# Approximation by Lupas-Type Operators and Szász-Mirakyan-Type Operators

Authors: Hee Sun Jung; Ryozi Sakai
Journal: Journal of Applied Mathematics
 (2012)

---

## Abstract

Lupas-type operators and Szász-Mirakyan-type operators are the modifications of Bernstein polynomials to infinite intervals. In this paper, we investigate the convergence of Lupas-type operators and Szász-Mirakyan-type operators on[
0
,
∞
).


---

## Body

## 1. Introduction and Main Results

Forf
∈
C
(
[
0,1
]
), Bernstein operator B
n
f
(
x
) is defined as follows: Let(1.1)
p
n
,
k
(
x
)
=
(
n
k
)
x
k
(
1
-
x
)
n
-
k
,
0
⩽
k
⩽
n
,
and then we define(1.2)
(
B
n
f
)
(
x
)
=
∑
k
=
0
n
p
n
,
k
(
x
)
f
(
k
n
)
.
Derriennic [1] gave a modified operator of B
n
f such as(1.3)
(
M
n
*
f
)
(
x
)
=
(
n
+
1
)
∑
k
=
0
n
p
n
,
k
(
x
)
∫
0
1
p
n
,
k
(
t
)
f
(
t
)
d
t
,
and obtained the result that for f
∈
C
(
2
)
(
[
0,1
]
),(1.4)
lim
⁡
n
→
∞
(
(
M
n
*
f
)
(
x
)
-
f
(
x
)
)
=
(
1
-
2
x
)
f
′
(
x
)
+
x
(
1
-
x
)
f
′′
(
x
)
.
Lupas investigated a family of linear positive operators which mapped the class of all bounded and continuous functions on [
0
,
∞
) into C
[
0
,
∞
) such that(1.5)
(
L
n
f
)
(
x
)
=
∑
k
=
0
∞
(
n
+
k
-
1
k
)
x
k
(
1
+
x
)
n
+
k
f
(
k
n
)
,
x
∈
[
0
,
∞
)
.
Moreover, Sahai and Prasad [2] modified Lupas operators as follows: Let f be integrable on [
0
,
∞
) and let n be a positive integer. Then we define(1.6)
(
M
n
[
f
]
)
(
x
)
=
(
n
-
1
)
∑
k
=
0
∞
P
n
,
k
(
x
)
∫
0
∞
P
n
,
k
(
y
)
f
(
y
)
d
y
,
x
∈
[
0
,
∞
)
,
where(1.7)
P
n
,
k
(
x
)
=
(
n
+
k
-
1
k
)
x
k
(
1
+
x
)
n
+
k
.
In this paper, we assume that n is a positive integer. Then they obtained the following;Theorem 1.1 (see [2], Theorem 1).
Iff is integrable on [
0
,
∞
) and admits its (
r
+
1
)th and (
r
+
2
)th derivatives, which are bounded at a point x
∈
[
0
,
∞
), and f
(
r
)
(
x
)
=
O
(
x
α
) (α is a positive integer ⩾
2) as x
→
∞, then
(1.8)
lim
⁡
n
→
∞
n
(
(
M
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
)
=
(
r
+
1
)
(
1
-
2
x
)
f
(
r
+
1
)
(
x
)
+
x
(
1
-
x
)
f
(
r
+
2
)
(
x
)
.Theorem1.1 holds only for bounded x
⩽
K, so it does not mean the norm convergence on [
0
,
∞
). In this paper, we improve Theorem 1.1 with respect to the norm convergence on [
0
,
∞
).Let0
<
p
⩽
∞ and let w be a positive weight, that is, w
(
x
)
⩾
0 for x
∈
ℝ. For a function g on [
0
,
∞
), we define the norm by
(1.9)
‖
g
‖
L
p
(
[
0
,
∞
)
)
:
=
{
(
∫
[
0
,
∞
)
|
g
(
t
)
|
p
d
t
)
1
/
p
,
0
<
p
<
∞
sup
⁡
[
0
,
∞
)
|
g
(
t
)
|
,
p
=
∞
.
For convenience, for nonnegative integers n
⩾
2, r, and n
-
r
-
2
⩾
0, we let
(1.10)
A
n
,
r
:
=
(
n
-
1
)
!
(
n
-
2
)
!
(
n
-
r
-
2
)
!
(
n
+
r
-
1
)
!
.
Then we have the following results:Theorem 1.2.
Let0
<
p
⩽
∞. Let α and r be nonnegative integers and n
-
r
-
2
⩾
0. Let f
∈
C
(
r
+
1
)
(
[
0
,
∞
)
) satisfy
(1.11)
|
f
(
r
)
(
x
)
|
⩽
O
(
1
)
(
x
+
1
)
α
,
|
f
(
r
+
1
)
(
x
)
|
⩽
O
(
1
)
(
x
+
1
)
α
+
2
.
Then we have uniformly for f and n,
(1.12)
|
A
n
,
r
(
M
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
=
O
(
1
n
1
/
3
)
(
x
+
1
)
α
+
2
.
In particular, if ∥
(
x
+
1
)
α
+
2
w
(
x
)
∥
L
p
(
[
0
,
∞
)
)
<
∞, then we have uniformly for n,
(1.13)
‖
(
A
n
,
r
(
M
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
)
w
(
x
)
‖
L
p
(
[
0
,
∞
)
)
=
O
(
1
n
1
/
3
)
.Remark 1.3.
(a) We see that for nonnegative integersn
⩾
2, r, and n
-
r
-
2
⩾
0,
(1.14)
(
n
-
1
)
!
(
n
-
2
)
!
(
n
-
r
-
2
)
!
(
n
+
r
-
1
)
!
⟶
1
as
n
⟶
∞
.
(b) The following weight is useful.(1.15)
w
λ
(
x
)
=
1
1
+
x
λ
{
λ
>
1
p
+
α
+
2
,
0
<
p
<
∞
,
λ
⩾
α
+
2
,
p
=
∞
.Let(1.16)
ψ
(
x
)
:
=
1
1
+
x
.Theorem 1.4.
Letr and β be nonnegative integers and n
-
r
-
2
⩾
0. Let f
∈
C
(
r
+
2
)
(
[
0
,
∞
)
) satisfy
(1.17)
‖
f
(
r
+
1
)
(
x
)
ψ
2
β
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
<
∞
,
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
<
∞
.
Then we have uniformly for f and n,
(1.18)
|
A
n
,
r
(
M
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
ψ
2
β
+
2
(
x
)
⩽
O
(
1
n
)
(
‖
f
(
r
+
1
)
(
x
)
ψ
2
β
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
)
.Let us define the weighted modulus of smoothness by(1.19)
ω
k
(
f
;
η
;
t
)
:
=
sup
⁡
0
⩽
h
⩽
t
‖
Δ
h
k
f
(
⋅
)
η
(
⋅
)
‖
L
∞
(
[
0
,
∞
)
)
,
t
⩾
0
,
k
=
1,2
,
where
(1.20)
Δ
h
1
f
(
x
)
=
f
(
x
+
h
)
-
f
(
x
)
,
Δ
h
2
f
(
x
)
=
f
(
x
)
-
2
f
(
x
+
h
)
+
f
(
x
+
2
h
)
.Theorem 1.5.
Letβ and r be nonnegative integers and n
-
r
-
2
⩾
0. Let f
∈
C
r
(
[
0
,
∞
)
). Then we have uniformly for f and n,
(1.21)
‖
(
A
n
,
r
(
M
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
)
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
(
1
n
ω
1
(
f
(
r
)
;
ψ
2
β
+
1
;
1
n
)
+
ω
2
(
f
(
r
)
;
ψ
2
β
;
1
n
)
)
.The Szász-Mirakyan operators are also generalizations of Bernstein polynomials on infinite intervals. They are defined by:(1.22)
S
n
(
f
)
(
x
)
=
∑
k
=
0
∞
S
n
,
k
(
x
)
f
(
k
n
)
,
where
(1.23)
S
n
,
k
(
x
)
=
e
-
n
x
(
n
x
)
k
k
!
.In [3], the class of Szász-Mirakyan operators S
n
;
r
,
q
(
f
;
x
) was defined as follows:
(1.24)
S
n
;
r
,
q
(
f
;
x
)
:
=
1
A
r
(
n
x
)
∑
k
=
0
∞
(
n
x
)
r
k
(
r
k
)
!
f
(
r
k
n
+
q
)
,
x
∈
[
0
,
∞
)
,
where q
>
0 and
(1.25)
A
r
(
t
)
=
∑
k
=
0
∞
t
r
k
(
r
k
)
!
,
t
∈
[
0
,
∞
)
.Theorem 1.6 (see [3]).
Letq
>
0 and r
∈
ℕ be fixed numbers. Then there exists M
q
,
r
=
const. >
0 depending only on q and r such that, for every uniformly continuous and bounded function f
(
x
)
e
-
q
x on [
0
,
∞
), the following inequalities hold;(a)
(1.26)
‖
(
S
n
;
q
,
r
(
f
;
x
)
-
f
(
x
)
)
φ
(
x
)
e
-
q
x
‖
L
∞
(
[
0
,
∞
)
)
⩽
M
q
,
r
1
n
+
q
(
‖
f
′
(
x
)
e
-
q
x
‖
L
∞
(
[
0
,
∞
)
)
+
‖
f
′′
(
x
)
e
-
q
x
‖
L
∞
(
[
0
,
∞
)
)
)
;
(b)
(1.27)
‖
(
S
n
;
q
,
r
(
f
;
x
)
-
f
(
x
)
)
φ
(
x
)
e
-
q
x
‖
L
∞
(
[
0
,
∞
)
)
⩽
M
q
,
r
1
n
+
q
(
δ
n
,
q
ω
1
(
f
;
e
-
q
x
;
δ
n
,
q
)
+
ω
2
(
f
;
e
-
q
x
;
δ
n
,
q
)
)
,
where δ
n
,
q
:
=
(
n
+
q
)
-
1
/
2. (c)
for every fixedx
∈
[
0
,
∞
), we have for every continuous f with f
(
j
)
(
x
)
e
-
q
x, j
=
0,1
,
2, bounded on [
0
,
∞
),
(1.28)
lim
⁡
n
→
∞
n
(
S
n
;
q
,
r
(
f
;
x
)
-
f
(
x
)
)
=
-
q
x
f
′
(
x
)
+
x
2
f
′′
(
x
)
.Now, we modify the Szász-Mirakyan operators as follows: letf be integrable on [
0
,
∞
), then we define(1.29)
(
Q
n
,
β
[
f
]
)
(
x
)
=
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
(
y
)
f
(
y
)
d
y
,
x
∈
[
0
,
∞
)
,
where β is a nonnegative integer. Then we have the following results:Theorem 1.7.
Letα, β and r be nonnegative integers. Let f
∈
C
(
r
+
1
)
(
[
0
,
∞
)
) satisfies
(1.30)
|
f
(
r
)
(
x
)
|
⩽
O
(
1
)
e
β
x
(
x
+
1
)
α
,
|
f
(
r
+
1
)
(
x
)
|
⩽
O
(
1
)
e
β
x
(
x
+
1
)
α
+
2
.
Then one has uniformly for f and n,
(1.31)
|
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
=
O
(
1
n
1
/
3
)
e
β
x
(
x
+
1
)
α
+
2
.
In particular, let 0
<
p
⩽
∞. If one supposes ∥
e
β
x
(
x
+
1
)
α
+
2
w
(
x
)
∥
L
p
(
[
0
,
∞
)
)
<
∞, then one has uniformly for f and n,
(1.32)
‖
(
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
)
w
(
x
)
‖
L
p
(
[
0
,
∞
)
)
=
O
(
1
n
1
/
3
)
.Remark 1.8.
(a) We note that for nonnegative integersβ and r,
(1.33)
(
n
+
β
n
)
r
⟶
1
as
n
⟶
∞
.
(b) The following weight is useful.(1.34)
w
λ
,
β
(
x
)
=
e
-
β
x
w
λ
(
x
)
,
where w
λ
(
x
) is defined in Remark 1.3.Theorem 1.9.
Letβ, γ, and r be nonnegative integers. Let f
∈
C
(
r
+
2
)
(
[
0
,
∞
)
) satisfies
(1.35)
‖
f
(
r
+
1
)
(
x
)
e
-
β
x
ψ
2
γ
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
<
∞
,
‖
f
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
<
∞
.
Then one has uniformly for f and n,
(1.36)
|
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
|
⩽
O
(
1
n
)
(
‖
f
(
r
+
1
)
(
x
)
e
-
β
x
ψ
2
γ
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
f
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
)
.Theorem 1.10.
Letβ, γ, and r be nonnegative integers. Then one has for f
∈
C
r
(
[
0
,
∞
)
),
(1.37)
‖
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
(
1
n
ω
1
(
f
;
e
-
β
x
ψ
2
γ
+
1
(
x
)
;
1
n
)
+
ω
2
(
f
;
e
-
β
x
ψ
2
γ
(
x
)
;
1
n
)
)
.


## 2. Proofs of Results

First, we will prove results for Lupas-type operators such as Theorems1.2, 1.4, and 1.5. To prove theorems, we need some lemmas.Lemma 2.1.
Letm and r be nonnegative integers and n
>
m
+
r
+
1. Let
(2.1)
T
n
,
m
,
r
(
x
)
:
=
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞
P
n
-
r
,
k
+
r
(
y
)
(
y
-
x
)
m
d
y
.
Then(i)
T
n
,
0
,
r
(
x
)
=
1,
(ii)
(2.2)
T
n
,
1
,
r
(
x
)
=
(
r
+
1
)
(
1
+
2
x
)
(
n
-
r
+
2
)
,
T
n
,
2
,
r
(
x
)
=
2
(
n
-
1
)
x
(
1
+
x
)
(
n
-
r
-
2
)
(
n
-
r
-
3
)
+
(
r
+
1
)
(
r
+
2
)
(
1
+
2
x
)
2
(
n
-
r
-
2
)
(
n
-
r
-
3
)
;
(iii)
form
⩾
1,
(2.3)
(
n
-
m
-
r
-
2
)
T
n
,
m
+
1
,
r
(
x
)
=
x
(
1
+
x
)
(
T
n
,
m
,
r
′
(
x
)
+
2
m
T
n
,
m
-
1
,
r
(
x
)
)
+
(
m
+
r
+
1
)
(
1
+
2
x
)
T
n
,
m
,
r
,
where n
>
m
+
r
+
2;
(iv)
form
⩾
0,
(2.4)
T
n
,
m
,
r
(
x
)
=
O
(
1
n
[
(
m
+
1
)
/
2
]
)
q
n
,
m
,
r
(
x
)
,
where q
n
,
m
,
r
(
x
) is a polynomial of degree ⩽
m such that the coefficients are bounded independently of n and they are positive for n
>
m
+
r
+
1.Proof.
(i), (ii), and (iii) have been proved in [2, Lemma 1]. So we may show only the part of (2.4). For m
=
1,2, (2.4) holds. Let us assume (2.4) for m
(
⩾
2
). We note
(2.5)
T
n
,
m
,
r
′
(
x
)
=
O
(
1
n
[
(
m
+
1
)
/
2
]
)
q
m
,
r
′
(
x
)
,
q
m
,
r
′
(
x
)
∈
P
m
-
1
.
So, we have by the assumption of induction,
(2.6)
(
n
-
m
-
r
-
2
)
T
n
,
m
+
1
,
r
(
x
)
=
x
(
1
+
x
)
(
T
′
n
,
m
,
r
(
x
)
+
2
m
T
n
,
m
-
1
,
r
(
x
)
)
+
(
m
+
r
+
1
)
(
1
+
2
x
)
T
n
,
m
,
r
⩽
x
(
1
+
x
)
(
C
1
n
[
(
m
+
1
)
/
2
]
q
m
,
r
′
(
x
)
+
2
m
1
n
[
m
/
2
]
q
m
-
1
,
r
(
x
)
)
+
(
m
+
r
+
1
)
(
1
+
2
x
)
1
n
[
(
m
+
1
)
/
2
]
q
m
,
r
(
x
)
.
Here, if m is even, then
(2.7)
[
m
+
1
2
]
+
1
=
m
2
+
1
=
m
+
2
2
=
[
m
+
2
2
]
,
[
m
2
]
+
1
=
m
2
+
1
=
m
+
2
2
=
[
m
+
2
2
]
,
and if m is odd, then
(2.8)
[
m
+
1
2
]
+
1
=
m
+
1
2
+
1
=
[
m
+
2
2
]
,
[
m
2
]
+
1
=
m
-
1
2
+
1
=
m
+
1
2
=
[
m
+
2
2
]
.
Hence, we have
(2.9)
T
n
,
m
+
1
,
r
(
x
)
=
O
(
1
n
[
(
m
+
2
)
/
2
]
)
q
n
,
m
+
1
,
r
(
x
)
,
and here we see that q
n
,
m
+
1
,
r
(
x
) is a polynomial of degree ⩽
m
+
1 such that the coefficients of q
n
,
m
+
1
,
r
(
x
) are bounded independently of n. Moreover, we see from (2.6) that the coefficients of q
n
,
m
+
1
,
r
(
x
) are positive for n
>
m
+
r
+
2.Lemma 2.2 (see [2, Lemma 2]).
Letr be a nonnegative integer and n
-
r
-
2
⩾
0. Then one has for f
∈
C
r
(
[
0
,
∞
)
):
(2.10)
(
M
n
[
f
]
)
(
r
)
(
x
)
=
(
n
-
r
-
1
)
!
(
n
+
r
-
1
)
!
(
n
-
1
)
!
(
n
-
2
)
!
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞
P
n
-
r
,
k
+
r
(
y
)
f
(
r
)
(
y
)
d
y
.Let(2.11)
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
=
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞
P
n
-
r
,
k
+
r
(
y
)
f
(
r
)
(
y
)
d
y
.
Then we have(2.12)
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
=
A
n
,
r
(
M
n
[
f
]
)
(
r
)
(
x
)
,
where A
n
,
r is defined by (1.10).Proof of Theorem1.2.
Let|
y
-
x
|
⩽
1. By the second inequality in (1.11),
(2.13)
|
f
(
r
)
(
y
)
-
f
(
r
)
(
x
)
|
=
|
y
-
x
|
|
f
(
r
+
1
)
(
ξ
)
|
⩽
O
(
1
)
|
y
-
x
|
|
ξ
α
+
2
|
⩽
O
(
1
)
|
y
-
x
|
(
x
+
1
)
α
+
2
.
Let ɛ
:
=
n
-
γ
,
0
<
γ
<
1,
(2.14)
|
(
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
)
|
=
|
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
×
(
∫
|
y
-
x
|
<
ɛ
P
n
-
r
,
k
+
r
(
y
)
|
f
(
r
)
(
y
)
-
f
(
r
)
(
x
)
|
d
y
+
∫
|
y
-
x
|
⩾
ɛ
P
n
-
r
,
k
+
r
(
y
)
|
f
(
r
)
(
y
)
-
f
(
r
)
(
x
)
|
d
y
)
∑
k
=
0
∞
|
=
:
A
+
B
.
First, we see by (2.13) and Lemma 2.1,
(2.15)
A
=
O
(
1
)
|
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
|
y
-
x
|
<
ɛ
P
n
-
r
,
k
+
r
(
y
)
|
y
-
x
|
(
x
+
1
)
α
+
2
d
y
|
⩽
O
(
1
)
ɛ
|
T
n
,
0
,
r
(
x
)
|
(
x
+
1
)
α
+
2
=
O
(
1
)
ɛ
(
x
+
1
)
α
+
2
.
Next, we estimate B. By the first inequality in (1.11),
(2.16)
B
⩽
C
|
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
P
n
-
r
,
k
+
r
(
y
)
(
|
f
(
r
)
(
y
)
|
+
|
f
(
r
)
(
x
)
|
)
d
y
|
⩽
C
|
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
P
n
-
r
,
k
+
r
(
y
)
(
(
y
+
1
)
α
+
(
x
+
1
)
α
)
d
y
|
.
Here, using
(2.17)
(
y
+
1
)
α
=
(
(
y
-
x
)
+
x
+
1
)
α
=
∑
i
=
0
α
(
α
i
)
(
y
-
x
)
i
(
x
+
1
)
α
-
i
and the notation:
(2.18)
〈
i
〉
=
{
1
,
(
i
:
odd
)
0
,
(
i
:
even
)
,
we have
(2.19)
B
⩽
C
|
(
n

-

r
-
1
)
|
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
P
n
-
r
,
k
+
r
(
y
)
×
(
∑
i
=
1
α
(
α
i
)
(
y
-
x
)
i
(
x
+
1
)
α
-
i
+
2
(
x
+
1
)
α
)
d
y
⩽
C
|
(
n
-
r
-
1
)
|
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
P
n
-
r
,
k
+
r
(
y
)
×
(
∑
i
=
1
α
(
α
i
)
(
y
-
x
)
i
|
y
-
x
ɛ
|
〈
i
〉
(
x
+
1
)
α
-
i
)
d
y
+
C
|
(
n
-
r
-
1
)
|
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
P
n
-
r
,
k
+
r
(
y
)
(
y
-
x
ɛ
)
2
(
x
+
1
)
α
d
y
:
=
B
1
+
B
2
.
Then, we obtain
(2.20)
B
1
⩽
C
(
∑
i
=
1
α
(
α
i
)
|
T
n
,
i
+
〈
i
〉
,
r
|
(
1
ɛ
)
〈
i
〉
(
x
+
1
)
α
-
i
)
⩽
C
∑
i
=
1
α
(
α
i
)
O
(
n
γ
〈
i
〉
n
[
(
i
+
〈
i
〉
+
1
)
/
2
]
)
|
q
n
,
i
+
〈
i
〉
,
r
(
x
)
|
(
x
+
1
)
α
-
i
⩽
O
(
1
n
[
(
i
+
〈
i
〉
+
1
)
/
2
]
-
γ
〈
i
〉
)
(
x
+
1
)
α
+
〈
i
〉
⩽
O
(
1
n
1
-
γ
)
(
x
+
1
)
α
+
1
.
Here, we used the following that for i
⩾
1,
(2.21)
[
i
+
〈
i
〉
+
1
2
]
-
γ
〈
i
〉
⩾
1
-
γ
,
because
(2.22)
[
i
+
〈
i
〉
+
1
2
]
-
γ
〈
i
〉
=
{
i
+
1
2
-
γ
,
i
:
odd
,
i
2
,
i
:
even
.
And we know that
(2.23)
B
2
⩽
C
|
T
n
,
2
,
r
(
x
)
|
(
1
ɛ
)
2
x
α
⩽
O
(
1
n
[
3
/
2
]
)
|
q
n
,
2
,
r
(
x
)
|
(
1
ɛ
)
2
x
α
⩽
O
(
n
2
γ
n
[
3
/
2
]
)
|
q
n
,
2
,
r
(
x
)
|
x
α
⩽
O
(
1
n
1
-
2
γ
)
(
x
+
1
)
α
+
2
.
Thus, we obtain
(2.24)
B
⩽
O
(
1
n
1
-
2
γ
)
(
x
+
1
)
α
+
2
.
Therefore, we have uniformly on n,
(2.25)
|
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
⩽
O
(
1
n
γ
)
(
x
+
1
)
α
+
2
+
O
(
1
n
1
-
2
γ
)
(
x
+
1
)
α
+
2
.
Here, if we let γ
=
1
/
3, then we have
(2.26)
|
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
=
O
(
1
n
1
/
3
)
(
x
+
1
)
α
+
2
,
that is, (1.12) is proved. So, we also have a norm convergence (1.13).Proof of Theorem1.4.
We know that forf
∈
C
(
r
+
2
)
(
[
0
,
∞
)
),
(2.27)
f
(
r
)
(
t
)
=
f
(
r
)
(
x
)
+
f
(
r
+
1
)
(
x
)
(
t
-
x
)
+
∫
x
t
(
t
-
u
)
f
(
r
+
2
)
(
u
)
d
u
,
(2.28)
|
∫
x
t
(
t
-
u
)
f
(
r
+
2
)
(
u
)
d
u
|
⩽
C
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
(
(
1
+
x
)
2
β
+
(
1
+
t
)
2
β
)
(
t
-
x
)
2
,
where ψ
(
t
)
=
1
/
(
1
+
x
). Then we obtain from (2.10) and (2.27),
(2.29)
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
=
f
(
r
)
(
x
)
+
f
(
r
+
1
)
(
x
)
T
n
,
1
,
r
(
x
)
+
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞
P
n
-
r
,
k
+
r
(
y
)
∫
x
y
(
y
-
u
)
f
(
r
+
2
)
(
u
)
d
u
d
y
and from (2.28),
(2.30)
|
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞
P
n
-
r
,
k
+
r
(
y
)
∫
x
y
(
y
-
u
)
f
(
r
+
2
)
(
u
)
d
u
d
y
|
⩽
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
×
(
(
1
+
x
)
2
β
|
T
n
,
2
,
r
(
x
)
|
+
|
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞
P
n
-
r
,
k
+
r
(
y
)
(
1
+
y
)
2
β
(
y
-
x
)
2
d
y
|
)
.
Using (
1
+
y
)
2
β
⩽
C
(
(
y
-
x
)
2
β
+
(
1
+
x
)
2
β
), we have
(2.31)
|
(
n

-


r
-

1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞

P
n
-
r
,
k
+
r
(
y
)
(
1

+

y
)
2
β
(
y

-

x
)
2
d
y
|
⩽
C
(
|
T
n
,
2
β
+
2
,
r
(
x
)
|

+

(
1

+

x
)
2
β
|
T
n
,
2
,
r
(
x
)
|
)
.
Therefore, we have
(2.32)
|
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
ψ
2
β
+
2
(
x
)
⩽
|
f
(
r
+
1
)
(
x
)
ψ
2
β
+
1
(
x
)
|
|
T
n
,
1
,
r
(
x
)
|
ψ
(
x
)
+
C
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
(
1
+
x
)
2
β
|
T
n
,
2
,
r
(
x
)
|
ψ
2
β
+
2
(
x
)
+
C
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
|
T
n
,
2
β
+
2
,
r
(
x
)
|
ψ
2
β
+
2
(
x
)
⩽
O
(
1
n
)
|
f
(
r
+
1
)
(
x
)
ψ
2
β
+
1
(
x
)
|
|
g
n
,
1
,
r
(
x
)
|
ψ
(
x
)
+
O
(
1
n
)
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
(
1
+
x
)
2
β
|
g
n
,
2
,
r
(
x
)
|
ψ
2
β
+
2
(
x
)
+
O
(
1
n
β
+
1
)
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
|
g
n
,
2
β
+
2
,
r
(
x
)
|
ψ
2
β
+
2
(
x
)
.
Since we know that for x
∈
[
0
,
∞
),
(2.33)
|
g
n
,
1
,
r
(
x
)
|
ψ
(
x
)
⩽
C
,
(
1
+
x
)
2
β
|
g
n
,
2
,
r
(
x
)
|
ψ
2
β
+
2
(
x
)
⩽
C
,
|
g
n
,
2
β
+
2
,
r
(
x
)
|
ψ
2
β
+
2
(
x
)
⩽
C
,
we have
(2.34)
|
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
ψ
2
β
+
2
(
x
)
⩽
O
(
1
n
)
(
‖
f
(
r
+
1
)
(
x
)
ψ
2
β
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
f
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
)
.Lemma 2.3.
Letr and β be nonnegative integers and n
-
r
-
2
⩾
0. Let f
∈
C
(
r
)
(
[
0
,
∞
)
) satisfies
(2.35)
‖
f
(
r
)
ψ
2
β
‖
L
∞
(
[
0
,
∞
)
)
<
∞
.
Then one has uniformly for n, f and x
∈
[
0
,
∞
),
(2.36)
|
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
|
ψ
2
β
(
x
)
⩽
C
‖
f
(
r
)
ψ
2
β
‖
L
∞
(
[
0
,
∞
)
)
.Proof.
Using(
1
+
y
)
2
β
⩽
C
(
(
y
-
x
)
2
β
+
(
1
+
x
)
2
β
), we have
(2.37)
|
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞
P
n
-
r
,
k
+
r
(
y
)
(
1
+
y
)
2
β
d
y
|
⩽
C
(
ψ
-
2
β
(
x
)
+
|
T
n
,
2
β
,
r
(
x
)
|
)
.
The assumption (2.35) means
(2.38)
|
f
(
r
)
(
y
)
|
⩽
C
(
1
+
y
)
2
β
.
Then we can obtain by (2.10),
(2.39)
|
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
|
⩽
C
|
(
n
-
r
-
1
)
∑
k
=
0
∞
P
n
+
r
,
k
(
x
)
∫
0
∞
P
n
-
r
,
k
+
r
(
y
)
(
1
+
y
)
2
β
d
y
|
‖
f
(
r
)
ψ
2
β
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
(
ψ
-
2
β
(
x
)
+
|
T
n
,
2
β
,
r
(
x
)
|
)
‖
f
(
r
)
ψ
2
β
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
(
ψ
-
2
β
(
x
)
+
O
(
1
n
β
)
|
q
n
,
2
β
,
r
(
x
)
|
)
‖
f
(
r
)
ψ
2
β
‖
L
∞
(
[
0
,
∞
)
)
.
Consequently, since |
q
n
,
2
β
,
r
(
x
)
|
ψ
2
β
(
x
) is uniformly bounded on [
0
,
∞
), we have the result.The Steklov function[
f
]
h
(
x
) for f
∈
C
(
[
0
,
∞
)
) is defined as follows:(2.40)
[
f
]
h
(
x
)
:
=
4
h
2
∬
0
h
/
2
[
2
f
(
x
+
s
+
t
)
-
f
(
x
+
2
(
s
+
t
)
)
]
d
s
d
t
,
x
⩾
0
,
h
>
0
.Then for the Steklov function[
f
]
h
(
x
) with respect to f
∈
C
(
[
0
,
∞
)
), we have the following properties.Lemma 2.4 (cf.[4]).
Letf
(
x
)
∈
C
(
[
0
,
∞
)
) and η
(
x
) be a positive and nonincreasing function on [
0
,
∞
). Then (i) [
f
]
h
(
x
)
∈
C
2
(
[
0
,
∞
)
);
(ii)(2.41)
‖
(
[
f
]
h
(
x
)
-
f
(
x
)
)
η
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
ω
2
(
f
;
η
;
h
2
)
;
(iii)(2.42)
‖
[
f
]
h
′
(
x
)
η
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
4
h
ω
1
(
f
;
η
;
h
2
)
η
(
x
)
η
(
x
+
(
h
/
2
)
)
+
1
h
ω
1
(
f
;
η
;
h
)
η
(
x
)
η
(
x
+
h
)
;
(iv)(2.43)
‖
[
f
]
h
′′
(
x
)
η
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
4
h
2
[
2
ω
2
(
f
;
η
;
h
2
)
+
1
4
ω
2
(
f
;
η
;
h
)
]
.Proof.
(i) Forf
∈
C
(
[
0
,
∞
)
), we have the Steklov functions [
f
]
h
′
(
x
) and [
f
]
h
′′
(
x
) as follows. We note
(2.44)
[
f
]
h
(
x
)
=
4
h
2
∫
0
h
/
2
(
∫
x
x
+
h
/
2
2
f
(
u
+
t
)
d
u
-
∫
x
x
+
h
1
2
f
(
u
+
2
t
)
d
u
)
d
t
,
x
⩾
0
,
h
>
0
.
Then, we can see from (2.44),
(2.45)
[
f
]
h
′
(
x
)
=
4
h
2
∫
0
h
/
2
[
2
(
f
(
x
+
h
2
+
t
)
-
f
(
x
+
t
)
)
-
1
2
(
f
(
x
+
h
+
2
t
)
-
f
(
x
+
2
t
)
)
]
d
t
=
4
h
2
∫
0
h
/
2
[
2
Δ
h
/
2
1
f
(
x
+
t
)
-
1
2
Δ
h
1
f
(
x
+
2
t
)
]
d
t
.
Similarly to (2.44), we know
(2.46)
[
f
]
h
′
(
x
)
=
4
h
2
[
2
∫
x
x
+
h
/
2
(
f
(
u
+
h
2
)
-
f
(
u
)
)
d
u
-
1
4
∫
x
x
+
h
(
f
(
u
+
h
)
-
f
(
u
)
)
d
u
]
.
Therefore, we have from (2.46),
(2.47)
[
f
]
h
′′
(
x
)
=
4
h
2
[
2
(
f
(
x
+
h
)
-
2
f
(
x
+
h
2
)
+
f
(
x
)
)
-
1
4
(
f
(
x
+
2
h
)
-
2
f
(
x
+
h
)
+
f
(
x
)
)
]
=
4
h
2
[
2
Δ
h
/
2
2
f
(
x
)
-
1
4
Δ
h
2
f
(
x
)
]
.
Therefore, (i) is proved.
(ii) We easily see from (2.44) that
(2.48)
|
(
f
(
x
)
-
[
f
]
h
(
x
)
)
η
(
x
)
|
=
|
4
h
2
∬
0
h
/
2
Δ
s
+
t
2
f
(
x
)
η
(
x
)
d
s
d
t
|
⩽
ω
2
(
f
;
η
;
h
2
)
.
(iii) From (2.46), we have
(2.49)
|
[
f
]
h
′
(
x
)
η
(
x
)
|
⩽
|
4
h
2
∫
0
h
/
2
2
(
Δ
h
/
2
1
f
(
x
+
t
)
η
(
x
+
t
)
)
η
(
x
)
η
(
x
+
t
)
d
t
|
+
|
4
h
2
∫
0
h
/
2
1
2
(
Δ
h
1
f
(
x
+
2
t
)
η
(
x
+
2
t
)
)
η
(
x
)
η
(
x
+
2
t
)
d
t
|
⩽
4
h
ω
1
(
f
;
η
;
h
2
)
η
(
x
)
η
(
x
+
h
/
2
)
+
1
h
ω
1
(
f
;
η
;
h
)
η
(
x
)
η
(
x
+
h
)
.
(iv) From (2.47), we have
(2.50)
|
[
f
]
h
′′
(
x
)
η
(
x
)
|
⩽
4
h
2
[
2
ω
2
(
f
;
η
;
h
2
)
+
1
4
ω
2
(
f
;
η
;
h
)
]
.Proof of Theorem1.5.
We know that forf
(
x
)
∈
C
r
(
[
0
,
∞
)
),
(2.51)
[
f
]
h
(
r
)
(
x
)
=
[
f
(
r
)
]
h
(
x
)
,
[
f
]
h
(
r
+
1
)
(
x
)
=
[
f
(
r
)
]
h
′
(
x
)
,
[
f
]
h
(
r
+
2
)
(
x
)
=
[
f
(
r
)
]
h
′′
(
x
)
.
Then, we have
(2.52)
‖
(
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
)
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
‖
(
M
̃
n
[
f
-
[
f
]
h
]
)
(
r
)
(
x
)
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
(
(
M
̃
n
[
[
f
]
h
]
)
(
r
)
(
x
)
-
[
f
(
r
)
]
h
(
x
)
)
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
[
[
f
(
r
)
]
h
(
x
)
-
f
(
r
)
(
x
)
]
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
.
From (2.51) and (2.41) of Lemma 2.4,
(2.53)
‖
(
M
̃
n
[
f
-
[
f
]
h
]
)
(
r
)
(
x
)
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
‖
[
f
(
r
)
(
x
)
-
[
f
]
h
(
r
)
(
x
)
]
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
=
‖
[
f
(
r
)
(
x
)
-
[
f
(
r
)
]
h
(
x
)
]
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
ω
2
(
f
(
r
)
;
ψ
2
β
+
2
;
h
)
.
Here, we suppose 0
<
h
⩽
1 and then we know that
(2.54)
ψ
(
x
)
ψ
(
x
+
h
)
⩽
2
,
ψ
(
x
)
ψ
(
x
+
h
/
2
)
⩽
2
.
From Theorem 1.4, (2.51), (2.42), and (2.43) of Lemma 2.4, we have
(2.55)
‖
(
(
M
̃
n
[
[
f
]
h
]
)
(
r
)
(
x
)
-
[
f
]
h
(
r
)
(
x
)
)
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
O
(
1
n
)
(
‖
[
f
]
h
(
r
+
1
)
(
x
)
ψ
2
β
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
[
f
]
h
(
r
+
2
)
(
x
)
ψ
2
β
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
)
⩽
O
(
1
n
)
(
1
h
ω
1
(
f
(
r
)
;
ψ
2
β
+
1
;
h
)
+
1
h
2
ω
2
(
f
(
r
)
;
ψ
2
β
;
h
)
)
.
Therefore, we have
(2.56)
‖
(
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
)
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
O
(
1
n
)
(
1
h
ω
1
(
f
(
r
)
;
ψ
2
β
+
1
;
h
)
+
1
h
2
ω
2
(
f
(
r
)
;
ψ
2
β
;
h
)
)
+
ω
2
(
f
(
r
)
;
ψ
2
β
+
2
;
h
)
.
If we let h
=
1
/
n, then
(2.57)
‖
(
(
M
̃
n
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
)
ψ
2
β
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
(
1
n
ω
1
(
f
(
r
)
;
ψ
2
β
+
1
;
1
n
)
+
ω
2
(
f
(
r
)
;
ψ
2
β
;
1
n
)
)
,
because ω
2
(
f
(
r
)
;
ψ
2
β
+
2
;
1
/
n
)
⩽
ω
2
(
f
(
r
)
;
ψ
2
β
;
1
/
n
).From now on, we will prove Theorems1.7, 1.9, and 1.10, which are the results for the Szász-Mirakyan operators, analogously to the case of Lupas-type operators.Lemma 2.5.
Letr be a nonnegative integer. Then one has for f
∈
C
r
(
[
0
,
∞
)
),
(2.58)
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
=
(
n
+
β
)
(
n
n
+
β
)
r
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
f
(
r
)
(
y
)
d
y
,
x
∈
[
0
,
∞
)
.Proof.
We know that(2.59)
S
n
,
k
(
r
)
(
x
)
=
∑
i
=
0
r
(
r
i
)
(
e
-
n
y
)
(
r
-
i
)
(
(
n
y
)
k
)
(
i
)
k
!
=
∑
i
=
0
r
(
r
i
)
(
-
1
)
r
(
-
1
)
i
n
r
S
n
,
k
-
i
(
x
)
,
S
n
,
k
(
r
)
(
x
)
=
∑
i
=
0
r
(
r
i
)
(
e
-
n
y
)
(
i
)
(
(
n
y
)
k
)
(
r
-
i
)
k
!
=
∑
i
=
0
r
(
r
i
)
(
-
1
)
i
n
r
S
n
,
k
-
r
+
i
(
x
)
.
Therefore, we have
(2.60)
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
=
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
r
)
(
x
)
∫
0
∞
S
n
+
β
,
k
(
y
)
f
(
y
)
d
y
=
(
n
+
β
)
∑
i
=
0
r
∑
k
=
i
∞
(
r
i
)
(
-
1
)
r
(
-
1
)
i
n
r
S
n
,
k
-
i
(
x
)
∫
0
∞
S
n
+
β
,
k
(
y
)
f
(
y
)
d
y
=
(
n
+
β
)
(
n
n
+
β
)
r
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
∑
i
=
0
r
(
r
i
)
(
-
1
)
r
(
-
1
)
i
(
n
+
β
)
r
S
n
+
β
,
k
+
i
(
y
)
f
(
y
)
d
y
=
(
n
+
β
)
(
n
n
+
β
)
r
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
(
-
1
)
r
S
n
+
β
,
k
+
r
(
r
)
(
y
)
f
(
y
)
d
y
=
(
n
+
β
)
(
n
n
+
β
)
r
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
f
(
r
)
(
y
)
d
y
.Lemma 2.6.
Leta, b, and m be nonnegative integers. 							(2.61)
R
n
,
m
,
r
(
a
,
b
;
x
)
:
=
(
n
+
b
)
∑
k
=
0
∞
S
n
+
a
,
k
(
x
)
∫
0
∞
S
n
+
b
,
k
+
r
(
y
)
(
y
-
x
)
m
d
y
.
Then one has(i)
R
n
,
0
,
r
(
a
,
b
;
x
)
=
1 and R
n
,
1
,
r
(
a
,
b
;
x
)
=
(
(
a
-
b
)
x
+
r
+
1
)
/
(
n
+
b
);
(ii)
Form
⩾
1
(2.62)
(
n
+
b
)
R
n
,
m
+
1
,
r
(
a
,
b
;
x
)
=
x
R
n
,
m
,
r
(
1
)
(
a
,
b
;
x
)
+
(
(
a
-
b
)
x
+
m
+
r
+
1
)
R
n
,
m
,
r
(
a
,
b
;
x
)
+
2
x
m
R
n
,
m
-
1
,
r
(
a
,
b
;
x
)
;
(iii)
(2.63)
R
n
,
m
,
r
(
a
,
b
;
x
)
=
O
(
1
n
[
(
m
+
1
)
/
2
]
)
g
n
,
m
,
r
(
a
,
b
;
x
)
,
where g
n
,
m
,
r
(
a
,
b
;
x
) is a polynomial of degree ⩽
m such that the coefficients of g
n
,
m
,
r
(
a
,
b
;
x
) are bounded independently of n.Proof.
LetR
n
,
m
,
r
(
x
)
:
=
R
n
,
m
,
r
(
a
,
b
;
x
). Then (i)
(2.64)
R
n
,
0
,
r
(
x
)
=
(
n
+
b
)
∑
k
=
0
∞
S
n
+
a
,
k
(
x
)
∫
0
∞
S
n
+
b
,
k
+
r
(
y
)
d
y
=
1
,
R
n
,
1
,
r
(
x
)
=
(
n
+
b
)
∑
k
=
0
∞
S
n
+
a
,
k
(
x
)
∫
0
∞
S
n
+
b
,
k
+
r
(
y
)
(
y
-
x
)
d
y
=
∑
k
=
0
∞
S
n
+
a
,
k
(
x
)
k
+
r
+
1
n
+
b
-
x
=
∑
k
=
0
∞
S
n
+
a
,
k
(
x
)
k
n
+
b
+
r
+
1
n
+
b
-
x
=
(
n
+
a
)
x
n
+
b
+
r
+
1
n
+
b
-
x
=
(
a
-
b
)
x
+
r
+
1
n
+
b
.
(ii) Usingx
S
n
,
k
(
1
)
(
x
)
=
(
k
-
n
x
)
S
n
,
k
(
x
), we obtain
(2.65)
x
(
R
n
,
m
,
r
(
1
)
(
x
)
+
m
R
n
,
m
-
1
,
r
(
x
)
)
=
(
n
+
b
)
∑
k
=
0
∞
x
S
n
+
a
,
k
(
1
)
(
x
)
∫
0
∞
S
n
+
b
,
k
+
r
(
y
)
(
y
-
x
)
m
d
y
=
(
n
+
b
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
(
k
-
(
n
+
a
)
x
)
S
n
+
b
,
k
+
r
(
y
)
(
y
-
x
)
m
d
y
.
Here, we see
(2.66)
(
k
-
(
n
+
a
)
x
)
S
n
+
b
,
k
+
r
(
y
)
=
(
(
k
+
r
-
(
n
+
b
)
y
)
-
(
r
+
(
a
-
b
)
x
)
+
(
n
+
b
)
(
y
-
x
)
)
S
n
+
b
,
k
+
r
(
y
)
=
y
S
n
+
b
,
k
+
r
(
1
)
(
y
)
-
(
r
+
(
a
-
b
)
x
)
S
n
+
b
,
k
+
r
(
y
)
+
(
n
+
b
)
S
n
+
b
,
k
+
r
(
y
)
(
y
-
x
)
.
Then substituting (2.66) for (2.65), we consider the following;
(2.67)
x
(
R
n
,
m
,
r
(
1
)
(
x
)
+
m
R
n
,
m
-
1
,
r
(
x
)
)
=
(
n
+
b
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
(
y
S
n
+
b
,
k
+
r
(
1
)
(
y
)
-
(
r
+
(
a
-
b
)
x
)
S
n
+
b
,
k
+
r
(
y
)
+
(
n
+
b
)
S
n
+
b
,
k
+
r
(
y
)
(
y
-
x
)
)
(
y
-
x
)
m
d
y
:
=
∫
1
+
∫
2
+
∫
3
.
Then, we have
(2.68)
∫
1
=
(
n
+
b
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
y
S
n
+
b
,
k
+
r
(
1
)
(
y
)
(
y
-
x
)
m
d
y
=
(
n
+
b
)
∑
k
=
0
∞
S
n
+
a
,
k
(
x
)
∫
0
∞
S
n
+
b
,
k
+
r
(
1
)
(
y
)
(
y
-
x
)
m
+
1
d
y
+
x
(
n
+
b
)
∑
k
=
0
∞
S
n
+
a
,
k
(
x
)
∫
0
∞
S
n
+
b
,
k
+
r
(
1
)
(
y
)
(
y
-
x
)
m
d
y
=
-
(
m
+
1
)
R
n
,
m
,
r
(
x
)
-
x
m
R
n
,
m
-
1
,
r
(
x
)
.
Here the last equation follows by parts of integration. Furthermore, we have
(2.69)
∫
1
+
∫
2
=
-
(
r
+
(
a
-
b
)
x
)
R
n
,
m
,
r
(
x
)
+
(
n
+
b
)
R
n
,
m
+
1
,
r
(
x
)
.
Therefore, we have
(2.70)
(
n
+
b
)
R
n
,
m
+
1
,
r
(
x
)
=
x
R
n
,
m
,
r
(
1
)
(
x
)
+
(
(
a
-
b
)
x
+
m
+
r
+
1
)
R
n
,
m
,
r
(
x
)
+
2
x
m
R
n
,
m
-
1
,
r
(
x
)
.
(iii) It is proved by the same method as the proof of Lemma2.1 (iv).Proof of Theorem1.7.
Let|
y
-
x
|
⩽
1. By the second inequality in (1.30),
(2.71)
|
f
(
r
)
(
y
)
-
f
(
r
)
(
x
)
|
=
|
y
-
x
|
|
f
(
r
+
1
)
(
ξ
)
|
⩽
C
|
y
-
x
|
|
e
β
ξ
ξ
α
+
2
|
⩽
C
|
y
-
x
|
e
β
x
(
x
+
1
)
α
+
2
.
Let ɛ
=
n
-
γ, 0
<
γ
<
1,
(2.72)
|
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
=
|
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
(
∫
|
y
-
x
|
<
ɛ
S
n
+
β
,
k
+
r
(
y
)
|
f
(
r
)
(
y
)
-
f
(
r
)
(
x
)
|
d
y
+
∫
|
y
-
x
|
⩾
ɛ
S
n
+
β
,
k
+
r
(
y
)
|
f
(
r
)
(
y
)
-
f
(
r
)
(
x
)
|
d
y
)
∑
k
=
0
∞
∑
k
=
0
∞
|
=
:
A
+
B
.
First, we see that by (2.71) and Lemma 2.6(i),
(2.73)
A
⩽
C
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
|
y
-
x
|
<
ɛ
S
n
+
β
,
k
+
r
(
y
)
|
y
-
x
|
d
y
e
β
x
(
x
+
1
)
α
+
2
⩽
C
ɛ
e
β
x
(
x
+
1
)
α
+
2
⩽
O
(
1
n
γ
)
e
β
x
(
x
+
1
)
α
+
2
.
Next, to estimate B, we split it into two parts:
(2.74)
B
=
|
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
|
y
-
x
|
⩾
ɛ
S
n
+
β
,
k
+
r
(
y
)
|
f
(
r
)
(
y
)
-
f
(
r
)
(
x
)
|
d
y
|
⩽
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
|
y
-
x
|
⩾
ɛ
S
n
+
β
,
k
+
r
(
y
)
(
|
f
(
r
)
(
y
)
|
+
|
f
(
r
)
(
x
)
|
)
d
y
⩽
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
|
y
-
x
|
⩾
ɛ
S
n
+
β
,
k
+
r
(
y
)
(
e
β
y
(
y
+
1
)
α
+
e
β
x
(
x
+
1
)
α
)
d
y
=
:
B
1
+
B
2
.
First, we estimate
(2.75)
B
1
=
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
+
β
,
k
+
r
(
y
)
e
β
y
(
y
+
1
)
α
d
y
.
Then, using the following facts:
(2.76)
S
n
+
β
,
k
+
r
(
y
)
e
β
y
=
S
n
,
k
+
r
(
y
)
(
n
+
β
n
)
k
+
r
,
(2.77)
(
n
+
β
n
)
k
+
r
S
n
,
k
(
x
)
=
(
n
+
β
n
)
r
S
n
+
β
,
k
(
x
)
e
β
x
,
(2.78)
(
y
+
1
)
α
=
(
(
y
-
x
)
+
x
+
1
)
α
=
∑
i
=
0
α
(
α
i
)
(
y
-
x
)
i
(
x
+
1
)
α
-
i
,
we have
(2.79)
B
1
⩽
C
e
β
x
(
n
+
β
)
(
n
+
β
n
)
r
∑
k
=
0
∞
S
n
+
β
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
,
k
+
r
(
y
)
(
y
+
1
)
α
d
y
=
C
e
β
x
n
(
n
+
β
n
)
r
+
1
∑
k
=
0
∞
S
n
+
β
,
k
(
x
)
×
∫
|
y
-
x
|
>
ɛ
S
n
,
k
+
r
(
y
)
∑
i
=
1
α
(
α
i
)
(
y
-
x
)
i
(
x
+
1
)
α
-
i
d
y
+
C
e
β
x
n
(
n
+
β
n
)
r
∑
k
=
0
∞
S
n
+
β
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
,
k
+
r
(
y
)
(
x
+
1
)
α
d
y
=
:
C
e
β
x
(
B
11
+
B
12
)
.
Then, using (2.18) and Lemma 2.6, we have
(2.80)
B
11
=
n
∑
k
=
0
∞
S
n
+
β
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
,
k
+
r
(
y
)
∑
i
=
1
α
(
α
i
)
(
y
-
x
)
i
(
x
+
1
)
α
-
i
d
y
⩽
n
∑
k
=
0
∞
S
n
+
β
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
,
k
+
r
(
y
)
∑
i
=
1
α
(
α
i
)
(
y
-
x
)
i
|
y
-
x
ɛ
|
〈
i
〉
(
x
+
1
)
α
-
i
d
y
⩽
n
∑
k
=
0
∞
S
n
+
β
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
,
k
+
r
(
y
)
∑
i
=
1
α
(
α
i
)
(
y
-
x
)
i
+
〈
i
〉
(
1
ɛ
)
〈
i
〉
(
x
+
1
)
α
-
i
d
y
⩽
∑
i
=
1
α
(
α
i
)
R
n
,
i
+
〈
i
〉
r
(
β
,
0
;
x
)
(
1
ɛ
)
〈
i
〉
(
x
+
1
)
α
-
i
=
∑
i
=
1
α
(
α
i
)
O
(
〈
i
〉
n
[
(
i
+
〈
i
〉
+
1
)
/
2
]
)
q
n
,
i
+
〈
i
〉
,
r
(
β
,
0
;
x
)
(
x
+
1
)
α
-
i
.
Then by (2.21) we have
(2.81)
B
11
⩽
O
(
1
n
1
-
γ
)
(
x
+
1
)
α
+
1
.
For B
12, we have
(2.82)
B
12
=
n
∑
k
=
0
∞
S
n
+
β
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
,
k
+
r
(
y
)
(
x
+
1
)
α
d
y
⩽
n
∑
k
=
0
∞
S
n
+
β
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
,
k
+
r
(
y
)
(
y
-
x
ɛ
)
2
d
y
(
x
+
1
)
α
⩽
R
n
,
2
,
r
(
β
,
0
;
x
)
(
1
ɛ
)
2
(
x
+
1
)
α
=
O
(
n
2
γ
n
[
3
/
2
]
)
q
n
,
2
,
r
(
β
,
0
;
x
)
(
x
+
1
)
α
=
O
(
1
n
1
-
2
γ
)
(
x
+
1
)
α
+
2
.
From (2.81), (2.82) and (2.79), we have
(2.83)
B
1
⩽
O
(
1
n
1
-
γ
)
e
β
x
(
x
+
1
)
α
+
2
.
We estimate
(2.84)
B
2
=
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
+
β
,
k
+
r
(
y
)
e
β
x
(
x
+
1
)
α
d
y
.
Then we can estimate B
2 by the same method as B
12,
(2.85)
B
2
=
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
+
β
,
k
+
r
(
y
)
e
β
x
(
x
+
1
)
α
d
y
⩽
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
|
y
-
x
|
>
ɛ
S
n
+
β
,
k
+
r
(
y
)
(
y
-
x
ɛ
)
2
d
y
e
β
x
(
x
+
1
)
α
⩽
R
n
,
2
,
r
(
0
,
β
;
x
)
(
1
ɛ
)
2
e
β
x
(
x
+
1
)
α
,
so we have
(2.86)
B
2
⩽
O
(
n
2
γ
n
[
3
/
2
]
)
q
n
,
2
,
r
(
0
,
β
;
x
)
e
β
x
(
x
+
1
)
α
⩽
O
(
1
n
1
-
2
γ
)
e
β
x
(
x
+
1
)
α
+
2
.
Consequently, we obtain from (2.83) and (2.86),
(2.87)
B
⩽
(
O
(
1
n
1
-
γ
)
+
O
(
1
n
1
-
2
γ
)
)
e
β
x
(
x
+
1
)
α
+
2
⩽
O
(
1
n
1
-
2
γ
)
e
β
x
(
x
+
1
)
α
+
2
.
Therefore, from (2.73) and (2.87), we have uniformly on n,
(2.88)
|
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
⩽
O
(
1
n
γ
)
e
β
x
x
α
+
2
+
O
(
1
n
1
-
2
γ
)
e
β
x
x
α
+
2
.
Here, if we let γ
=
1
/
3, then we have
(2.89)
|
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
=
O
(
1
n
1
/
3
)
e
β
x
x
α
+
2
,
that is, (1.31) is proved. So, we also have a norm convergence (1.32).Lemma 2.7.
Letm and b be nonnegative integers. Let
(2.90)
U
n
,
m
,
r
(
b
;
x
)
:
=
(
n
+
b
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
b
,
k
+
r
(
y
)
(
y
-
x
)
m
e
b
y
d
y
.
Then one has
(2.91)
U
n
,
m
,
r
(
b
;
x
)
=
(
n
+
b
n
)
r
+
1
e
b
x
R
n
,
m
,
r
(
b
,
0
;
x
)
.Proof.
From (2.76) and (2.77) we have
(2.92)
S
n
+
b
,
k
+
r
(
x
)
e
b
x
=
S
n
,
k
+
r
(
x
)
(
n
+
b
n
)
k
+
r
,
(2.93)
(
n
+
b
n
)
k
+
r
S
n
,
k
(
x
)
=
(
n
+
b
n
)
r
S
n
+
b
,
k
(
x
)
e
b
x
.
We have from (2.92), (2.93), and noting (2.61),
(2.94)
U
n
,
m
,
r
(
b
;
x
)
=
(
n
+
b
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
b
,
k
+
r
(
y
)
(
y
-
x
)
m
e
b
y
d
y
=
(
n
+
b
)
∑
k
=
0
∞
(
n
+
b
n
)
k
+
r
S
n
,
k
(
x
)
∫
0
∞
S
n
,
k
+
r
(
y
)
(
y
-
x
)
m
d
y
=
(
n
+
b
)
(
n
+
b
n
)
r
e
b
x
∑
k
=
0
∞
S
n
+
b
,
k
(
x
)
∫
0
∞
S
n
,
k
+
r
(
y
)
(
y
-
x
)
m
d
y
=
(
n
+
b
n
)
r
+
1
e
b
x
R
n
,
m
,
r
(
b
,
0
;
x
)
.Proof of Theorem1.9.
We prove this theorem, similarly to the proof of Theorem1.4. Using (2.93) and (2.27), we have for f
∈
C
(
r
+
2
)
(
[
0
,
∞
)
),
(2.95)
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
=
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
f
(
r
)
(
y
)
d
y
=
f
(
r
)
(
x
)
+
f
(
r
+
1
)
(
x
)
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
(
y
-
x
)
d
y
+
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
(
∫
x
y
(
y
-
u
)
f
(
r
+
2
)
(
u
)
d
u
)
d
y
=
f
(
r
)
(
x
)
+
f
(
r
+
1
)
(
x
)
R
n
,
1
,
r
(
0
,
β
;
x
)
+
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
(
∫
x
y
(
y
-
u
)
f
(
r
+
2
)
(
u
)
d
u
)
d
y
.
We estimate the last term. We note the given condition:
(2.96)
‖
f
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
<
∞
.
Using the inequality (
1
+
t
)
2
γ
⩽
C
(
(
1
+
x
)
2
γ
+
(
t
-
x
)
2
γ
) and Lemma 2.7, we have
(2.97)
∑
:
=
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
(
e
β
y
ψ
-
2
γ
(
y
)
+
e
β
x
ψ
-
2
γ
(
x
)
)
(
y
-
x
)
2
d
y
⩽
C
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
×
(
e
β
y
ψ
-
2
γ
(
x
)
+
e
β
y
(
(
y
-
x
)
2
γ
+
e
β
x
ψ
-
2
γ
(
x
)
)
(
y
-
x
)
2
d
y
⩽
C
(
ψ
-
2
γ
(
x
)
U
n
,
2
,
r
(
β
;
x
)
+
U
n
,
2
γ
+
2
,
r
(
β
;
x
)
+
e
β
x
ψ
-
2
γ
(
x
)
R
n
,
2
,
r
(
0
,
β
;
x
)
)
⩽
C
(
e
β
x
ψ
-
2
γ
(
x
)
R
n
,
2
,
r
(
β
,
0
;
x
)
(
n
+
β
n
)
r
+
1
+
e
β
x
R
n
,
2
γ
+
2
,
r
(
β
,
0
;
x
)
(
n
+
β
n
)
r
+
1
+
e
β
x
ψ
-
2
γ
(
x
)
R
n
,
2
,
r
(
0
,
β
;
x
)
)
.
Then, we can estimate as follows:
(2.98)
|
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
(
∫
x
y
(
y
-
u
)
f
(
r
+
2
)
(
u
)
d
u
)
d
y
|
⩽
‖
f
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
∑
⩽
C
‖
f
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
e
β
x
×
(
ψ
-
2
γ
(
x
)
R
n
,
2
,
r
(
β
,
0
;
x
)
(
n
+
β
n
)
r
+
1
+
R
n
,
2
γ
+
2
,
r
(
β
,
0
;
x
)
(
n
+
β
n
)
r
+
1
+
ψ
-
2
γ
(
x
)
R
n
,
2
,
r
(
0
,
β
;
x
)
(
n
+
β
n
)
r
+
1
)
.
Then, we have by (iv) of Lemma 2.6,
(2.99)
|
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
|
e
-
β
x
ψ
2
γ
+
2
(
x
)
⩽
‖
f
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
×
(
O
(
1
n
)
ψ
2
(
x
)
g
n
,
2
,
r
(
β
,
0
;
x
)
+
O
(
1
n
γ
)
g
n
,
2
γ
+
2
,
r
(
β
,
0
;
x
)
ψ
2
γ
+
2
(
x
)
+
O
(
1
n
)
ψ
2
(
x
)
g
n
,
2
,
r
(
0
,
β
;
x
)
)
+
O
(
1
n
)
|
f
(
r
+
1
)
(
x
)
e
-
β
x
ψ
2
γ
+
1
(
x
)
g
n
,
1
,
r
(
0
,
β
;
x
)
ψ
(
x
)
|
.
Consequently, we have
(2.100)
|
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
|
⩽
O
(
1
n
)
(
‖
f
(
r
+
1
)
(
x
)
e
-
β
x
ψ
2
γ
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
f
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
)
,
since we know that |
g
n
,
2
,
r
(
β
,
0
;
x
)
ψ
2
(
x
)
|
,
|
g
n
,
2
,
r
(
0
,
β
;
x
)
ψ
2
(
x
)
|
,
|
g
n
,
2
γ
+
2
,
r
(
β
,
0
;
x
)
ψ
2
γ
+
2
(
x
)
|, and |
g
n
,
1
,
r
(
0
,
β
;
x
)
ψ
(
x
)
| are uniformly bounded on [
0
,
∞
).Theorem 2.8.
Letβ and γ be nonnegative integers and r
>
0 be a positive integer. Then one has uniformly for n, f and x
∈
[
0
,
∞
),
(2.101)
|
[
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
(
x
)
|
⩽
C
n
+
β
n
‖
f
(
r
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
.Proof.
Using(
1
+
y
)
2
γ
⩽
C
(
(
1
+
x
)
2
γ
+
(
y
-
x
)
2
γ
), we have
(2.102)
|
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
e
β
y
(
1
+
y
)
2
γ
d
y
|
⩽
|
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
e
β
y
(
(
1
+
x
)
2
γ
+
(
y
-
x
)
2
γ
)
d
y
|
C
(
U
n
,
0
,
r
(
β
;
x
)
ψ
-
2
γ
(
x
)
+
U
n
,
2
γ
,
r
(
β
;
x
)
)
.
By Lemma 2.7 and (i) of Lemma 2.6, we know
(2.103)
U
n
,
0
,
r
(
β
;
x
)
=
(
n
+
β
n
)
r
+
1
e
β
x
R
n
,
0
,
r
(
β
,
0
;
x
)
=
(
n
+
β
n
)
r
+
1
e
β
x
,
U
n
,
2
γ
,
r
(
β
;
x
)
=
(
n
+
β
n
)
r
+
1
e
β
x
R
n
,
2
γ
,
r
(
β
,
2
γ
;
x
)
=
O
(
1
n
γ
)
(
n
+
β
n
)
r
+
1
e
β
x
g
n
,
2
γ
,
r
(
β
,
2
γ
;
x
)
.
Therefore, we have
(2.104)
|
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
e
β
y
(
1
+
y
)
2
γ
d
y
|
⩽
C
(
n
+
β
n
)
r
+
1
e
β
x
(
ψ
-
2
γ
(
x
)
+
O
(
1
n
γ
)
g
n
,
2
γ
,
r
(
β
,
2
γ
;
x
)
)
.
Since |
g
n
,
2
γ
,
r
(
β
,
2
γ
;
x
)
ψ
2
γ
(
x
)
| is uniformly bounded on [
0
,
∞
), we have
(2.105)
|
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
|
⩽
|
(
n
+
β
)
∑
k
=
0
∞
S
n
,
k
(
x
)
∫
0
∞
S
n
+
β
,
k
+
r
(
y
)
e
β
y
(
1
+
y
)
2
γ
d
y
|
e
-
β
x
ψ
2
γ
(
x
)
×
‖
f
(
r
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
(
n
+
β
n
)
r
+
1
(
1
+
O
(
1
n
γ
)
)
‖
f
(
r
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
.
Therefore, we have the result.Proof of Theorem1.10.
We will prove this theorem by the same method as the proof of Theorem1.5. First, we split it as follows:
(2.106)
‖
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
‖
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
-
[
f
]
h
]
)
(
r
)
(
x
)
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
[
f
]
h
]
)
(
r
)
(
x
)
-
[
f
]
h
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
[
[
f
]
h
(
r
)
(
x
)
-
f
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
.
Then for the first term, we have, using Theorem 2.8 and (2.41),
(2.107)
‖
(
Q
n
,
β
[
f
-
[
f
]
h
]
)
(
r
)
(
x
)
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
n
+
β
n
‖
[
f
(
r
)
(
x
)
-
[
f
(
r
)
]
h
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
ω
2
(
f
(
r
)
;
e
-
β
x
ψ
2
γ
+
2
(
x
)
;
h
)
.
For the second term, we have from Theorem 1.9,
(2.108)
‖
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
h
]
)
(
r
)
(
x
)
-
f
h
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
O
(
1
n
)
(
‖
f
h
(
r
+
1
)
(
x
)
e
-
β
x
ψ
2
γ
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
+
‖
f
h
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
)
.
Here, we suppose 0
<
h
⩽
1 and then we know that (
e
-
β
x
ψ
(
x
)
)
/
(
e
-
β
(
x
+
h
)
ψ
(
x
+
h
)
) and e
-
β
x
/
e
-
β
(
x
+
h
) are uniformly bounded on [
0
,
∞
). Therefore, we have from (2.42) and (2.43) of Lemma 2.4,
(2.109)
‖
f
h
(
r
+
1
)
(
x
)
e
-
β
x
ψ
2
γ
+
1
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
1
h
ω
1
(
f
;
e
-
β
x
ψ
2
γ
+
1
(
x
)
;
h
)
,
‖
f
h
(
r
+
2
)
(
x
)
e
-
β
x
ψ
2
γ
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
C
1
h
2
ω
2
(
f
;
e
-
β
x
ψ
2
γ
(
x
)
;
h
)
.
Therefore, we have
(2.110)
‖
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
h
]
)
(
r
)
(
x
)
-
f
h
(
r
)
(
x
)
]
e
-
β
x
ψ
2
(
x
)
‖
L
∞
(
[
0
,
∞
)
)
⩽
O
(
1
n
)
(
1
h
ω
1
(
f
;
e
-
β
x
ψ
2
+
1
(
x
)
;
h
)
+
1
h
2
ω
2
(
f
;
e
-
β
x
ψ
2
γ
(
x
)
;
h
)
)
.
Consequently, we have
(2.111)
‖
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
⩽
O
(
1
n
)
(
1
h
ω
1
(
f
;
e
-
β
x
ψ
2
γ
+
1
(
x
)
;
h
)
+
1
h
2
ω
2
(
f
;
e
-
β
x
ψ
2
γ
(
x
)
;
h
)
)
+
C
ω
2
(
f
(
r
)
;
e
-
β
x
ψ
2
γ
+
2
(
x
)
(
x
)
;
h
)
.
If we let h
=
1
/
n, then
(2.112)
‖
[
(
n
+
β
n
)
r
(
Q
n
,
β
[
f
]
)
(
r
)
(
x
)
-
f
(
r
)
(
x
)
]
e
-
β
x
ψ
2
γ
+
2
(
x
)
‖
⩽
C
(
1
n
ω
1
(
f
;
e
-
β
x
ψ
2
γ
+
1
(
x
)
;
1
n
)
+
ω
2
(
f
;
e
-
β
x
ψ
2
γ
(
x
)
;
1
n
)
)
,
since ω
2
(
f
(
r
)
;
e
-
β
x
ψ
2
γ
+
2
(
x
)
;
1
/
n
)
⩽
ω
2
(
f
;
e
-
β
x
ψ
2
γ
(
x
)
;
1
/
n
).


## 3. Conclusion

In this paper, Lupas-type operators and Szász-Mirakyan-type operators are treated and the various weighted norm convergence on[
0
,
∞
) of these operators are investigated. Moreover, this paper proves theorems on degree of approximation of f
∈
C
r
(
[
0
,
∞
)
) by these operators using the modulus of smoothness of f.



---

*Source: 546784-2012-03-18.xml*