# Strong Inequalities for Hermite-Fejér Interpolations and Characterization ofK-Functionals

Authors: Gongqiang You
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The works of Smale and Zhou (2003, 2007), Cucker and Smale (2002), and Cucker and Zhou (2007) indicate that approximation operators serve as cores of many machine learning algorithms. In this paper we study the Hermite-Fejér interpolation operator which has this potential of
applications. The interpolation is defined by zeros of the Jacobi polynomials with parameters−
1
<
α, β
<
0. Approximation rate is obtained for continuous functions. Asymptotic expression of the K-functional associated with the interpolation operators is given.


---

## Body

## 1. Introduction

Zhou and Jetter [1] used Bernstein-Durrmeyer operators for studying support vector machine classification algorithms. This work initiates the direction of applying more linear operators from approximation theory to learning theory. We will follow this direction and study Hermite-Fejér interpolation operator. It would be interesting to derive explicit learning rates by means of these operators for some specific learning algorithms.LetP
n
(
α
,
β
) denote the Jacobi polynomial of order n. Let {
x
k
}
k
=
1
n be the zeros of P
n
(
α
,
β
). We assume that α
,
β
>
-
1. For any continuous function f on [
-
1,1
], the Hermite-Fejér interpolation H
n
(
α
,
β
)
(
f
,
·
) is a polynomial of order 2
n
-
1 that satisfies
(1)
H
n
(
α
,
β
)
(
f
,
x
k
)
=
f
(
x
k
)
,
H
n
(
α
,
β
)
′
(
f
,
x
k
)
=
0
,
for any k
=
1,2
,
…
,
n. Let ∥
·
∥ be the norm on C
[
-
1,1
] (∀
f
∈
C
[
-
1,1
], ∥
f
∥
∶
=
max
⁡
{
|
f
(
x
)
|
,
-
1
≤
x
≤
1
}). Without introducing ambiguity we also use ∥
·
∥ to denote the norm on C
2
π (which is the totality of the continuous functions on ℝ with period 2
π, and in this case ∥
f
∥
∶
=
max
⁡
{
|
f
(
x
)
|
,
x
∈
ℝ
}).One has (see e.g., [2])
(2)
lim
⁡
n
→
∞
∥
H
n
(
α
,
β
)
(
f
)
-
f
∥
=
0
,
∀
f
∈
C
[
-
1,1
]
,
if and only if α
,
β
<
0.Define(3)
w
(
α
,
β
)
(
t
)
=
|
sin
t
2
|
2
α
+
1
|
cos
⁡
t
2
|
2
β
+
1
.
Denote by  F
^ the conjugate function of F, and write g
(
t
)
=
f
(
cos
⁡
t
). When -
1
<
α, β
<
0, one has (see [3, 4])
(4)
∥
H
n
(
α
,
β
)
(
f
)
-
f
∥
=
𝒪
(
1
n
)
⇔
g
∈
Lip1
,
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
=
𝒪
(
1
)
,
∥
H
n
(
α
,
β
)
(
f
)
-
f
∥
=
o
(
1
n
)
⇔
f
is
a
constant
.
So when -
1
<
α, β
<
0, H
n
(
α
,
β
) is saturation with {
1
/
n
} and the saturation class is
(5)
S
(
α
,
β
)
=
{
f
∈
C
[
-
1,1
]
∣
g
(
t
)
=
f
(
cos
⁡
t
)
,
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
g
∈
Lip1,
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
=
𝒪
(
1
)
}
.Note that for allα
,
β
∈
(
-
1
,
-
1
/
2
), the associated classes S
(
α
,
β
) are identical ([4, Theorem  6]). See [2, 5–7] for related works.Denote thatΔ
t
h
(
x
)
=
h
(
x
+
t
/
2
)
-
h
(
x
-
t
/
2
) and recursively Δ
t
k
h
(
x
)
=
Δ
t
Δ
t
k
-
1
h
(
x
) with Δ
t
1
=
Δ
t. Write φ
(
x
)
=
1
-
x
2. Define
(6)
ω
φ
(
f
,
δ
)
=
max
⁡
0
<
t
≤
δ
∥
Δ
φ
t
f
∥
.
For any f
∈
C
[
-
1,1
], one has [8]
(7)
inf
⁡
h
∈
C
2
[
-
1,1
]
(
∥
f
-
h
∥
+
δ
∥
φ
h
′
∥
)
≍
ω
φ
(
f
,
δ
)
.
Let f
∈
C
[
-
1,1
] and h
(
θ
)
=
H
(
cos
⁡
θ
). We use the following definition of K-functional from [9]:
(8)
K
α
,
β
(
f
,
δ
)
=
inf
⁡
H
∈
C
2
[
-
1,1
]
(
∥
f
-
H
∥
+
δ
∥
h
′
∥
+
δ
∥
h
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
.We cite the following three Theorems from [9].Theorem 1.
Let-
1
<
α, β
<
-
1
/
2 be fixed. Then there is a constant 1
<
A
<
∞ such that for all f
∈
C
[
-
1,1
] and all n
≥
1, one has
(9)
max
⁡
k
≥
n
∥
H
k
(
α
,
β
)
(
f
)
-
f
∥
≍
max
⁡
n
≤
k
≤
A
n
∥
H
k
(
α
,
β
)
(
f
)
-
f
∥
max
⁡
k
≥
n
∥
H
k
(
α
,
β
)
(
f
)
-
f
∥
≍
1
n
∥
∫
1
/
n
1
‍
Δ
t
φ
2
f
t
2
d
t
∥
+
ω
φ
(
f
,
1
n
)
,
in which the symbol≍ does not rely on n and f.Theorem 2.
Let-
1
<
α, β
<
-
1
/
2 be fixed. Then the following relation holds:
(10)
K
α
,
β
(
f
,
δ
)
≍
δ
∥
∫
δ
1
‍
Δ
φ
u
2
f
u
2
d
u
∥
+
ω
φ
(
f
,
δ
)
,
iiiiii
∀
f
∈
C
[
-
1,1
]
,
δ
>
0
.Theorem 3.
Let-
1
<
α, β
<
0 be fixed. Then there is a constant 1
<
A
<
∞ such that, for all f
∈
C
[
-
1,1
] and all n
≥
1, one has
(11)
max
⁡
k
≥
n
∥
H
k
(
α
,
β
)
(
f
)
-
f
∥
≍
max
⁡
n
≤
k
≤
A
n
∥
H
k
(
α
,
β
)
(
f
)
-
f
∥
max
⁡
k
≥
n
∥
H
k
(
α
,
β
)
(
f
)
-
f
∥
≍
K
α
,
β
(
f
,
1
n
)
.
Here the symbol≍ does not rely on n and f.TheK-functional K
α
,
β
(
f
,
δ
) for all -
1
<
α, β
<
0 is characterized by the followingTheorem 4.
Let-
1
<
α, β
<
0 and -
1
<
α
0, β
0
<
-
1
/
2 be fixed. Then, for all δ
>
0 and all f
∈
C
[
-
1,1
], the following holds:
(12)
K
α
,
β
(
f
,
δ
)
≍
K
α
0
,
β
0
(
f
,
δ
)
+
J
α
,
β
,
1
(
f
,
δ
)
+
J
α
,
β
,
2
(
f
,
δ
)
,
in which(13)
J
α
,
β
,
1
(
f
,
δ
)
=
max
⁡
0
≤
x
≤
1
-
δ
2
δ
(
1
-
x
)
α
+
1
/
2
×
|
∫
-
1
x
‍
f
(
t
)
-
f
(
1
)
t
-
1
∫
-
1
x
‍
f
(
t
)
-
f
(
1
)
t
-
1
×
(
1
-
α
+
β
-
(
1
+
α
+
β
)
t
)
(
1
-
t
)
α
(
1
+
t
)
β
d
t
|
,
(14)
J
α
,
β
,
2
(
f
,
δ
)
=
max
⁡
-
1
+
δ
2
≤
x
≤
0
δ
(
1
+
x
)
β
+
1
/
2
×
|
∫
x
1
‍
f
(
t
)
-
f
(
-
1
)
t
+
1
(
1
+
α
-
β
+
(
1
+
α
+
β
)
t
)
∫
x
1
‍
f
(
t
)
-
f
(
-
1
)
t
+
1
×
(
1
-
t
)
α
(
1
+
t
)
β
d
t
|
.Theorems3 and 4 give
(15)
K
-
1
/
2
,
-
1
/
2
(
f
,
δ
)
≍
δ
∥
∫
δ
1
‍
Δ
t
φ
f
t
2
d
t
∥
+
δ
max
⁡
0
≤
x
≤
1
-
δ
2
|
∫
-
1
x
‍
f
(
t
)
-
f
(
1
)
t
-
1
(
1
-
t
2
)
-
1
/
2
d
t
|
+
δ
max
⁡
-
1
+
δ
2
≤
x
≤
0
|
∫
x
1
‍
f
(
t
)
-
f
(
-
1
)
t
+
1
(
1
-
t
2
)
-
1
/
2
d
t
|
+
ω
φ
(
f
,
δ
)
.Moreover, if-
1
<
α, β
<
-
1
/
2, then, due to |
f
(
1
)
-
f
(
t
)
|
≤
C
ω
φ
(
f
,
1
-
t
) (see (2.6) of [9]), we have
(16)
J
α
,
β
,
1
(
f
,
δ
)
+
J
α
,
β
,
2
(
f
,
δ
)
≤
C
ω
φ
(
f
,
δ
)
.Theorem4 will be proved in Section 3. In Section 2 we will discuss some properties of Jacobi polynomials and make some remarks concerning the conjugate function g
′
w
(
α
,
β
)
^.


## 2. Estimates for Jacobi Polynomials and Conjugate Functions

Chapter 8 of [10] gave the following.Lemma 5.
LetC
>
0 be fixed. For all θ
∈
[
C
/
n
,
π
-
C
/
n
] and n
≥
C
/
π
+
1, one has
(17)
P
n
(
α
,
β
)
(
cos
⁡
θ
)
=
(
π
n
w
(
α
,
β
)
(
θ
)
)
-
1
/
2
{
cos
⁡
(
N
θ
+
r
)
+
𝒪
(
n
sin
θ
)
-
1
}
,
in whichN
=
n
+
(
α
+
β
+
1
)
/
2 and r
=
-
(
α
+
1
/
2
)
π
/
2. If 0
≤
θ
≤
π, then there exists a θ
′ making P
n
(
α
,
β
)
(
cos
⁡
θ
′
)
=
0 and |
θ
-
θ
′
|
≍
1
/
n. Moreover, let M
=
{
θ
∣
θ
=
arccos
x
,
P
n
(
α
,
β
)
′
(
x
)
=
0
}. Then
(18)
{
P
n
(
α
,
β
)
(
cos
⁡
θ
)
}
2
w
(
α
,
β
)
(
θ
)
≍
1
n
,
∀
θ
∈
M
,
and, for0
≤
θ
≤
π, one has  θ
′
∈
M with |
θ
-
θ
′
|
≍
1
/
n.Define(19)
G
(
θ
,
t
)
=
P
n
(
α
,
β
)
(
cos
⁡
(
θ
+
t
)
)
w
(
α
,
β
)
(
θ
+
t
)
G
(
θ
,
t
)
=
-
P
n
(
α
,
β
)
(
cos
⁡
(
θ
-
t
)
)
w
(
α
,
β
)
(
θ
-
t
)
,and we have(20)
G
n
(
α
,
β
)
(
θ
)
=
1
2
π
∫
0
π
‍
G
(
θ
,
t
)
tan
(
t
/
2
)
d
t
.ForG
n
(
α
,
β
) we have a conclusion similar to (17) (see [11]) as follows.Lemma 6.
For all fixedα
,
β
>
-
1 and a
>
0, there is C
>
0 such that
(21)
max
⁡
a
/
n
≤
θ
≤
π
-
a
/
n
|
(
w
(
α
,
β
)
(
θ
)
)
-
1
/
2
G
n
(
α
,
β
)
(
θ
)
|
≤
C
n
,
∀
n
≥
a
π
+
1
.Denote byΠ
n the set of n-order algebraic polynomials and by π
n the set of n-order trigonometric polynomials. Denote further that
(22)
Δ
n
(
α
,
β
)
(
f
,
x
)
=
H
n
(
α
,
β
)
(
f
,
x
)
-
f
(
x
)
.The following identity can be found in [9, 12] (see Lemma  5 of [12] and its proof).Lemma 7.
Let-
1
<
α, β
<
0 be fixed. Then for all T
∈
Π
2
n
-
1
(23)
Δ
n
(
α
,
β
)
(
T
,
x
)
=
4
π
C
n

{
P
n
(
α
,
β
)
2
(
x
)
t
′
w
(
α
,
β
)
^
(
θ
)

-

t
′
(
θ
)
P
n
(
α
,
β
)
(
x
)
G
n
(
α
,
β
)
(
θ
)
}
.For our purpose, we need the following result, which improves the estimate of [4] (see Lemma  4 of [4]).Lemma 8.
Let-
1
<
α, β
<
0 be fixed. If g
∈
C
2
π
2 and ∥
g
′
w
(
α
,
β
)
^
/
w
(
α
,
β
)
∥
<
∞, then for all 0
<
δ
≤
π
/
20, the following  holds:
(24)
∥
1
w
(
α
,
β
)
∫
δ
π
‍
Δ
2
t
(
g
′
w
(
α
,
β
)
)
tan
t
/
2
d
t
∥
[
δ
,
π
-
δ
]
≤
C
(
∥
g
′
∥
+
∥
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
.Proof.
WriteF
′
(
θ
)
=
g
′
(
θ
)
w
(
α
,
β
)
(
θ
). So |
F
^
′
/
w
(
α
,
β
)
|
≤
C and for δ
≤
θ
≤
π
/
20
(25)
1
w
(
α
,
β
)
(
θ
)
|
F
^
(
θ
+
δ
)
-
F
^
(
θ
)
δ
|
=
1
w
(
α
,
β
)
|
1
δ
∫
0
δ
‍
F
^
′
(
θ
+
t
)
d
t
|
≤
C
∥
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
.
On the other hand,(26)
F
^
(
θ
+
δ
)
-
F
^
(
θ
)
δ
=
-
1
2
π
δ
{
∫
0
π
‍
Δ
2
t
F
(
θ
+
δ
)
tan
t
/
2
d
t
-
∫
0
π
‍
Δ
2
t
F
(
θ
)
tan
t
/
2
d
t
}
=
-
1
2
π
δ
∫
0
π
‍
d
t
∫
0
δ
‍
Δ
2
t
F
′
(
θ
+
u
)
tan
t
/
2
d
u
=
-
1
2
π
δ
∫
0
δ
‍
d
u
∫
δ
π
‍
Δ
2
t
F
′
(
θ
+
u
)
tan
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
.
Through integration by parts,(27)
-
1
2
π
δ
∫
0
δ
‍
d
u
∫
δ
π
‍
Δ
2
t
F
′
(
θ
+
u
)
tan
t
/
2
d
t
=
-
1
2
π
δ
∫
0
δ
‍
d
u
∫
δ
π
‍
[
Δ
t
2
F
(
θ
+
u
)
]
′
tan
t
/
2
d
t
=
-
1
4
π
δ
∫
0
δ
‍
d
u
∫
δ
π
‍
Δ
t
2
F
(
θ
+
u
)
sin
2
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
.
Moreover, we have(28)
-
1
4
π
δ
∫
0
δ
‍
d
u
∫
δ
π
‍
Δ
t
2
F
(
θ
+
u
)
sin
2
t
/
2
d
t
=
-
1
4
π
∫
δ
π
‍
Δ
t
2
F
(
θ
)
sin
2
t
/
2
d
t
-
1
4
π
δ
∫
0
δ
‍
d
u
∫
δ
π
‍
Δ
t
2
(
F
(
θ
+
u
)
-
F
(
θ
)
)
sin
2
t
/
2
d
t
,
but(29)
-
1
4
π
δ
∫
0
δ
‍
d
u
∫
δ
π
‍
Δ
t
2
(
F
(
θ
+
u
)
-
F
(
θ
)
)
sin
2
t
/
2
d
t
=
-
1
4
π
δ
∫
0
δ
‍
d
u
∫
δ
π
/
2
‍
Δ
t
2
(
F
(
θ
+
u
)
-
F
(
θ
)
)
sin
2
t
/
2
d
t
-
1
4
π
δ
∫
0
δ
‍
d
u
∫
π
/
2
π
‍
Δ
t
2
(
F
(
θ
+
u
)
-
F
(
θ
)
)
sin
2
t
/
2
d
t
=
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
,
-
1
4
π
∫
δ
π
‍
Δ
t
2
F
(
θ
)
sin
2
t
/
2
d
t
=
-
1
2
π
{
∫
δ
π
‍
Δ
2
t
F
′
(
θ
)
tan
t
/
2
d
t
-
Δ
δ
2
F
(
θ
)
tan
δ
/
2
}
=
-
1
2
π
∫
δ
π
‍
Δ
2
t
F
′
(
θ
)
tan
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
.
So we have(30)
F
^
(
θ
+
δ
)
-
F
^
(
θ
)
δ
=
-
1
2
π
∫
δ
π
‍
Δ
2
t
F
′
(
θ
)
tan
t
/
2
d
t
F
^
(
θ
+
δ
)
-
F
^
(
θ
)
δ
=
+
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
.
Thus, ifδ
≤
θ
≤
π
/
20, then
(31)
|
1
w
(
α
,
β
)
(
θ
)
∫
δ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
|
≤
2
π
w
(
α
,
β
)
(
θ
)
|
F
^
(
θ
+
δ
)
-
F
^
(
θ
)
δ
|
+
C
∥
g
′
∥
≤
C
(
∥
g
′
∥
+
∥
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
.
Ifπ
/
20
≤
θ
≤
π
/
2, then w
(
α
,
β
)
(
θ
)
≍
1 and
(32)
1
w
(
α
,
β
)
(
θ
)
∫
θ
/
2
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
𝒪
(
∥
g
′
∥
)
.
In this case, for0
≤
t
≤
θ
/
2, we have
(33)
w
(
α
,
β
)
(
θ
±
t
)
-
w
(
α
,
β
)
(
θ
)
t
=
𝒪
(
1
)
,
(34)
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
=
g
′
(
θ
+
t
)
w
(
α
,
β
)
(
θ
+
t
)
-
g
′
(
θ
-
t
)
w
(
α
,
β
)
(
θ
-
t
)
=
Δ
2
t
(
g
′
(
θ
)
)
w
(
α
,
β
)
(
θ
)
+
g
′
(
θ
+
t
)
×
[
w
(
α
,
β
)
(
θ
+
t
)
-
w
(
α
,
β
)
(
θ
)
]
-
g
′
(
θ
-
t
)
[
w
(
α
,
β
)
(
θ
-
t
)
-
w
(
α
,
β
)
(
θ
)
]
;
therefore,(35)
1
w
(
α
,
β
)
(
θ
)
∫
δ
θ
/
2
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
∫
δ
θ
/
2
‍
Δ
2
t
g
′
(
θ
)
tan
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
)
.
Letδ
→
0 in (35) and, with (32), we obtain
(36)
∫
0
θ
/
2
‍
Δ
2
t
g
′
(
θ
)
tan
t
/
2
d
t
=
𝒪
(
∥
g
′
∥
+
∥
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
.
Forπ
/
20
≤
θ
≤
π
/
2, it is clear that if
(37)
∫
δ
θ
/
2
‍
Δ
2
t
g
′
(
θ
)
tan
t
/
2
d
t
=
𝒪
(
∥
g
′
∥
+
∥
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
,
iiiiiiiiiiiiiiii
π
20
≤
θ
≤
π
2
,
then (31) and (37) imply that
(38)
|
1
w
(
α
,
β
)
(
θ
)
∫
δ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
|
≤
C
(
∥
g
′
∥
+
∥
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
,
δ
≤
θ
≤
π
2
.
On the other hand, it is easy to see that the above estimate also holds forπ
/
2
≤
θ
≤
π
-
δ. Thus, the desired inequality is obtained.
Next, we are going to prove (37). This time we define
(39)
F
(
θ
)
=
∫
0
θ
/
2
‍
Δ
2
t
g
(
θ
)
tan
t
/
2
d
t
.
Simple calculation shows that(40)
F
(
θ
+
δ
)
-
F
(
θ
)
δ
=
1
δ
(
∫
0
(
θ
+
δ
)
/
2
‍
Δ
2
t
g
(
θ
+
δ
)
tan
t
/
2
d
t
-
∫
0
θ
/
2
‍
Δ
2
t
g
(
θ
)
tan
t
/
2
d
t
)
=
1
δ
∫
0
θ
/
2
‍
∫
0
δ
‍
Δ
2
t
g
′
(
θ
+
u
)
tan
t
/
2
d
u
d
t
+
𝒪
(
∥
g
′
∥
)
.
With (36), we obtain
(41)
F
(
θ
+
δ
)
-
F
(
θ
)
δ
=
𝒪
(
∥
g
′
∥
+
∥
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
,
iiiiiiiiiiiiiiii
π
20
≤
θ
≤
π
2
.
Hence,(42)
1
δ
∫
0
θ
/
2
‍
∫
0
δ
‍
Δ
2
t
g
′
(
θ
+
u
)
tan
t
/
2
d
u
d
t
=
𝒪
(
∥
g
′
∥
+
∥
g
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
,
iiiiiiiiiiiiiiiii
π
20
≤
θ
≤
π
2
.
Similar to the case ofδ
≤
θ
≤
π
/
20, we have
(43)
1
δ
∫
0
θ
/
2
‍
∫
0
δ
‍
Δ
2
t
g
′
(
θ
+
u
)
tan
t
/
2
d
u
d
t
=
1
2
δ
∫
0
δ
‍
∫
δ
θ
/
2
‍
Δ
2
t
2
g
(
θ
+
u
)
sin
2
t
/
2
d
t
d
u
+
𝒪
(
∥
g
′
∥
)
=
1
2
δ
∫
0
δ
‍
∫
δ
θ
/
2
‍
Δ
2
t
2
g
(
θ
)
sin
2
t
/
2
d
t
d
u
+
1
2
δ
∫
0
δ
‍
∫
δ
θ
/
2
‍
Δ
2
t
2
(
g
(
θ
+
u
)
-
g
(
θ
)
)
sin
2
t
/
2
d
t
d
u
+
𝒪
(
∥
g
′
∥
)
=
1
2
∫
δ
θ
/
2
‍
Δ
2
t
2
g
(
θ
)
sin
2
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
)
,
which obviously implies (37).In what follows, we will give an estimate of the conjugate function defined in the saturation classS
(
α
,
β
).Lemma 9.
Let-
1
<
α, β
<
0 be fixed. Then, for all 0
<
δ
≤
θ
≤
π
/
2 and even g
∈
C
2
π
2, one has
(44)
1
w
(
α
,
β
)
(
θ
)
∫
δ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
2
∫
δ
sin
θ
‍
Δ
t
2
g
(
θ
)
t
2
d
t
+
2
w
(
α
,
β
)
(
θ
)
×
∫
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
d
t
+
𝒪
(
∥
g
′
∥
)
=
2
∫
δ
sin
θ
‍
Δ
t
2
g
(
θ
)
t
2
d
t
-
2
w
(
α
,
β
)
(
θ
)
×
∫
θ
π
‍
(
g
(
t
)
-
g
(
0
)
)
(
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
)
′
d
t
+
𝒪
(
∥
g
′
∥
)
,
in which𝒪 does not depend on ∥
g
′
∥ and δ. Moreover, let t
n
∈
π
n be the best approximation of g. Then, for all 1
/
n
≤
θ
≤
π
-
1
/
n, one gets
(45)
1
w
(
α
,
β
)
(
θ
)
∫
1
/
n
π
‍
Δ
2
t
(
t
n
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
1
w
(
α
,
β
)
(
θ
)
∫
1
/
n
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
)
.Proof.
Since(46)
1
w
(
α
,
β
)
(
θ
)
∫
δ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
1
w
(
α
,
β
)
(
θ
)
∫
δ
sin
θ
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
+
1
w
(
α
,
β
)
(
θ
)
∫
sin
θ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
,
∫
δ
sin
θ
‍
|
w
(
α
,
β
)
(
θ
±
t
)
-
w
(
α
,
β
)
(
θ
)
|
tan
t
/
2
d
t
=
𝒪
(
w
(
α
,
β
)
(
θ
)
)
,
we have(47)
1
w
(
α
,
β
)
(
θ
)
∫
δ
sin
θ
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
∫
δ
sin
θ
‍
Δ
2
t
g
′
(
θ
)
tan
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
)
=
∫
δ
sin
θ
‍
[
Δ
t
2
g
(
θ
)
]
′
tan
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
)
.
Integrating by parts, we obtain(48)
∫
δ
sin
θ
‍
Δ
2
t
g
′
(
θ
)
tan
t
/
2
d
t
=
2
∫
δ
sin
θ
‍
Δ
t
2
g
(
θ
)
t
2
d
t
+
𝒪
(
∥
g
′
∥
)
.
Therefore,(49)
1
w
(
α
,
β
)
(
θ
)
∫
δ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
2
∫
δ
sin
θ
‍
Δ
t
2
g
(
θ
)
t
2
d
t
+
1
w
(
α
,
β
)
(
θ
)
×
∫
sin
θ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
+
𝒪
(
∥
g
′
∥
)
.
To deal with the second term of the above estimate, we note that, ifπ
/
20
≤
θ
≤
π
/
2, then w
(
α
,
β
)
(
θ
)
≍
1, and
(50)
1
w
(
α
,
β
)
(
θ
)
∫
sin
θ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
𝒪
(
∥
g
′
∥
)
,
2
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
d
t
=
𝒪
(
∥
g
′
∥
)
.
Thus,(51)
1
w
(
α
,
β
)
(
θ
)
∫
sin
θ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
2
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
d
t
+
𝒪
(
∥
g
′
∥
)
.
Ifδ
≤
θ
≤
π
/
20, rewrite the previous term as
(52)
1
w
(
α
,
β
)
(
θ
)
∫
sin
θ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
1
w
(
α
,
β
)
(
θ
)
∫
sin
θ
2
θ
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
+
1
w
(
α
,
β
)
(
θ
)
∫
2
θ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
.
Obviously, we get(53)
1
w
(
α
,
β
)
(
θ
)
∫
sin
θ
2
θ
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
𝒪
(
∥
g
′
∥
)
.
On the other hand, we have(54)
∫
2
θ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
∫
3
θ
π
+
θ
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
tan
(
t
-
θ
)
/
2
d
t
+
∫
θ
π
-
θ
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
tan
(
t
+
θ
)
/
2
d
t
=
2
∫
3
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
sin
t
cos
⁡
θ
-
cos
⁡
t
d
t
+
∫
θ
3
θ
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
tan
(
t
+
θ
)
/
2
d
t
-
∫
π
-
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
tan
⁡
(
t
+
θ
)
/
2
d
t
+
∫
π
π
+
θ
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
tan
⁡
(
t
-
θ
)
/
2
d
t
.
Since(55)
sin
2
θ
2
∫
3
θ
π
‍
w
(
α
,
β
)
(
t
)
sin
t
|
cos
⁡
θ
-
cos
⁡
t
|
(
1
-
cos
⁡
t
)
d
t
=
𝒪
(
w
(
α
,
β
)
(
θ
)
)
,
we obtain(56)
2
∫
3
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
sin
t
cos
⁡
θ
-
cos
⁡
t
d
t
=
2
∫
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
d
t
+
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
.
Moreover, the following estimates hold:(57)
∫
θ
3
θ
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
tan
(
t
+
θ
)
/
2
d
t
=
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
,
∫
π
-
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
tan
(
t
+
θ
)
/
2
d
t
=
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
,
∫
π
π
+
θ
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
tan
(
t
-
θ
)
/
2
d
t
=
𝒪
(
∥
g
′
∥
w
(
α
,
β
)
(
θ
)
)
.
Consequently, for all0
<
δ
≤
θ
≤
π
/
2, we get
(58)
1
w
(
α
,
β
)
(
θ
)
∫
δ
π
‍
Δ
2
t
(
g
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
t
/
2
d
t
=
2
∫
δ
sin
θ
‍
Δ
t
2
g
(
θ
)
t
2
d
t
+
2
w
(
α
,
β
)
(
θ
)
×
∫
θ
π
‍
g
′
(
t
)
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
d
t
+
𝒪
(
∥
g
′
∥
)
,
which proves the first assertion of the lemma.
The second estimate can be obtained from Lemma8, the first estimate, and integration by parts.Lemma 10.
There exists an absolute constantC
>
0 such that, for all even g
∈
C
2
π
2,
(59)
|
∫
θ
π
‍
Δ
t
2
g
(
0
)
t
2
d
t
|
≤
C
(
∥
g
′
∥
+
max
⁡
δ
≤
η
≤
π
/
2
|
∫
δ
π
‍
Δ
t
2
g
(
η
)
t
2
d
t
|
)
,
iiiiiiiiiiiiiiiiiiiiiiiiii
∀
0
≤
δ
≤
θ
≤
π
2
.Proof.
We may assume thatδ
≤
θ
<
1 and 1
/
θ, 1
/
δ
∈
ℕ. Thus, by [13, 14], for Fejér mean σ
1
/
θ
g of g, we have
(60)
θ
|
∫
θ
π
‍
Δ
t
2
g
(
0
)
sin
2
(
t
/
2
)
d
t
|
≤
θ
∥
∫
θ
π
‍
Δ
t
2
g
sin
2
(
t
/
2
)
d
t
∥
≤
C
∥
σ
1
/
θ
g
-
g
∥
≤
C
θ
∥
g
′
^
∥
;
therefore,(61)
|
∫
θ
π
‍
Δ
t
2
g
(
0
)
t
2
d
t
|
≤
C
(
∥
∫
0
π
‍
Δ
t
2
g
t
2
d
t
∥
+
∥
g
′
∥
)
.
ForT
∈
π
1
/
δ, we have
(62)
|
∫
θ
π
‍
Δ
t
2
T
(
0
)
t
2
d
t
|
≤
C
(
∥
∫
δ
π
‍
Δ
t
2
T
t
2
d
t
∥
+
∥
T
′
∥
+
δ
∥
T
′′
∥
)
≤
C
(
∥
∫
δ
π
‍
Δ
t
2
T
t
2
d
t
∥
+
∥
T
′
∥
)
.
Consequently, forT
∈
π
1
/
δ with ∥
T
-
g
∥
≤
C
δ
∥
g
′
∥ and ∥
T
′
∥
≤
C
∥
g
′
∥, we obtain
(63)
|
∫
θ
π
‍
Δ
t
2
g
(
0
)
t
2
d
t
|
≤
C
(
1
θ
∥
g
-
T
∥
+
∥
∫
δ
π
‍
Δ
t
2
T
t
2
d
t
∥
+
∥
g
′
∥
)
≤
C
(
1
θ
∥
g
-
T
∥
+
1
δ
∥
T
-
g
∥
+
∥
∫
δ
π
‍
Δ
t
2
g
t
2
d
t
∥
+
∥
g
′
∥
)
≤
C
(
∥
∫
δ
π
‍
Δ
t
2
g
t
2
d
t
∥
+
∥
g
′
∥
)
.
We may assume that(64)
∥
∫
δ
π
‍
Δ
t
2
g
t
2
d
t
∥
=
max
⁡
δ
≤
η
≤
π
/
2
|
∫
δ
π
‍
Δ
t
2
g
(
η
)
t
2
d
t
|
.
Otherwise, chooseh
(
η
)
=
g
(
η
)  (
|
η
|
≤
π
/
20
) and h
(
η
)
=
g
(
2
π
)
(
π
/
10
≤
|
η
|
≤
2
π
) to make h
∈
C
2
π
2 even and ∥
h
′
∥
≤
C
∥
g
′
∥. Then
(65)
∥
∫
δ
π
‍
Δ
t
2
h
t
2
d
t
∥
≤
C
(
max
⁡
δ
≤
η
≤
π
/
2
|
∫
δ
π
‍
Δ
t
2
h
(
η
)
t
2
d
t
|
+
∥
h
′
∥
)
∥
∫
δ
π
‍
Δ
t
2
h
t
2
d
t
∥
≤
C
(
max
⁡
δ
≤
η
≤
π
/
2
|
∫
δ
π
‍
Δ
t
2
g
(
η
)
t
2
d
t
|
+
∥
g
′
∥
)
.
We conclude that(66)
|
∫
θ
π
‍
Δ
t
2
g
(
0
)
t
2
d
t
|
≤
C
(
|
∫
θ
π
‍
Δ
t
2
h
(
0
)
t
2
d
t
|
+
∥
g
′
∥
)
≤
C
(
∥
∫
δ
π
‍
Δ
t
2
h
t
2
d
t
∥
+
∥
g
′
∥
)
≤
C
(
max
⁡
δ
≤
η
≤
π
/
2
|
∫
δ
π
‍
Δ
t
2
g
(
η
)
t
2
d
t
|
+
∥
g
′
∥
)
,
which gives the desired inequality.


## 3. Proof of Theorem4

We need to prove the following Lemma before Theorem4.Lemma 11.
Given-
1
<
α, β
<
0, there is C
>
0 such that, for all f
∈
C
[
-
1,1
] and all n
≥
1, one has
(67)
1
n
∥
φ
H
n
(
α
,
β
)
′
(
f
)
∥
≤
C
K
α
,
β
(
f
,
1
n
)
,
∥
f
-
H
n
(
α
,
β
)
(
f
)
∥
+
1
n
∥
φ
H
n
(
α
,
β
)
′
(
f
)
∥
≍
K
α
,
β
(
f
,
1
n
)
.Proof.
Denote thatT
=
H
n
(
α
,
β
)
(
f
). Then from Theorem 3, we have
(68)
∥
f
-
T
∥
≤
C
K
α
,
β
(
f
,
1
n
)
.
Next, let us estimate(
1
/
n
)
∥
φ
T
′
∥.
We know thatT
′
(
x
)
φ
(
x
)
=
-
t
n
′
(
θ
) with θ
=
arccos
x and t
n
(
θ
)
=
T
(
cos
⁡
θ
). But (1) tells that T
′
(
x
k
)
=
0. Hence, t
n
′
(
θ
k
)
=
0 for θ
k
=
arccos
x
k. Lemma 5 tells that, for each 0
≤
θ
≤
π, there is θ
k satisfying |
θ
-
θ
k
|
≤
C
/
n. Assume that |
t
n
′
(
θ
)
|
=
∥
t
n
′
∥. Then ∥
t
n
′
∥
=
|
t
n
′
(
θ
)
-
t
n
′
(
θ
k
)
| and further
(69)
∥
t
n
′
∥
≤
C
n
∥
t
n
′
′
∥
.
We may assume thatn
=
2
m. Thus, the Bernstein inequality for trigonometric polynomials yields
(70)
∥
t
n
′
′
∥
≤
∑
j
=
1
m
‍
∥
t
2
j
′
′
-
t
2
j
-
1
′
′
∥
≤
C
∑
j
=
0
m
‍
2
2
j
∥
f
-
H
2
j
(
α
,
β
)
(
f
)
∥
≤
C
n
max
⁡
1
≤
m
≤
n
m
∥
f
-
H
m
(
α
,
β
)
(
f
)
∥
.
But(71)
1
n
max
⁡
1
≤
m
≤
n
m
∥
f
-
H
m
(
α
,
β
)
(
f
)
∥
≤
C
n
max
⁡
1
≤
m
≤
n
(
m
K
α
,
β
(
f
,
1
m
)
)
≤
C
K
α
,
β
(
f
,
1
n
)
.
So, we have(72)
1
n
∥
φ
T
′
∥
≤
C
K
α
,
β
(
f
,
1
n
)
.
Combining this inequality with (68), we get
(73)
∥
f
-
T
∥
+
1
n
∥
φ
T
′
∥
≤
C
K
α
,
β
(
f
,
1
n
)
.
Now we need only to prove(74)
K
α
,
β
(
f
,
1
n
)
≤
C
(
∥
f
-
T
∥
+
1
n
∥
φ
T
′
∥
)
.
Obviously,(75)
K
α
,
β
(
f
,
1
n
)
≤
∥
f
-
T
∥
+
1
n
∥
φ
T
′
∥
+
1
n
∥
t
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
.
Lemma5 tells the following. Let a
>
0 be fixed, then for θ
∈
M
∩
{
θ
∣
a
/
n
≤
θ
≤
π
-
a
/
n
},
(76)
1
n
≍
{
P
n
(
α
,
β
)
(
cos
⁡
θ
)
}
2
w
(
α
,
β
)
(
θ
)
.
Thus, for thoseθ,
(77)
1
n
|
t
′
w
(
α
,
β
)
^
(
θ
)
w
(
α
,
β
)
(
θ
)
|
≍
{
P
n
(
α
,
β
)
(
cos
⁡
θ
)
}
2
|
t
′
w
(
α
,
β
)
^
(
θ
)
|
.
SinceΔ
n
(
α
,
β
)
(
T
)
=
0, following (23), (18), and Lemma 6, we have, for those θ,
(78)
1
n
|
t
′
w
(
α
,
β
)
^
(
θ
)
w
(
α
,
β
)
(
θ
)
|
≤
1
n
∥
φ
T
′
∥
.
Ifa
/
n
≤
θ
≤
π
-
a
/
n and θ
∉
M, then there is a θ
′
∈
M
∩
{
θ
∣
a
/
n
≤
θ
≤
π
-
a
/
n
} satisfying |
θ
-
θ
′
|
≤
C
/
n (see Lemma 5). Hence, from Bernstein inequality ∥
t
′′
∥
≤
C
n
∥
t
′
∥, we have
(79)
1
n
∫
1
/
n
sin
θ
‍
Δ
u
2
t
(
θ
)
sin
2
(
u
/
2
)
d
u
=
1
n
∫
1
/
n
sin
θ
′
‍
Δ
u
2
t
(
θ
′
)
sin
2
(
u
/
2
)
d
u
+
1
n
∫
sin
θ
′
sin
θ
‍
Δ
u
2
t
(
θ
)
sin
2
(
u
/
2
)
d
u
+
1
n
∫
1
/
n
sin
θ
′
‍
Δ
u
2
(
t
(
θ
)
-
t
(
θ
′
)
)
sin
2
(
u
/
2
)
d
u
=
1
n
∫
1
/
n
sin
θ
′
‍
Δ
u
2
t
(
θ
′
)
sin
2
(
u
/
2
)
d
+
𝒪
(
1
n
∥
φ
T
′
∥
)
.
Moreover,(80)
1
n
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
=
1
n
w
(
α
,
β
)
(
θ
′
)
∫
θ
′
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
+
𝒪
(
1
n
∥
φ
T
′
∥
)
.
Consequently, (78) obtained from Lemma 9 holds for all θ
∈
[
1
/
n
,
π
-
1
/
n
]. Finally, from (78) and (75) we obtain (74).Proof of Theorem4.
Firstly, we prove that(81)
K
α
,
β
(
f
,
δ
)
≤
C
(
K
α
0
,
β
0
(
f
,
δ
)
+
J
α
,
β
,
1
(
f
,
δ
)
+
J
α
,
β
,
2
(
f
,
δ
)
)
.
LetT
=
H
n
(
α
0
,
β
0
)
(
f
), t
(
θ
)
=
T
(
cos
⁡
θ
), and g
(
θ
)
=
f
(
cos
⁡
θ
). From Theorems 1 and 2 we conclude that, for some n
≤
n
′
≤
A
n,
(82)
K
α
,
β
(
f
,
1
n
)
≤
C
∥
f
-
H
n
′
(
α
,
β
)
(
f
)
∥
≤
C
(
∥
f
-
H
n
(
α
0
,
β
0
)
(
f
)
∥
+
∥
H
n
(
α
0
,
β
0
)
(
f
)
-
H
n
′
(
α
,
β
)
(
H
n
(
α
0
,
β
0
)
(
f
)
)
∥
)
.
We know thatH
n
(
α
0
,
β
0
)
(
f
)
-
H
n
′
(
α
,
β
)
(
H
n
(
α
0
,
β
0
)
(
f
)
)
∈
Π
2
n
′
-
1, so (see [14, page 43])
(83)
∥
H
n
(
α
0
,
β
0
)
(
f
)
-
H
n
′
(
α
,
β
)
(
H
n
(
α
0
,
β
0
)
(
f
)
)
∥
≤
C
∥
H
n
(
α
0
,
β
0
)
(
f
)
-
H
n
′
(
α
,
β
)
(
H
n
(
α
0
,
β
0
)
(
f
)
)
∥
[
-
1
+
1
/
n
2
,
1
-
1
/
n
2
]
.
Thus, (18) and Lemmas 6 and 7 imply that
(84)
∥
H
n
(
α
0
,
β
0
)
(
f
)
-
H
n
′
(
α
,
β
)
(
H
n
(
α
0
,
β
0
)
(
f
)
)
∥
≤
C
n
(
∥
t
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
[
1
/
n
,
π
-
1
/
n
]
+
∥
t
′
∥
)
.
Consequently, by Lemma11, we get
(85)
K
α
,
β
(
f
,
1
n
)
≤
C
(
K
α
0
,
β
0
(
f
,
1
n
)
+
1
n
∥
t
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
[
1
/
n
,
π
-
1
/
n
]
)
.
Obviously,(86)
∥
t
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
[
1
/
n
,
π
-
1
/
n
]
≤
C
∥
1
w
(
α
,
β
)
∫
1
/
n
π
‍
Δ
2
u
(
t
′
w
(
α
,
β
)
)
tan
(
u
/
2
)
d
u
∥
[
1
/
n
,
π
-
1
/
n
]
+
C
∥
t
′
∥
.
If1
/
n
≤
θ
≤
π
/
2, then Lemma 9 implies that
(87)
1
w
(
α
,
β
)
(
θ
)
∫
1
/
n
π
‍
Δ
2
u
(
t
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
(
u
/
2
)
d
u
=
2
∫
1
/
n
sin
θ
‍
Δ
u
2
t
(
θ
)
u
2
d
u
-
2
w
(
α
,
β
)
(
θ
)
×
∫
θ
π
‍
(
t
(
u
)
-
t
(
0
)
)
(
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
)
′
d
u
+
𝒪
(
∥
t
′
∥
)
.
Since(88)
1
n
|
∫
1
/
n
sin
θ
‍
Δ
u
2
t
(
θ
)
u
2
d
u
|
≤
C
K
α
0
,
β
0
(
f
,
1
n
)
,
1
n
|
2
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
(
t
(
u
)
-
t
(
0
)
)
(
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
)
′
d
u
|
≤
C
J
α
,
β
,
1
(
f
,
1
n
)
,
therefore, for1
/
n
≤
θ
≤
π
/
2, we have
(89)
1
n
|
1
w
(
α
,
β
)
(
θ
)
∫
1
/
n
π
‍
Δ
2
u
(
t
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
(
u
/
2
)
d
u
|
≤
C
(
K
α
0
,
β
0
(
f
,
1
n
)
+
J
α
,
β
,
1
(
f
,
1
n
)
)
.
In the same way, ifπ
/
2
≤
θ
≤
π
-
1
/
n, we obtain
(90)
1
n
|
1
w
(
α
,
β
)
(
θ
)
∫
1
/
n
π
‍
Δ
2
u
(
t
′
(
θ
)
w
(
α
,
β
)
(
θ
)
)
tan
(
u
/
2
)
d
u
|
≤
C
(
K
α
0
,
β
0
(
f
,
1
n
)
+
J
α
,
β
,
2
(
f
,
1
n
)
)
.
Consequently,(91)
1
n
∥
t
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
[
1
/
n
,
π
-
1
/
n
]
≤
C
(
K
α
0
,
β
0
(
f
,
1
n
)
+
J
α
,
β
,
1
(
f
,
1
n
)
+
J
α
,
β
,
2
(
f
,
1
n
)
)
.
Therefore,(92)
K
α
,
β
(
f
,
1
n
)
≤
C
(
K
α
0
,
β
0
(
f
,
1
n
)
+
J
α
,
β
,
1
(
f
,
1
n
)
+
J
α
,
β
,
2
(
f
,
1
n
)
)
.
Next, we prove that(93)
K
α
0
,
β
0
(
f
,
δ
)
+
J
α
,
β
,
1
(
f
,
δ
)
+
J
α
,
β
,
2
(
f
,
δ
)
≤
C
K
α
,
β
(
f
,
δ
)
.
Firstly, we assume that-
1
/
2
<
α
<
0 and 0
<
1
/
n
≤
θ
≤
π
/
2. Thus, for T
∈
Π
n and t
(
θ
)
=
T
(
cos
⁡
θ
), we have
(94)
max
⁡
1
/
n
≤
θ
≤
π
/
2
|
1
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
|
≤
max
⁡
1
/
n
≤
θ
≤
π
/
2
|
1
w
(
α
,
β
)
(
θ
)
∫
0
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
|
+
C
∥
t
′
∥
≤
C
|
1
w
(
α
,
β
)
(
1
/
n
)
∫
0
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
|
+
C
∥
t
′
∥
≤
C
max
⁡
1
/
n
≤
θ
≤
π
/
2
|
∫
1
/
n
sin
θ
‍
Δ
u
2
t
(
θ
)
u
2
d
u
+
1
w
(
α
,
β
)
(
θ
)
iiiiiiiiiiiiiiiiiiiiiiii
×
∫
θ
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
|
+
C
∥
t
′
∥
.
ForH
∈
C
2
[
-
1,1
], denote that h
(
θ
)
=
H
(
cos
⁡
θ
) and let T be the best approximation of H. Following from (94), Lemmas 8 and 9, we have
(95)
1
n
max
⁡
1
/
n
≤
θ
≤
π
/
2
|
1
w
(
α
,
β
)
(
θ
)
(
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
)
′
×
∫
θ
π
‍
(
h
(
u
)
-
h
(
0
)
)
(
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
)
′
d
u
|
≤
C
n
(
∥
h
′
∥
+
∥
h
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
.
Thus,(96)
J
α
,
β
,
1
(
f
,
1
n
)
≤
C
(
∥
f
-
H
∥
+
1
n
(
∥
h
′
∥
+
∥
h
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
)
.
Therefore,(97)
J
α
,
β
,
1
(
f
,
1
n
)
≤
C
K
α
,
β
(
f
,
1
n
)
.
Next, suppose thatα
=
-
1
/
2. Clearly, we have
(98)
(
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
)
′
=
-
(
β
+
1
)
cos
⁡
2
β
+
1
t
2
(
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
)
′
=
-
1
2
cos
⁡
2
β
+
3
t
2
sin
-
2
t
2
,
andw
(
α
,
β
)
(
θ
)
≍
1 for δ
≤
θ
≤
π
/
2. Hence, let H
=
H
n
(
α
,
β
)
(
f
), h
(
θ
)
=
H
(
cos
⁡
θ
), and n
=
1
/
δ. From Lemma 9 we obtain
(99)
|
1
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
(
h
(
t
)
-
h
(
0
)
)
(
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
)
′
d
t
|
≤
C
(
|
∫
θ
π
‍
Δ
t
2
h
(
0
)
t
2
d
t
|
+
∥
h
′
∥
)
,
forδ
≤
θ
≤
π
/
2.
On the other hand, sinceα
=
-
1
/
2, we have
(100)
max
⁡
δ
≤
η
≤
π
/
2
|
∫
δ
π
‍
Δ
t
2
h
(
η
)
t
2
d
t
|
+
∥
h
′
∥
≍
max
⁡
δ
≤
η
≤
π
/
2
|
1
w
(
α
,
β
)
(
η
)
∫
δ
π
‍
Δ
2
t
(
h
′
(
η
)
w
(
α
,
β
)
(
η
)
)
tan
t
/
2
d
t
|
+
∥
h
′
∥
.
We know that, forg
(
θ
)
=
f
(
cos
⁡
θ
),
(101)
J
α
,
β
,
1
(
f
,
δ
)
≍
max
⁡
δ
≤
θ
≤
π
/
2
|
1
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
(
g
(
t
)
-
g
(
0
)
)
(
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
)
′
d
t
|
.
Lemma10 shows that the last three estimates imply that
(102)
J
α
,
β
,
1
(
f
,
δ
)
≤
C
(
∥
f
-
H
∥
|
1
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
(
h
(
t
)
-
h
(
0
)
)
(
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
)
′
d
t
|
+
δ
max
⁡
δ
≤
θ
≤
π
/
2
|
1
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
(
h
(
t
)
-
h
(
0
)
)
(
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
)
′
×
(
w
(
α
,
β
)
(
t
)
sin
t
1
-
cos
⁡
t
)
′
d
t
|
)
+
C
δ
∥
h
′
∥
≤
C
(
∥
f
-
H
∥
∥
h
′
∥
+
δ
max
⁡
δ
≤
η
≤
π
/
2
|
1
w
(
α
,
β
)
(
η
)
∫
δ
π
‍
Δ
2
t
(
h
′
(
η
)
w
(
α
,
β
)
(
η
)
)
tan
t
/
2
d
t
|
+
δ
∥
h
′
∥
)
≤
C
(
∥
f
-
H
∥
+
δ
∥
h
′
∥
+
δ
∥
h
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
[
δ
,
π
-
δ
]
)
.
So by Lemma7 (see (78)) and Lemma 11 we obtain again
(103)
J
α
,
β
,
1
(
f
,
δ
)
≤
C
K
α
,
β
(
f
,
δ
)
.
If-
1
<
α
<
-
1
/
2, then, due to |
f
(
1
)
-
f
(
t
)
|
≤
C
ω
φ
(
f
,
1
-
t
) (see (2.6) of [9]), we have
(104)
J
α
,
β
,
1
(
f
,
δ
)
≤
C
ω
φ
(
f
,
δ
)
≤
C
inf
⁡
H
∈
C
2
[
-
1,1
]
(
∥
f
-
H
∥
+
δ
∥
h
′
∥
)
≤
C
K
α
,
β
(
f
,
δ
)
.
Thus, (103) holds for all -
1
<
α
<
0.
In the same way one can verify that(105)
J
α
,
β
,
2
(
f
,
δ
)
≤
C
K
α
,
β
(
f
,
δ
)
.
To complete the proof we have to verify that(106)
K
α
0
,
β
0
(
f
,
δ
)
≤
C
K
α
,
β
(
f
,
δ
)
,
where-
1
<
α
0, β
0
<
-
1
/
2 as given by this theorem. Write g
(
θ
)
=
f
(
cos
⁡
θ
). Theorem 2 shows that
(107)
K
α
0
,
β
0
(
f
,
δ
)
≍
max
⁡
δ
≤
θ
≤
π
-
δ
|
∫
δ
sin
θ
‍
Δ
u
2
g
(
θ
)
u
2
d
u
|
+
ω
(
g
,
δ
)
.
Clearly, we have(108)
ω
(
g
,
δ
)
≤
C
inf
⁡
H
∈
C
2
(
∥
f
-
H
∥
+
δ
∥
h
′
∥
)
≤
C
K
α
,
β
(
f
,
δ
)
,
whereh
(
θ
)
=
H
(
cos
⁡
θ
). We need only to prove that
(109)
max
⁡
δ
≤
θ
≤
π
-
δ
|
∫
δ
sin
θ
‍
Δ
u
2
g
(
θ
)
u
2
d
u
|
≤
C
K
α
,
β
(
f
,
δ
)
.
Finally, lett
∈
π
n be even and n
=
1
/
δ. Notice that, for α
=
-
1
/
2,
(110)
|
1
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
|
≤
C
|
∫
θ
π
‍
Δ
u
2
t
(
0
)
u
2
d
u
|
+
C
∥
t
′
∥
.
It follows from Lemmas10 and 9 that
(111)
max
⁡
1
/
n
≤
θ
≤
π
/
2
|
∫
1
/
n
sin
θ
‍
Δ
u
2
t
(
θ
)
u
2
d
u
|
≤
C
max
⁡
1
/
n
≤
θ
≤
π
/
2
|
∫
1
/
n
sin
θ
‍
Δ
u
2
t
(
θ
)
u
2
d
u
+
1
w
(
α
,
β
)
(
θ
)
×
∫
θ
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
|
+
C
∥
t
′
∥
.
If-
1
/
2
<
α
<
0, then (94) implies (111). If -
1
<
α
<
-
1
/
2, then
(112)
max
⁡
1
/
n
≤
θ
≤
π
/
2
|
1
w
(
α
,
β
)
(
θ
)
∫
θ
π
‍
t
′
(
u
)
w
(
α
,
β
)
(
u
)
sin
u
1
-
cos
⁡
u
d
u
|
≤
C
∥
t
′
∥
.
Thus, (111) is valid for all -
1
<
α
<
0. Consequently, now let t
∈
π
n be the best approximation of h
(
θ
)
=
H
(
cos
⁡
θ
) and H
∈
C
2
[
-
1,1
]. From (111) and  Lemmas 8 and 9, we have
(113)
1
n
max
⁡
1
/
n
≤
θ
≤
π
/
2
|
∫
1
/
n
sin
θ
‍
Δ
u
2
t
(
θ
)
u
2
d
u
|
≤
C
n
(
∥
h
′
∥
+
∥
h
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
.
When we use Lemma9 for π
-
θ, the above is true for π
/
2
≤
θ
≤
π
-
1
/
n instead of 1
/
n
≤
θ
≤
π
/
2. Thus,
(114)
1
n
max
⁡
1
/
n
≤
θ
≤
π
-
1
/
n
|
∫
1
/
n
sin
θ
‍
Δ
u
2
h
(
θ
)
u
2
d
u
|
≤
C
n
(
∥
h
′
∥
+
∥
h
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
.
Therefore,(115)
1
n
max
⁡
1
/
n
≤
θ
≤
π
-
1
/
n
|
∫
1
/
n
sin
θ
‍
Δ
u
2
g
(
θ
)
u
2
d
u
|
≤
C
∥
f
-
H
∥
+
C
n
(
∥
h
′
∥
+
∥
h
′
w
(
α
,
β
)
^
w
(
α
,
β
)
∥
)
,
which gives (109).



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*Source: 781068-2014-01-16.xml*