0
(−1)
r
1
+l
2
2j+1−l−r
1
(r
1
+ l)!n
2l−2 j
γ
2l+2r
1
n

r
1
+ l
2 j

r
1
−1

i=0


2r
1
i

(−1)
i

r
1
−i

2r
1
+2l−2 j
G
∗
n,2j
f (x)
+
r
1

l=0
(−1)
2r
1
−l
γ
4r

1
−2l
n

2r
1

!
2
l
l!

2r
1
−2l

!
G
∗
n,2j
f (x).
(2.33)
It is easy to see, by the method of induction, that
p
(v)
n,k
(x; γ) = p
n,k
(x; γ)
[v/2]


i=0
v!(−1)
i
(v −2i)!(2i)!!
1
γ
2v−2i
n

k
n
−x

v−2i
, v ∈ N. (2.34)
Therefore,
S
n,1
f (x) =
r
1

l=0
l
−1

j=0
(−1)
r

1
+l
2
2j+1−l−r
1

r
1
+ l

!n
2l−2 j
γ
2l+2r
1
n

r
1
+ l
2 j

r
1
−1

i=0

2r
1

i

(−1)
i

r
1
−i

2r
1
+2l−2 j
G
∗
n,2j
f (x)
+

F
∗
n
f (x)

(2r
1
)
.
(2.35)
G. Nowak and A. Sikorska-Nowak 9
Consequently, from (2.29)



(F
∗
n
f )
(2r
1
)
(x) −n
2r
1
F
∗
n
Δ
2r
1
1/n
f (x)


≤
K
5
(r)
r
1
−1


j=0
r
1

l=j+1
n
2j
(nγ
n
)
2l
γ
2r
1
n


G
∗
n,2j
f (x)


+


S
n,2
f (x)



.
(2.36)
The condition nγ
r
1
+1
n
≥ cK(r) and the boundedness of the sequence (γ
n
)leadto



F
∗
n
f

(2r
1
)
(x) −n
2r
1
F
∗
n
Δ
2r

1
1/n
f (x)


≤
K
6
(r,c)
r
1
−1

j=0
γ
−2j
n


G
∗
n,2j
f (x)|+


S
n,2
f (x)



. (2.37)
Collecting the results we get estimate (2.25)forevenr, immediately.
Now, we will prove inequality (2.25)foroddr.Namely,letr
= 2r
2
+1,r
2
∈ N, x ∈R.
Then
n
r
F
∗
n

Δ
r
2/n
f (x)

=
n
2r
2
+2
r
2

i=0
∞


k=−∞

2r
2
+1
i

(−1)
i
×

p
n,k−(2r
2
+1−2i)
(x; γ) − p
n,k+(2r
2
+1−2i)
(x; γ)


(k+1)/n
k/n
f (t)dt.
(2.38)
It is easy to see that
p
n,k−(2r

2
+1−2i)
(x; γ) − p
n,k+(2r
2
+1−2i)
(x; γ)
= p
n,k
(x; γ)
∞

l=1
(−1)
l+1
l!
[(l−1)/2]

j=0

l
2 j +1

2
2j+2−l

k
n
−x


2j+1
n
2j+1−2l
γ
2l
n

2r
2
+1−2i

2j−2l+1
.
(2.39)
Consequently,
n
r
F
∗
n
(Δ
r
2/n
f (x)) =
2r
2
+1

l=1
n

2r
2
+2
[(l
−1)/2]

j=0
n
2j−2l
γ
−2l
n
(−1)
l+1
2
2j+2−l
(2 j +1)!(l −2 j −1)!
×
r
2

i=0

2r
2
+1
i

(−1)
i

(2r
2
+1−2i)
2l−2 j−1
G
∗
n,2j+1
f (x)
+
∞

l=2r
2
+2
n
2r
2
+2
[(l
−1)/2]

j=0
n
2j−2l
γ
−2l
n
(−1)
l+1
2

2j+2−l
(2 j +1)!(l −2 j −1)!
×
r
2

i=0

2r
2
+1
i

(−1)
i
(2r
2
+1−2i)
2l−2 j−1
G
∗
n,2j+1
f (x)
= S
∗
n,1
f (x)+S
∗
n,2
f (x).

(2.40)
Some simple calculation, Stirling formula and (2.5)give


w
σ
1
S
∗
n,2
f


p
≤ K
7
(σ,σ
1
,c,r)


w
σ
f


p
(2.41)
10 Journal of Inequalities and Applications
for n

∈ N such that (16r
2
)/(nγ
n
) < 1.Next,inviewof(2.25) and the equality
r
2

i=0

2r
2
+1
i

(−1)
i

r
2
−i+1/2

2s−1
=
⎧
⎨
⎩
0if0<s<r
2
+1,


2r
2
+1

!/2ifs = r
2
+1
(2.42)
we obtain
S
∗
n,1
f (x) =
r
2

l=0
l
−1

j=0
(−1)
r
2
+l
2
2j+1−l−r
2
(2 j +1)!


l + r
2
−2j

!
n
2j−2l
γ
−2l−2r
2
−2
n
×
r
2

i=0

2r
2
+1
i

(−1)
i

2r
2
+1−2i


2r
2
+2l−2 j+1
×G
∗
n,2j+1
f (x)+2
2r
2
+1
(F
∗
n
f )
(2r
2
+1)
(x) .
(2.43)
Using (2.40) and the condition nγ
r
2
+3/2
n
≥ cK(r), we have


(F
∗

n
f )
(2r
2
+1)
(x) −(n/2)
2r
2
+1
F
∗
n
Δ
2r
2
+1
2/n
f (x)


≤
K
8
(r,c)
r
2
−1

j=0
1

γ
2j+1
n


G
∗
n,2j+1
f (x)


+


S
∗
n,2
f (x)


.
(2.44)
Applying (2.5), we get (2.25)foroddr. Therefore, inequality (2.25)isproved.
Now we will prove (2.26). Let r
= 2r
1
, r
1
∈ N. A simple calculation and the equality
p

n,k
(t −(r
1
−i)/n;γ) = p
n,k+r
1
−i
(t;γ)give
n
r

F
n

Δ
r
1/n
f (x)

=
n
2r
1
+1
r
1
−1

i=0
∞


k=−∞

2r
1
i

(−1)
i

p
n,k−(r
1
−i)
(x; γ)+p
n,k+(r
1
−i)
(x; γ)

×

∞
−∞
p
n,k
(t;γ) f (t)dt + n
2r
1
+1

∞

k=−∞

2r
1
r
1

(−1)
i
p
n,k
(x; γ)
×

∞
−∞
p
n,k
(t;γ) f (t)dt.
(2.45)
The estimate (2.26) follows now the same way as ( 2.25).

3. Main result
Theorem 3.1. Suppose that r
∈ N, (γ
n
) is a positive null sequence satisfying nγ
r/2+1

n
≥
cK(r) for all n ∈ N with some c>0 where K(r) = max
n∈N
{γ
r/2−1
n
}. Then there exists a
constant K>0, such that for all f
∈ L
p,σ
(R), σ
1
>σ>0, 1 ≤ p ≤∞,andforanarbitrary
positive number t
0
,
ω
r

F
∗
n
f ,t

σ
1
,p
≤ K


σ,σ
1
,r,c

1+t
2
0

ω
r
( f ,t)
σ,p
+ t
r


w
σ
f


p

0 <t≤ t
0

(3.1)
for all n ∈ N such that γ
2
n

≤ (σ
1
−σ)/(4σ(σ + σ
1
)) and nγ
n
> 16r
2
,and
ω
r


F
n
f ,t

σ
1
,p
≤ K

σ,σ
1
,r,c

1+t
2
0


ω
r
( f ,t)
σ,p
+ t
r


w
σ
f


p

0 <t≤ t
0

(3.2)
G. Nowak and A. Sikorska-Nowak 11
for all n
∈ N such that γ
n
≤ max{(σ
1
− σ)/(2
√
σ(σ + σ
1
));

√
σ
1
−σ/(
√
2(σ + σ
1
))} and
nγ
n
>r
2
/4.
Proof. Let σ
2
= (3σ
1
+ σ)/4. In view of the inequality
exp

−
σ
1
x
2
+ σ
2
(x + u)
2


≤
exp

σ
2
σ
1
σ
1
−σ
2
u
2

(u ∈ R) (3.3)
and the generalized Minkowski inequalit y it is easy to see that for 0 <h
≤ 1


w
σ
1
Δ
r
h
f


p
=






w
σ
1

h/2
−h/2
···

h/2
−h/2
f
(r)

◦
+ s
1
+ ···+ s
r

exp

σ
2

◦

+ s
1
+ ···+ s
r

2

×
exp

−
σ
2

◦
+ s
1
+ ···+ s
r

2

ds
1
···ds
r






p
≤ exp

r
2
4
σ
2
σ
1
σ
1
−σ
2

h
r


w
σ
2
f
(r)


p
,
(3.4)



w
σ
1
Δ
r
h
f


p
≤ 2
r
exp

r
2
4
σ
2
σ
1
σ
1
−σ
2




w
σ
2
f


p
.
(3.5)
Applying these inequalities, we get


w
σ
1
Δ
r
h
f


p
≤


w
σ
1
Δ
r

h

f − f
(δ,2r)



p
+


w
σ
1
Δ
r
h
f
(δ,2r)


p
≤ 2
r
exp

r
2
4
σ

2
σ
1
σ
1
−σ
2




w
σ
2
( f − f
(δ,2r)
)


p
+ h
r


w
σ
2
f
(r)
(δ,2r)



p

,
(3.6)
where f
(δ,2r)
(x)(δ>0, x ∈ R, r ∈N)isdefinedby(2.16).
Hence, applying this inequality for F
∗
n
f we have
ω
r

F
∗
n
f ,t

σ
1
,p
≤ 2
r
exp

r
2

4
σ
2
σ
1
σ
1
−σ
2




w
σ
2
F
∗
n

f − f
(δ,2r)



p
+ t
r



w
σ
2

F
∗
n
f
(δ,2r)

(r)


p

.
(3.7)
Hence,
w
σ
2
(F
∗
n
f
(δ,2r)
)
(r)

p

can be estimated by (2.5)forj = 0, (2.25), and (3.4). Let
σ
3
= (2σ
1
+ σ)/3, then


w
σ
2

F
∗
n
f
(δ,2r)

(r)


p
≤


w
σ
2

F

∗
n
f
(δ,2r)

(r)
−

n/a
r
)F
∗
n
Δ
r
a
r
/n
f
(δ,2r)



p
+ n
r


w
σ

2
F
∗
n
Δ
r
a
r
/n
f
(δ,2r)


p
≤ K

σ
2
,σ
3
,r,c




w
σ
3
f
(δ,2r)



p
+


w
σ
3
f
(r)
(δ,2r)


p

.
(3.8)
Using (2.5)forj
= 0and(3.7)wehave
ω
r

F
∗
n
f ,t

σ
1

,p
≤ K

σ,σ
1
,r,c



w
σ
3
( f − f
(δ,2r)
)


p
+ t
r


w
σ
3
f
(r)
(δ,2r)



p
+ t
r


w
σ
3
f
(δ,2r)


p

.
(3.9)
12 Journal of Inequalities and Applications
Consequently by (2.17), (2.18) and assuming now 0 <t
≤ t
0
,wehave
ω
r

F
∗
n
f ,t

σ

1
,p
≤ K

σ,σ
1
,r,c

1+t
r
0

ω
r
( f ;t)
σ,p
+ t
r


w
σ
f 
p

. (3.10)
Onthesamewaywecanprove(3.2)for

F
n

f , using (2.6)and(2.26). 
References
[1] W. Gawronski and U. Stadtm
¨
uller, “Approximation of continuous functions by generalized
Favard operators,” Journal of Approximation Theory, vol. 34, no. 4, pp. 384–396, 1982.
[2] J. Favard, “Sur les multiplicateurs d’interpolation,” Journal de Math
´
ematiques Pures et Appliqu
´
ees,
vol. 23, pp. 219–247, 1944.
[3] M. Becker, P. L. Butzer, and R. J. Nessel, “Saturation for Favard oper ators in weighted function
spaces,” Studia Mathematica, vol. 59, no. 2, pp. 139–153, 1976.
[4] M. Becker, “Inverse theorems for Favard operators in polynomial weight spaces,” Commenta-
tiones Mathematicae, vol. 22, no. 2, pp. 165–173, 1980/81.
[5] P. Pych-Taberska, “On the generalized Favard operators,” Functiones et Approximatio Commen-
tarii Mathematici, vol. 26, pp. 265–273, 1998.
[6] P. Pych-Taberska and G. Nowak, “Approximation properties of the generalized Favard-
Kantorovich operators,” Commentationes Mathematicae, vol. 39, pp. 139–152, 1999.
[7] G. Nowak and P. Pych-Taberska, “Some pr operties of the generalized Favard-Durrmeyer opera-
tors,” Functiones et Approximatio Commentarii Mathematici, vol. 29, pp. 103–112, 2001.
[8] G. A. Anastassiou, C. Cottin, and H. H. Gonska, “Global smoothness of approximating func-
tions,” Analysis, vol. 11, no. 1, pp. 43–57, 1991.
[9] G. Nowak, “Direct theorems for gener alized Favard-Kantorovich and Favard-Durrmeyer oper-
ators in exponential function spaces,” to appear in Ukrainian Mathematical Journal.
[10] W. Kratz and U. Stadtm
¨
uller, “On the uniform modulus of continuity of certain discrete approx-
imation operators,” Journal of Approximation Theory, vol. 54, no. 3, pp. 326–337, 1988.

Grzegorz Nowak: Higher School of Marketing and Management, Ostroroga 9a,
64-100 Leszno, Poland
Email address:
Aneta Sikorska-Nowak: Faculty of Mathematics and Computer Science,
Adam Mickiewicz University, Umultowska 87, 61-614 Pozna
´
n, Poland
Email address: