Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 607204, 20 pages
http://dx.doi.org/10.1155/2013/607204
Research Article
Some Bounded Linear Integral Operators and Linear Fredholm
Integral Equations in the Spaces 𝐻𝛼,𝛿,𝛾((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and
𝐻𝛼,𝛿((𝑎, 𝑏), 𝑋)
Esmet Özdemir,1 Ali M. Akhmedov,2 and Ö. Faruk Temizer1
1
2
İnönü Üniversitesi, Eğitim Fakültesi, A-Blok, Posta Kodu: 44280, Malatya, Turkey
Baku State University, Department of Mech. & Math., Z. Khalilov Street, 23, P.O. Box 370145, Baku, Azerbaijan
Correspondence should be addressed to İsmet Özdemir; ismet.ozdemir@inonu.edu.tr
Received 14 November 2012; Accepted 16 January 2013
Academic Editor: Juan Carlos Cortés López
Copyright © 2013 İsmet Özdemir et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The spaces 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), R) and 𝐻𝛼,𝛿 ((𝑎, 𝑏), R) were defined in ((Hüseynov (1981)), pages 271–277). Some singular integral
operators on Banach spaces were examined, (Dostanic (2012)), (Dunford (1988), pages 2419–2426 and (Plamenevskiy (1965)). The
solutions of some singular Fredholm integral equations were given in (Babolian (2011), Okayama (2010), and Thomas (1981)) by
numerical methods. In this paper, we define the sets 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋) by taking an arbitrary Banach space
𝑋 instead of R, and we show that these sets which are different from the spaces given in (Dunford (1988)) and (Plamenevskiy (1965))
are Banach spaces with the norms ‖ ⋅ ‖𝛼,𝛿,𝛾 and ‖ ⋅ ‖𝛼,𝛿 . Besides, the bounded linear integral operators on the spaces 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) ×
(𝑎, 𝑏), 𝑋) and 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋), some of which are singular, are derived, and the solutions of the linear Fredholm integral equations of
𝑏
𝑏
𝑏
the form 𝑓(𝑠) = 𝜙(𝑠) + 𝜆 ∫𝑎 𝐴(𝑠, 𝑡)𝑓(𝑡)𝑑𝑡, 𝑓(𝑠) = 𝜙(𝑠) + 𝜆 ∫𝑎 𝐴(𝑡, 𝑠)𝑓(𝑡)𝑑𝑡 and 𝑓(𝑠, 𝑡) = 𝜙(𝑠, 𝑡) + 𝜆 ∫𝑎 𝐴(𝑠, 𝑡)𝑓(𝑡, 𝑠)𝑑𝑡 are investigated
in these spaces by analytical methods.
1. Preliminaries, Background, and Notation
The approximate solution of the singular integral equation
𝑢 (𝑠) −
1
𝑘 (𝑠, 𝑡) 𝑢 (𝑡)
1
𝑝.𝑣. ∫
𝑑𝑡 = 𝑓 (𝑡)
𝜋
𝑡−𝑠
−1
(1)
was obtained in [1], where 𝑘(𝑠, 𝑡) is a real-valued kernel, 𝑓(𝑠)
is a given function, and 𝑢(𝑠) is the unknown function.
High-order numerical methods for the singular Fredholm integral equations of the form:
𝑏
𝜆𝑢 (𝑡) − 𝑝.𝑣. ∫ |𝑡 − 𝑠|
𝑎
𝑝−1
𝑘 (𝑡, 𝑠) 𝑢 (𝑠) 𝑑𝑠 = 𝑔 (𝑡) ,
problems, mathematical problems of radiative equilibrium,
and radiative heat transfer problems, [2].
The polar kernel of integral equations was introduced in
[3, 4]. This singular kernel is in the following form:
(2)
𝑎≤𝑡≤𝑏
were developed, where 𝜆 ≠ 0, 0 < 𝑝 < 1, 𝑘 and 𝑔 are given
functions, and 𝑢 is the unknown function. Equations of this
form often arise in practical applications such as Dirichlet
𝑘 (𝑥, 𝑦) =
𝑔 (𝑥, 𝑦)
𝛼
(𝑥 − 𝑦)
+ ℎ (𝑥, 𝑦) ,
0 < 𝛼 ≤ 1,
(3)
where the first term of this kernel is weakly singular and 𝑔
and ℎ are bounded on the square 𝑠 = [−1, 1] × [−1, 1] and
𝑔(𝑥, 𝑦) ≠ 0. With 𝑔 = 1, ℎ = 0, we have the special case of the
above kernel:
𝑘 (𝑥, 𝑦) =
1
𝛼,
(𝑥 − 𝑦)
0 < 𝛼 ≤ 1,
(4)
2
Abstract and Applied Analysis
see [5]. One of the weakly singular integral and integrodifferential equations with this kernel was given in [6–8]. The
solution of the singular integral equation of the form:
1
𝜇 (𝑥) 𝜙 (𝑥) + 𝜆 (𝑥) 𝑝.𝑣. ∫
−1
𝜙 (𝑦)
𝛼 𝑑𝑦
(𝑦 − 𝑥)
= 𝑓 (𝑥) ,
Theorem 5 (see [13, page 336]). Let 𝑋 be a Banach space with
the norm ‖ ⋅ ‖𝑋 and (𝑆, Σ, 𝜇) be a finite measure space. If 𝑔 :
𝑆 → 𝑋 is strongly measurable, then the scalar function
ℎ : 𝑆 󳨀→ R,
(5)
|𝑥| < 1, 0 < 𝛼 ≤ 1
was examined in [5] by numerical methods, where
𝜇(𝑥) ≠ 0, 𝜆(𝑥) ≠ 0 and 𝜇(𝑥), 𝜆(𝑥), 𝑓(𝑥) ∈ 𝐿2 [−1, 1] are
given functions and 𝜙(𝑥) is the unknown function to be
determined.
The integral operator defined by
𝑇
Definition 6 (see [11, page 88]). Let (𝑆, Σ, 𝜇) be a finite
measure space and 𝑋 be a Banach space, and then the
Bochner integral of a strongly measurable function 𝑔 :
𝑆 → 𝑋 is the strongly limit of the Bochner integral of an
approximating sequence (𝑓𝑛 ) of simple functions satisfying
(10). That is,
lim
∫ 𝑓𝑛 𝑑𝜇 = ∫ 𝑔𝑑𝜇.
𝑛
𝐸
(6)
0
was studied in [9].
An integral equation of the form:
𝑏
𝑓 (𝑠) = 𝜙 (𝑠) + 𝜆 ∫ 𝐴 (𝑠, 𝑡) 𝑓 (𝑡) 𝑑𝑡
𝑎
(7)
is called Fredholm integral equation of the second type. Here,
[𝑎, 𝑏] is a given interval, 𝑓 is a function on [𝑎, 𝑏] which is
unknown, and 𝜆 is a parameter. The kernel 𝐴 of the equation
is a given function on the square [𝑎, 𝑏] × [𝑎, 𝑏] and 𝜙 is a given
function on [𝑎, 𝑏].
Now, we may give some required definitions and theorems.
Definition 1 (see [10, page 41]). Let 𝑋 be a Banach space and
(𝑆, Σ, 𝜇) be a finite measure space. Then 𝑓 : 𝑆 → 𝑋 is called
measurable simple function if there exist 𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ∈ 𝑋
and 𝐸1 , 𝐸2 , . . . , 𝐸𝑛 ∈ Σ such that 𝑓 = ∑𝑛𝑖=1 𝑥𝑖 𝜒𝐸𝑖 , where
1, 𝑤 ∈ 𝐸𝑖
𝜒𝐸𝑖 (𝑤) = {
0, 𝑤 ∉ 𝐸𝑖 .
(8)
Definition 2 (see [11, page 88]). The Bochner integral of a
simple function 𝑓 given by Definition 1 with respect to 𝜇 on
𝐸 ∈ Σ is defined by
𝑛
∫ 𝑓𝑑𝜇 = ∑𝑥𝑖 𝜇 (𝐸𝑖 ∩ 𝐸) .
𝐸
(11)
is Σ-measurable.
𝐴 : 𝐿 2 (0, 𝑇) 󳨀→ 𝐿 2 (0, 𝑇) ,
(𝐴𝑓) (𝑥) = ∫ 𝑘 (𝑥 − 𝑦) 𝑓 (𝑦) 𝑑𝑦
󵄩
󵄩
ℎ (𝑡) = 󵄩󵄩󵄩𝑔 (𝑡)󵄩󵄩󵄩𝑋
(9)
𝑖=1
(12)
𝐸
Theorem 7 (see [12, page 202]). Let (𝑆, Σ, 𝜇) be a finite
measure space, 𝑋 be a Banach space with the norm ‖ ⋅ ‖𝑋 , and
𝑔 : 𝑆 → 𝑋 be a strongly measurable function. If ∫𝐸 𝑔𝑑𝜇 exists,
then
󵄩
󵄩󵄩
󵄩󵄩∫ 𝑔𝑑𝜇󵄩󵄩󵄩 ≤ ∫ 󵄩󵄩󵄩𝑔󵄩󵄩󵄩 𝑑𝜇.
(13)
󵄩󵄩
󵄩󵄩
󵄩 󵄩𝑋
󵄩𝑋
󵄩 𝐸
𝐸
Theorem 8 (see [12, page 203]). Let (𝑆, Σ, 𝜇) be a finite
measure space and 𝑋 be a Banach space with the norm ‖ ⋅ ‖𝑋 .
The Bochner integral ∫𝐸 𝑔𝑑𝜇 exists if and only if 𝑔 is strongly
measurable and ∫𝐸 ‖𝑔‖𝑋 𝑑𝜇 < ∞.
Theorem 9 (see [14, page 82]). Let 𝑋 be a real or complex
Banach space with the norm ‖ ⋅ ‖𝑋 . Let 𝑎, 𝑏, 𝛼, 𝛿 be the real
numbers satisfying −∞ < 𝑎, 𝑏 < ∞, 0 < 𝛼, 𝛿 < 1 and
𝑀 be a nonnegative constant. The set 𝐶0,𝛼,𝛿 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋)
consisting of all functions 𝐴 : (𝑎, 𝑏) × (𝑎, 𝑏) → 𝑋 fulfilling the
conditions:
‖𝐴 (𝑠, 𝑡)‖𝑋 ≤ 𝑀,
(14)
‖𝐴 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) − 𝐴 (𝑠, 𝑡)‖𝑋 ≤ 𝑀 (|Δ𝑠|𝛼 + |Δ𝑡|𝛿 )
for all 𝑠, 𝑡, 𝑠 + Δ𝑠, 𝑡 + Δ𝑡 ∈ (𝑎, 𝑏) is a linear space with the
algebraic operations:
(𝐴 + 𝐵) (𝑠, 𝑡) = 𝐴 (𝑠, 𝑡) + 𝐵 (𝑠, 𝑡) ,
(𝜆𝐴) (𝑠, 𝑡) = 𝜆𝐴 (𝑠, 𝑡) ,
𝜆 ∈ 𝐾, (𝐾 = R or 𝐾 = C) ,
(15)
and 𝐶0,𝛼,𝛿 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) is a Banach space with the norm
Definition 3 (see [12, page 201]). Let 𝑋 be a Banach space with
the norm ‖ ⋅ ‖𝑋 and (𝑆, Σ, 𝜇) be a finite measure space. Then
𝑔 : 𝑆 → 𝑋 is called strongly measurable if there exists a
sequence (𝑓𝑛 ) of 𝑋-valued simple functions defined on 𝑆 such
that
󵄩
󵄩󵄩
lim
󵄩𝑓 (𝑡) − 𝑔 (𝑡)󵄩󵄩󵄩𝑋 = 0; 𝑡 ∈ 𝑆 − 𝑁, 𝜇 (𝑁) = 0. (10)
𝑛 󵄩 𝑛
where
Theorem 4 (see [11, page 88]). All continuous functions 𝑓 :
𝑆 → 𝑋 are strongly measurable.
Again, let 𝑋 be a real or complex Banach space with
the norm ‖ ⋅ ‖𝑋 . Let 𝑎, 𝑏, 𝛼 be the real numbers with
‖𝐴‖0,𝛼,𝛿 = max { sup ‖𝐴 (𝑠, 𝑡)‖𝑋 , 𝑚𝐴 } ,
𝑠,𝑡 ∈ (𝑎,𝑏)
𝑚𝐴 =
sup
𝑠,𝑡,𝑠+Δ𝑠,𝑡+Δ𝑡 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0∨Δ𝑡 ≠ 0
‖𝐴 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) − 𝐴 (𝑠, 𝑡)‖𝑋
|Δ𝑠|𝛼 + |Δ𝑡|𝛿
(16)
. (17)
Abstract and Applied Analysis
3
−∞ < 𝑎,𝑏 < ∞, 0 < 𝛼 < 1 and 𝑀 also be a nonnegative
constant. Let 𝐴 be an 𝑋-valued function defined on (𝑎, 𝑏) such
that
2. The Main Results
(18)
2.1. The Spaces 𝐻𝛼,𝛿,𝛾 ((𝑎,𝑏) × (𝑎,𝑏),𝑋) and 𝐻𝛼,𝛿 ((𝑎,𝑏),𝑋). In
this section, we determine essentially the spaces 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏)×
(𝑎, 𝑏), 𝑋) and 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋).
for all 𝑠, 𝑠 + Δ𝑠 ∈ (𝑎, 𝑏) with Δ𝑠 ≥ 0. By 𝐶0,𝛼 ((𝑎, 𝑏), 𝑋), we
denote the set of all functions 𝐴 satisfying (18). 𝐶0,𝛼 ((𝑎, 𝑏), 𝑋)
is a linear space with the algebraic operations:
Theorem 11. Let 𝑋 be a real or complex Banach space with the
norm ‖ ⋅ ‖𝑋 , 𝑎, 𝑏, 𝛼, 𝛿, 𝛾 be real numbers with −∞ < 𝑎, 𝑏 <
∞, 0 < 𝛼, 𝛼 + 𝛾 < 1, 0 < 𝛾 < 𝛿 < 1, and 𝑀 be a nonnegative
constant. Then the set 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) of all functions
‖𝐴 (𝑠)‖𝑋 ≤ 𝑀,
‖𝐴 (𝑠 + Δ𝑠) − 𝐴 (𝑠)‖𝑋 ≤ 𝑀(Δ𝑠)𝛼
(𝐴 + 𝐵) (𝑠) = 𝐴 (𝑠) + 𝐵 (𝑠) ,
(𝜆𝐴) (𝑠) = 𝜆𝐴 (𝑠) ,
𝜆 ∈ 𝐾, (𝐾 = R or 𝐾 = C) ,
‖𝐴 (𝑠, 𝑡)‖𝑋 ≤
{
{
= max { sup ‖𝐴 (𝑠)‖𝑋 ,
{𝑠 ∈ (𝑎,𝑏)
{
𝐶 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and 𝐶 ((𝑎, 𝑏), 𝑋) are called a
Hölder space. The functions spaces which are similar to
𝐶0,𝛼,𝛿 ((𝑎, 𝑏)×(𝑎, 𝑏), 𝑋) and 𝐶0,𝛼 ((𝑎, 𝑏), 𝑋) were investigated in
[15, pages 2419–2426], [16, pages 25–51], and [17, pages 18–33].
The class 𝐺1 (ℎ, 𝑒) of the functions 𝑘 satisfying the equalities
󵄨−𝑚 󵄨
󵄨
󵄨
(21)
𝑘 (𝑥, 𝑦) = 𝑂 (󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 ℎ (󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨))
𝑘 (𝑥󸀠 , 𝑦) − 𝑘 (𝑥󸀠󸀠 , 𝑦) = 𝑂 (
󵄨
󵄨
ℎ (𝑝𝑦 (𝑥󸀠 , 𝑥󸀠󸀠 )) 𝑒 (󵄨󵄨󵄨󵄨𝑥󸀠 − 𝑥󸀠󸀠 󵄨󵄨󵄨󵄨)
)
𝑝𝑦𝑚 (𝑥󸀠 , 𝑥󸀠󸀠 ) 𝑒 (𝑝𝑦 (𝑥󸀠 − 𝑥󸀠󸀠 ))
(22)
was introduced in [18], where functions ℎ, 𝑒 : (0, ∞) →
(0, ∞) are increasing,
󵄨 󵄨
󵄨
󵄨
𝑝𝑦 (𝑥󸀠 , 𝑥󸀠󸀠 ) = min {󵄨󵄨󵄨󵄨𝑦 − 𝑥󸀠 󵄨󵄨󵄨󵄨 , 󵄨󵄨󵄨󵄨𝑦 − 𝑥󸀠󸀠 󵄨󵄨󵄨󵄨}
(23)
and 𝑚 is a natural number.
Theorem 10 (see [14, page 16]). Let 𝑇 : 𝑋 → 𝑋 be a
bounded linear operator mapping a Banach space 𝑋 into itself
with ‖𝑇‖ < 1 and 𝐼 : 𝑋 → 𝑋 denote the identity operator.
Then 𝐼 − 𝑇 has a bounded inverse operator on 𝑋 which is given
by the Neumann series:
∞
(𝐼 − 𝑇)−1 = ∑ 𝑇𝑘
𝑀
,
(𝑡 − 𝑎)𝛼
𝑀 (|Δ𝑠|𝛿 + (Δ𝑡)𝛾 )
(𝑡 − 𝑎)𝛼+𝛾
(27)
,
(28)
Δ𝐴 = 𝐴 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) − 𝐴 (𝑠, 𝑡)
0,𝛼
and
‖Δ𝐴‖𝑋 ≤
(20)
}
‖𝐴 (𝑠 + Δ𝑠) − 𝐴 (𝑠)‖𝑋 }
sup
.
𝛼
}
}
(Δ𝑠)
𝑠+Δ𝑠 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0
}
0,𝛼,𝛿
(26)
satisfying the inequalities
and 𝐶0,𝛼 ((𝑎, 𝑏), 𝑋) is a Banach space with the norm:
‖𝐴‖0,𝛼
𝐴 : (𝑎, 𝑏) × (𝑎, 𝑏) 󳨀→ 𝑋,
󳨀→ 𝐴 (𝑠, 𝑡)
(𝑠, 𝑡)
(19)
(24)
𝑘=0
for all 𝑠, 𝑡, 𝑠 + Δ𝑠, 𝑡 + Δ𝑡 ∈ (𝑎, 𝑏) with Δ𝑡 ≥ 0 is a linear
space with the usual algebraic operations addition and scalar
multiplication defined by
(𝐴 + 𝐵) (𝑠, 𝑡) = 𝐴 (𝑠, 𝑡) + 𝐵 (𝑠, 𝑡) ,
(𝜆𝐴) (𝑠, 𝑡) = 𝜆𝐴 (𝑠, 𝑡) ,
𝜆 ∈ 𝐾, (𝐾 = R or 𝐾 = C) ,
(29)
and 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) is a Banach space with the norm
‖𝐴‖𝛼,𝛿,𝛾 = max {𝑝𝐴 , 𝑘𝐴 } ,
(30)
where 𝑝𝐴 and 𝑘𝐴 are defined by
𝑝𝐴 =
𝑘𝐴 =
sup (𝑡 − 𝑎)𝛼 ‖𝐴 (𝑠, 𝑡)‖𝑋 ,
(31)
𝑠,𝑡 ∈ (𝑎,𝑏)
sup
𝑠,𝑡,𝑠+Δ𝑠,𝑡+Δ𝑡 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0∨Δ𝑡 ≠ 0
(𝑡 − 𝑎)𝛼+𝛾 ‖Δ𝐴‖𝑋
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
.
(32)
Proof. Let 𝑋 be a Banach space with the norm ‖ ⋅ ‖𝑋 . It is
known that the set A = {𝐴 | 𝐴 : (𝑎, 𝑏) × (𝑎, 𝑏) → 𝑋} is a
linear space with operations in 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋). Also,
it is obvious that 𝐴 + 𝐵, 𝜆𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋). On the
other hand, since
𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋)
= {𝐴 | 𝐴 : (𝑎, 𝑏) × (𝑎, 𝑏) 󳨀→ 𝑋, 𝐴 satisifes (27) and (28)}
⊂ A,
which satisfies
(33)
1
󵄩󵄩
󵄩
󵄩󵄩(𝐼 − 𝑇)−1 󵄩󵄩󵄩 ≤
󵄩
󵄩 1 − ‖𝑇‖ .
(25)
The iterated operators 𝑇𝑛 are defined by 𝑇0 = 𝐼 and 𝑇𝑛 =
𝑇𝑇𝑛−1 . The series in the right of (24) is convergent in the norm
on 𝐵(𝑋, 𝑋).
𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) is a linear space.
Furthermore, 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) is a normed space
with the norm ‖ ⋅ ‖𝛼,𝛿,𝛾 . Indeed, consider the following.
(N1) It is clear that ‖𝐴‖𝛼,𝛿,𝛾 ≥ 0 for all 𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) ×
(𝑎, 𝑏), 𝑋).
4
Abstract and Applied Analysis
(N2) If 𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and ‖𝐴‖𝛼,𝛿,𝛾 = 0, then
𝐴(𝑠, 𝑡) = 0 for all 𝑠, 𝑡 ∈ (𝑎, 𝑏), since
‖𝐴‖𝛼,𝛿,𝛾 = max {𝑝𝐴 , 𝑘𝐴} = 0.
(34)
So 𝐴 = 0. On the other hand, if 𝐴 = 0, it is found
that ‖𝐴‖𝛼,𝛿,𝛾 = 0 by (30), (31) and (32). Hence, the
proposition “‖𝐴‖𝛼,𝛿,𝛾 = 0 if and only if 𝐴 = 0” is true.
(N3) Let 𝜆 ∈ 𝐾, (𝐾 = R or 𝐾 = C). Since
𝑝(𝜆𝐴) =
𝑘(𝜆𝐴) =
sup (𝑡 − 𝑎)𝛼 ‖(𝜆𝐴) (𝑠, 𝑡)‖𝑋 = |𝜆| 𝑝𝐴 ,
which holds for all real numbers 𝑝𝐴 , 𝑝𝐵 , 𝑘𝐴 , 𝑘𝐵 ≥ 0, we obtain
‖𝐴 + 𝐵‖𝛼,𝛿,𝛾 ≤ ‖𝐴‖𝛼,𝛿,𝛾 + ‖𝐵‖𝛼,𝛿,𝛾 .
As a result, 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏)×(𝑎, 𝑏), 𝑋) is a normed space with
the norm ‖ ⋅ ‖𝛼,𝛿,𝛾 .
The space 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) is a Banach space
with respect to ‖ ⋅ ‖𝛼,𝛿,𝛾 . To see this we consider an arbitrary
Cauchy sequence (𝐴 𝑛 ) in 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋), and we
show that (𝐴 𝑛 ) converges to a function 𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) ×
(𝑎, 𝑏), 𝑋). Since (𝐴 𝑛 ) is Cauchy, for every 𝜀 > 0, there exists
𝑛0 (𝜀) ∈ N such that
󵄩
󵄩󵄩
󵄩󵄩𝐴 𝑛 − 𝐴 𝑚 󵄩󵄩󵄩𝛼,𝛿,𝛾 < 𝜀,
𝑠,𝑡 ∈ (𝑎,𝑏)
(𝑡 − 𝑎)
sup
𝑠,𝑡,𝑠+Δ𝑠,𝑡+Δ𝑡 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0∨Δ𝑡 ≠ 0
𝛼+𝛾
‖Δ (𝜆𝐴)‖𝑋
= |𝜆| 𝑘𝐴
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
(35)
(𝑚, 𝑛 > 𝑛0 (𝜀)) .
(41)
So from (41),
󵄩
󵄩
sup (𝑡 − 𝑎)𝛼 󵄩󵄩󵄩𝐴 𝑛 (𝑠, 𝑡) − 𝐴 𝑚 (𝑠, 𝑡)󵄩󵄩󵄩𝑋 < 𝜀,
by (31) and (32),
‖𝜆𝐴‖𝛼,𝛿,𝛾 = max {𝑝(𝜆𝐴) , 𝑘(𝜆𝐴) } = |𝜆| max {𝑝𝐴 , 𝑘𝐴}
= |𝜆| ‖𝐴‖𝛼,𝛿,𝛾 .
𝑠,𝑡 ∈ (𝑎,𝑏)
(36)
sup
󵄩
󵄩
(𝑡 − 𝑎)𝛼+𝛾 󵄩󵄩󵄩Δ (𝐴 𝑛 − 𝐴 𝑚 )󵄩󵄩󵄩𝑋
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
𝑠,𝑡,𝑠+Δ𝑠,𝑡+Δ𝑡 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0∨Δ𝑡 ≠ 0
< 𝜀,
(42)
(N4) If 𝐴, 𝐵 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋),
𝑝(𝐴+𝐵) =
≤
where
sup (𝑡 − 𝑎)𝛼 ‖(𝐴 + 𝐵) (𝑠, 𝑡)‖𝑋
𝑠,𝑡 ∈ (𝑎,𝑏)
Δ (𝐴 𝑛 − 𝐴 𝑚 ) = (𝐴 𝑛 − 𝐴 𝑚 ) (𝑠 + Δ𝑠, 𝑡 + Δ𝑡)
sup (𝑡 − 𝑎)𝛼 ‖𝐴 (𝑠, 𝑡)‖𝑋
𝑠,𝑡 ∈ (𝑎,𝑏)
− (𝐴 𝑛 − 𝐴 𝑚 ) (𝑠, 𝑡) .
(37)
+ sup (𝑡 − 𝑎)𝛼 ‖𝐵 (𝑠, 𝑡)‖𝑋
𝑠,𝑡 ∈ (𝑎,𝑏)
By (42), while Δ𝑠 ≠ 0 or Δ𝑡 ≠ 0, we have
= 𝑝𝐴 + 𝑝𝐵
󵄩
󵄩󵄩
𝛼
𝛼
󵄩󵄩(𝑡 − 𝑎) 𝐴 𝑛 (𝑠, 𝑡) − (𝑡 − 𝑎) 𝐴 𝑚 (𝑠, 𝑡)󵄩󵄩󵄩𝑋 < 𝜀,
by (31) and
𝑘(𝐴+𝐵) =
≤
sup
(𝑡 − 𝑎)𝛼+𝛾 ‖Δ (𝐴 + 𝐵)‖𝑋
sup
(𝑡 − 𝑎)𝛼+𝛾 ‖Δ𝐴‖𝑋
𝛿
𝑠,𝑡,𝑠+Δ𝑠,𝑡+Δ𝑡 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0∨Δ𝑡 ≠ 0
+
sup
󵄩
󵄩
(𝑡 − 𝑎)𝛼+𝛾 󵄩󵄩󵄩Δ (𝐴 𝑛 − 𝐴 𝑚 )󵄩󵄩󵄩𝑋
< 𝜀,
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
𝛾
|Δ𝑠| + (Δ𝑡)
𝑠,𝑡,𝑠+Δ𝑠,𝑡+Δ𝑡 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0∨Δ𝑡 ≠ 0
𝑠,𝑡,𝑠+Δ𝑠,𝑡+Δ𝑡 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0∨Δ𝑡 ≠ 0
(43)
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
(38)
(𝑡 − 𝑎)𝛼+𝛾 ‖Δ𝐵‖𝑋
(45)
for all 𝑠, 𝑡, 𝑠 + Δ𝑠, 𝑡 + Δ𝑡 ∈ (𝑎, 𝑏), (Δ𝑡 ≥ 0). We see by (44)
that ((𝑡 − 𝑎)𝛼 𝐴 𝑛 (𝑠, 𝑡)) is Cauchy in 𝑋. Since 𝑋 is complete,
there exists a unique 𝑥 ∈ 𝑋 such that
lim (𝑡 − 𝑎)𝛼 𝐴 𝑛 (𝑠, 𝑡) = 𝑥
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
(𝑚, 𝑛 > 𝑛0 (𝜀))
(44)
𝑛→∞
or
lim 𝐴 𝑛 (𝑠, 𝑡) =
𝑛→∞
1
𝑥.
(𝑡 − 𝑎)𝛼
(46)
= 𝑘𝐴 + 𝑘𝐵
The limit 𝑥 ∈ 𝑋 depends on the choice of 𝑠, 𝑡 ∈ (𝑎, 𝑏). This
defines a function:
by (32). Hence,
‖𝐴 + 𝐵‖𝛼,𝛿,𝛾 = max {𝑝(𝐴+𝐵) , 𝑘(𝐴+𝐵) }
≤ max {𝑝𝐴 + 𝑝𝐵 , 𝑘𝐴 + 𝑘𝐵 } .
(39)
From the inequality
max {𝑝𝐴 + 𝑝𝐵 , 𝑘𝐴 + 𝑘𝐵 } ≤ max {𝑝𝐴 , 𝑘𝐴} + max {𝑝𝐵 , 𝑘𝐵 }
(40)
𝐴 : (𝑎, 𝑏) × (𝑎, 𝑏) 󳨀→
𝑋,
󳨀→ 𝐴 (𝑠, 𝑡) ,
(𝑠, 𝑡)
(47)
lim 𝐴 𝑛 (𝑠, 𝑡) = 𝐴 (𝑠, 𝑡) .
(48)
where
𝑛→∞
Abstract and Applied Analysis
5
Now, we want to show that lim𝑛 → ∞ ‖𝐴 𝑛 − 𝐴‖𝛼,𝛿,𝛾 = 0 and
𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋). By (48), the continuity of the
norm gives together with (44) and (45) that
󵄩
󵄩
(𝑡 − 𝑎)𝛼 󵄩󵄩󵄩𝐴 𝑛 (𝑠, 𝑡) − 𝐴 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
󵄩󵄩
󵄩󵄩
= (𝑡 − 𝑎)𝛼 󵄩󵄩󵄩𝐴 𝑛 (𝑠, 𝑡) − lim 𝐴 𝑚 (𝑠, 𝑡)󵄩󵄩󵄩
𝑚→∞
󵄩𝑋
󵄩
(49)
󵄩
󵄩
= lim (𝑡 − 𝑎)𝛼 󵄩󵄩󵄩𝐴 𝑛 (𝑠, 𝑡) − 𝐴 𝑚 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
𝑚→∞
< 𝜀,
󵄩
󵄩
(𝑡 − 𝑎)𝛼+𝛾 󵄩󵄩󵄩Δ (𝐴 𝑛 − 𝐴)󵄩󵄩󵄩𝑋
for all 𝜀 > 0 and 𝑛 > 𝑛0 (𝜀). Therefore, we obtain from (54) and
(55) that there exists the nonnegative constant 𝐶 such that
‖𝐴 (𝑠, 𝑡)‖𝑋 ≤
‖𝐴 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) − 𝐴 (𝑠, 𝑡)‖𝑋 ≤
Hence, 𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋). This step completes the
proof.
(50)
≤𝐶
hold, where ‖𝐴‖0,𝛿,𝛾 is defined by (16).
(51)
Proof. If 𝐴 ∈ 𝐶0,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋), then by (16) and (17)
there exists the constant ‖𝐴‖0,𝛿,𝛾 satisfying the inequalities:
‖𝐴 (𝑠, 𝑡)‖𝑋 ≤
(52)
≤
(53)
(𝑡 − 𝑎)𝛼+𝛾
for all 𝑠, 𝑡, 𝑠 + Δ𝑠, 𝑡 + Δ𝑡 ∈ (𝑎, 𝑏) with Δ𝑡 ≥ 0 and 𝑛 ∈ N. By
(52) and (49),
󵄩
󵄩
‖𝐴 (𝑠, 𝑡)‖𝑋 = 󵄩󵄩󵄩𝐴 𝑛 (𝑠, 𝑡) + 𝐴 (𝑠, 𝑡) − 𝐴 𝑛 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
󵄩
󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝐴 𝑛 (𝑠, 𝑡)󵄩󵄩󵄩𝑋 + 󵄩󵄩󵄩𝐴 𝑛 (𝑠, 𝑡) − 𝐴 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
(54)
𝐶+𝜀
≤
.
(𝑡 − 𝑎)𝛼
By (53) and (50),
≤
(𝐶 + 𝜀) (|Δ𝑠|𝛿 + (Δ𝑡)𝛾 )
(𝑡 − 𝑎)𝛼+𝛾
,
‖𝐴 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) − 𝐴 (𝑠, 𝑡)‖𝑋
≤
(59)
‖𝐴‖0,𝛿,𝛾 (𝑏 − 𝑎)𝛼+𝛾 (|Δ𝑠|𝛿 + (Δ𝑡)𝛾 )
(𝑡 − 𝑎)𝛼+𝛾
for all 𝑠, 𝑡, 𝑠 + Δ𝑠, 𝑡 + Δ𝑡 ∈ (𝑎, 𝑏) with Δ𝑡 ≥ 0. By taking
(60)
it is obtained by (59) that 𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and
‖𝐴‖𝛼,𝛿,𝛾 ≤ 𝑀1 . That is,
as desired.
Example 13. Let us take R instead of 𝑋 and define the
function 𝐴 as
𝐴 : (𝑎, 𝑏) × (𝑎, 𝑏) 󳨀→ R,
𝐴 (𝑠, 𝑡) = 𝑠 + 𝑡.
‖𝐴 (𝑠, 𝑡)‖R = |𝐴 (𝑠, 𝑡)| ≤ 2 |𝑏| = 𝑀1 ,
‖𝐴 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) − 𝐴 (𝑠, 𝑡)‖R
= |𝐴 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) − 𝐴 (𝑠, 𝑡)|
𝛿
1−𝛿
= |Δ𝑠| |Δ𝑠|
(55)
(62)
Then, we have
≤ |Δ𝑠| + |Δ𝑡|
󵄩
󵄩
‖Δ𝐴‖𝑋 = 󵄩󵄩󵄩Δ𝐴 + Δ𝐴 𝑛 − Δ𝐴 𝑛 󵄩󵄩󵄩𝑋
󵄩
󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩Δ𝐴 𝑛 󵄩󵄩󵄩𝑋 + 󵄩󵄩󵄩Δ (𝐴 𝑛 − 𝐴)󵄩󵄩󵄩𝑋
(𝑡 − 𝑎)𝛼
𝐶0,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋) ⊂ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋) , (61)
󵄩
󵄩󵄩
󵄩󵄩𝐴 𝑛 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) − 𝐴 𝑛 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
𝐶 (|Δ𝑠|𝛿 + (Δ𝑡)𝛾 )
‖𝐴‖0,𝛿,𝛾 (𝑏 − 𝑎)𝛼
𝑀1 = max {‖𝐴‖0,𝛿,𝛾 (𝑏 − 𝑎)𝛼 , ‖𝐴‖0,𝛿,𝛾 (𝑏 − 𝑎)𝛼+𝛾 } ,
for all 𝑛 ∈ N. Thus, it is obtained by (51) that
𝐶
󵄩
󵄩󵄩
,
󵄩󵄩𝐴 𝑛 (𝑠, 𝑡)󵄩󵄩󵄩𝑋 ≤
(𝑡 − 𝑎)𝛼
(57)
‖𝐴‖𝛼,𝛿,𝛾 ≤ max {‖𝐴‖0,𝛿,𝛾 (𝑏 − 𝑎)𝛼 , ‖𝐴‖0,𝛿,𝛾 (𝑏 − 𝑎)𝛼+𝛾 } (58)
𝑠,𝑡 ∈ (𝑎,𝑏)
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
.
and the inequality
for all 𝑠, 𝑡, 𝑠 + Δ𝑠, 𝑡 + Δ𝑡 ∈ (𝑎, 𝑏) with Δ𝑡 ≥ 0 such that Δ𝑠 ≠ 0
or Δ𝑡 ≠ 0. Since ‖𝐴 𝑛 − 𝐴‖𝛼,𝛿,𝛾 < 𝜀 for all 𝑛 > 𝑛0 (𝜀) by (49)
and (50), we derive lim𝑛 → ∞ ‖𝐴 𝑛 − 𝐴‖𝛼,𝛿,𝛾 = 0. Furthermore,
since (𝐴 𝑛 ) is bounded, there exists the nonnegative constant
𝐶 such that ‖𝐴 𝑛 ‖𝛼,𝛿,𝛾 ≤ 𝐶 which yields
󵄩
󵄩
sup (𝑡 − 𝑎)𝛼 󵄩󵄩󵄩𝐴 𝑛 (𝑠, 𝑡)󵄩󵄩󵄩𝑋 ≤ 𝐶,
𝑠,𝑡,𝑠+Δ𝑠,𝑡+Δ𝑡 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0∨Δ𝑡 ≠ 0
(𝑡 − 𝑎)𝛼+𝛾
(56)
𝐶0,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋) ⊂ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋)
<𝜀
󵄩
󵄩
(𝑡 − 𝑎)𝛼+𝛾 󵄩󵄩󵄩Δ𝐴 𝑛 󵄩󵄩󵄩𝑋
𝐶 (|Δ𝑠|𝛿 + (Δ𝑡)𝛾 )
Theorem 12. The inclusion
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
󵄩
󵄩
(𝑡 − 𝑎)𝛼+𝛾 󵄩󵄩󵄩Δ𝐴 𝑛 − lim𝑚 → ∞ Δ𝐴 𝑚 󵄩󵄩󵄩𝑋
=
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
󵄩
󵄩
(𝑡 − 𝑎)𝛼+𝛾 󵄩󵄩󵄩Δ (𝐴 𝑛 − 𝐴 𝑚 )󵄩󵄩󵄩𝑋
= lim
𝑚→∞
|Δ𝑠|𝛿 + (Δ𝑡)𝛾
sup
𝐶
,
(𝑡 − 𝑎)𝛼
(63)
𝛾
1−𝛾
+ |Δ𝑡| |Δ𝑡|
≤ max {(𝑏 − 𝑎)1−𝛿 , (𝑏 − 𝑎)1−𝛾 } (|Δ𝑠|𝛿 + |Δ𝑡|𝛾 )
= 𝑀2 (|Δ𝑠|𝛿 + |Δ𝑡|𝛾 )
6
Abstract and Applied Analysis
for all 𝑠, 𝑡, 𝑠 + Δ𝑠, 𝑡 + Δ𝑡 ∈ (𝑎, 𝑏). Thus, we obtain by (63) that
𝐴 ∈ 𝐶0,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and ‖𝐴‖0,𝛿,𝛾 ≤ max{𝑀1 , 𝑀2 } =
𝑀. From Theorem 12, we conclude that 𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) ×
(𝑎, 𝑏), 𝑋) and
‖𝐴‖𝛼,𝛿,𝛾 ≤ max {𝑀(𝑏 − 𝑎)𝛼 , 𝑀(𝑏 − 𝑎)𝛼+𝛾 } .
Theorem 15. Let 𝑋 be a real or complex Banach space with the
norm ‖ ⋅ ‖𝑋 , 𝑎, 𝑏, 𝛼, 𝛿 be real numbers such that −∞ < 𝑎, 𝑏 <
∞, 0 < 𝛼, 𝛿, 𝛼 + 𝛿 < 1, and 𝑀 be a nonnegative constant. The
set 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋) of all functions satisfying
(64)
‖𝐴 (𝑠)‖𝑋 ≤
Theorem 14. The function 𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) is
continuous with respect to the Euclidean metric on (𝑎, 𝑏) ×
(𝑎, 𝑏) ⊂ R2 .
Proof. Let (𝑠0 , 𝑡0 ) ∈ (𝑎, 𝑏) × (𝑎, 𝑏) and 𝑑 be usual metric on
R2 . Then, we wish to show that the function 𝐴 : ((𝑎, 𝑏) ×
(𝑎, 𝑏), 𝑑) → (𝑋, ‖ ⋅ ‖𝑋 ) is continuous. It is clear that
lim
‖𝐴 (𝑠 + Δ𝑠) − 𝐴 (𝑠)‖𝑋 ≤
lim
(𝑠 − 𝑎)𝛼+𝛿
(𝐴 + 𝐵) (𝑠) = 𝐴 (𝑠) + 𝐵 (𝑠) ,
(65)
√(𝑠 − 𝑠0 )2 + (𝑡 − 𝑡0 )2 = 0.
(𝑠,𝑡) → (𝑠0 ,𝑡0 )
(71)
𝑀(Δ𝑠)𝛿
for all 𝑠, 𝑡, 𝑠 + Δ𝑠 ∈ (𝑎, 𝑏) with Δ𝑠 ≥ 0 is a linear space with the
usual algebraic operations addition and scalar multiplication
𝑑 ((𝑠, 𝑡) , (𝑠0 , 𝑡0 ))
(𝑠,𝑡) → (𝑠0 ,𝑡0 )
=
𝑀
,
(𝑠 − 𝑎)𝛼
(𝜆𝐴) (𝑠) = 𝜆𝐴 (𝑠) ,
𝜆 ∈ 𝐾, (𝐾 = R 𝑜𝑟 𝐾 = C) ,
(72)
and 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋) is a Banach space with the norm:
Besides, since
󵄨
󵄨
0 ≤ 󵄨󵄨󵄨𝑠 − 𝑠0 󵄨󵄨󵄨 ≤ 𝑑 ((𝑠, 𝑡) , (𝑠0 , 𝑡0 )) ,
󵄨
󵄨
0 ≤ 󵄨󵄨󵄨𝑡 − 𝑡0 󵄨󵄨󵄨 ≤ 𝑑 ((𝑠, 𝑡) , (𝑠0 , 𝑡0 )) ,
(66)
{
{
‖𝐴‖𝛼,𝛿 = max { sup (𝑠 − 𝑎)𝛼 ‖𝐴 (𝑠)‖𝑋 ,
{𝑠 ∈ (𝑎,𝑏)
{
we have by (65) that 𝑠 → 𝑠0 and 𝑡 → 𝑡0 . Thus, since the
equalities:
󵄩
󵄩󵄩
󵄩󵄩𝐴 (𝑠, 𝑡) − 𝐴 (𝑠0 , 𝑡0 )󵄩󵄩󵄩𝑋
(67)
󵄩
󵄩
= 󵄩󵄩󵄩𝐴 (𝑠0 + 𝑠 − 𝑠0 , 𝑡0 + 𝑡 − 𝑡0 ) − 𝐴 (𝑠0 , 𝑡0 )󵄩󵄩󵄩𝑋 ,
󵄩
󵄩󵄩
󵄩󵄩𝐴 (𝑠, 𝑡) − 𝐴 (𝑠0 , 𝑡0 )󵄩󵄩󵄩𝑋
󵄩
󵄩
= 󵄩󵄩󵄩𝐴 (𝑠0 , 𝑡0 ) − 𝐴 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
󵄩
󵄩
= 󵄩󵄩󵄩𝐴 (𝑠 + 𝑠0 − 𝑠, 𝑡 + 𝑡0 − 𝑡) − 𝐴 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
(68)
󵄩
󵄩
0 ≤ 󵄩󵄩󵄩𝐴 (𝑠, 𝑡) − 𝐴 (𝑠0 , 𝑡0 )󵄩󵄩󵄩𝑋
≤
𝛼+𝛾
(𝑡0 − 𝑎)
󵄩
󵄩
0 ≤ 󵄩󵄩󵄩𝐴 (𝑠, 𝑡) − 𝐴 (𝑠0 , 𝑡0 )󵄩󵄩󵄩𝑋 ≤
,
(𝑡 ≥ 𝑡0 )
𝐶0,𝛿 ((𝑎, 𝑏) , 𝑋) ⊂ 𝐻𝛼,𝛿 ((𝑎, 𝑏) , 𝑋)
𝛾
󵄨
󵄨𝛿
𝑀 (󵄨󵄨󵄨𝑠0 − 𝑠󵄨󵄨󵄨 + (𝑡0 − 𝑡) )
(𝑡 − 𝑎)𝛼+𝛾
󵄩󵄩
󵄩
󵄩󵄩𝐴 (𝑠, 𝑡) − 𝐴 (𝑠0 , 𝑡0 )󵄩󵄩󵄩𝑋 = 0.
(74)
and the inequality
(75)
hold, where ‖𝐴‖0,𝛿 is the norm in the space 𝐶0,𝛿 ((𝑎, 𝑏), 𝑋).
,
Since the proofs of Theorems 15–17 are completely similar
to that of Theorems 11–14, we leave them to the reader.
Lemma 18. Let 𝑋 be a real or complex Banach algebra with
the norm ‖ ⋅ ‖𝑋 and define 𝐵𝑛 by
𝐵𝑛 (𝑠, 𝑡) = 𝐴 1 (𝑠, 𝑡) 𝐴 2 (𝑠, 𝑡) ⋅ ⋅ ⋅ 𝐴 𝑛 (𝑠, 𝑡) ;
hold from (67) and (68), respectively. By (69), we have
lim
Theorem 17. The inclusion
‖𝐴‖𝛼,𝛿 ≤ max {‖𝐴‖0,𝛿 (𝑏 − 𝑎)𝛼 , ‖𝐴‖0,𝛿 (𝑏 − 𝑎)𝛼+𝛿 }
(𝑡 < 𝑡0 )
(69)
(𝑠,𝑡) → (𝑠0 ,𝑡0 )
where Δ𝐴 = 𝐴(𝑠 + Δ𝑠) − 𝐴(𝑠).
Theorem 16. The function 𝐴 ∈ 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋) is continuous
with respect to the Euclidean metric on (𝑎, 𝑏) ⊂ R.
hold, there exists the nonnegative constant 𝑀 by (28) that the
inequalities:
𝛾
󵄨𝛿
󵄨
𝑀 (󵄨󵄨󵄨𝑠 − 𝑠0 󵄨󵄨󵄨 + (𝑡 − 𝑡0 ) )
(73)
(𝑠 − 𝑎)𝛼+𝛿 ‖Δ𝐴‖𝑋 }
sup
},
(Δ𝑠)𝛿
𝑠,𝑠+Δ𝑠 ∈ (𝑎,𝑏)
Δ𝑠 ≠ 0
}
(70)
Hence, the function 𝐴 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏)×(𝑎, 𝑏), 𝑋) is continuous
at the arbitrary point (𝑠0 , 𝑡0 ) ∈ (𝑎, 𝑏)×(𝑎, 𝑏) which means that
it is continuous on (𝑎, 𝑏) × (𝑎, 𝑏).
𝑠, 𝑡 ∈ (𝑎, 𝑏) ,
(76)
where
𝐴 𝑖 ∈ 𝐻𝛼𝑖 ,𝛿𝑖 ,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋) ,
𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 + 𝛾 < 1
(𝑖 = 1, 2, . . . , 𝑛) ,
for 𝑛 ≥ 1.
(77)
Abstract and Applied Analysis
7
Then, if
𝛿 = min {𝛿1 , 𝛿2 , . . . , 𝛿𝑛 } ,
(𝑛 ≥ 1) ,
𝐷1 = 1,
𝐷𝑛 = max {(𝑏 − 𝑎)min{𝛿1 ,𝛿2 ,...,𝛿𝑛−1 }−𝛿 , (𝑏 − 𝑎)𝛿𝑛 −𝛿 } ,
(78)
(𝑛 ≥ 2) ,
𝐶𝑛 = 𝐷1 𝐷2 ⋅ ⋅ ⋅ 𝐷𝑛 ,
Furthermore,
󵄩
󵄩󵄩
󵄩󵄩𝐵𝑘+1 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩
= 󵄩󵄩󵄩𝐵𝑘 (𝑠, 𝑡) 𝐴 𝑘+1 (𝑠, 𝑡)󵄩󵄩󵄩𝑋 ≤ 󵄩󵄩󵄩𝐵𝑘 (𝑠, 𝑡)󵄩󵄩󵄩𝑋 󵄩󵄩󵄩𝐴 𝑘+1 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
󵄩
󵄩
󵄩󵄩 󵄩󵄩
󵄩󵄩𝐵𝑘 󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘 ,𝛿,𝛾 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾
≤
(𝑡 − 𝑎)𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘 (𝑡 − 𝑎)𝛼𝑘+1
(79)
≤
then 𝐵𝑛 ∈ 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and
󵄩󵄩 󵄩󵄩
󵄩󵄩𝐵𝑛 󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿,𝛾
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
≤ 2𝑛−1 𝐶𝑛 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑛 󵄩󵄩󵄩𝛼𝑛 ,𝛿𝑛 ,𝛾 .
(80)
Proof. We use the induction method. If 𝐴 1 ∈ 𝐻𝛼1 ,𝛿1 ,𝛾 ((𝑎, 𝑏) ×
(𝑎, 𝑏), 𝑋), then
𝛿 = min {𝛿1 } = 𝛿1 ,
𝐵1 (𝑠, 𝑡) = 𝐴 1 (𝑠, 𝑡) ,
𝐵1 ∈ 𝐻𝛼1 ,𝛿1 ,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋) ,
(81)
󵄩󵄩 󵄩󵄩
󵄩 󵄩
󵄩󵄩𝐵1 󵄩󵄩𝛼1 ,𝛿1 ,𝛾 = 𝐶1 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 .
Thus, Lemma is true for 𝑛 = 1.
Assume that if
𝐴 𝑖 ∈ 𝐻𝛼𝑖 ,𝛿𝑖 ,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋) ,
(𝑖 = 1, 2, . . . , 𝑘) ,
(82)
𝛿 = min {𝛿1 , 𝛿2 , . . . , 𝛿𝑘 } ,
then 𝐵𝑘 ∈ 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘 ,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and
󵄩󵄩 󵄩󵄩
󵄩󵄩𝐵𝑘 󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘 ,𝛿,𝛾
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
≤ 2𝑘−1 𝐶𝑘 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑘 󵄩󵄩󵄩𝛼𝑘 ,𝛿𝑘 ,𝛾 .
(83)
󵄩
󵄩󵄩
󵄩󵄩𝐵𝑘+1 󵄩󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1 ,𝛿,𝛾
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩
≤ 2 𝐶𝑘+1 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾 .
(84)
𝑘
(87)
for all 𝑠, 𝑡 ∈ (𝑎, 𝑏). Since 𝐷𝑘 = 1 for all 𝑘 ∈ N, 𝐶𝑘 = 1. So, we
have
󵄩 󵄩
󵄩
󵄩
2𝑘 𝐶𝑘+1 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾
󵄩󵄩
󵄩󵄩
(88)
󵄩󵄩𝐵𝑘+1 (𝑠, 𝑡)󵄩󵄩𝑋 ≤
(𝑡 − 𝑎)𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1
from (87). Additionally,
󵄩󵄩
󵄩
󵄩󵄩Δ𝐵𝑘+1 󵄩󵄩󵄩𝑋
󵄩
= 󵄩󵄩󵄩𝐵𝑘 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) 𝐴 𝑘+1 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡)
󵄩
−𝐵𝑘 (𝑠, 𝑡) 𝐴 𝑘+1 (𝑠, 𝑡)󵄩󵄩󵄩𝑋
󵄩
󵄩
= 󵄩󵄩󵄩(Δ𝐵𝑘 ) 𝐴 𝑘+1 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡) + 𝐵𝑘 (𝑠, 𝑡) Δ𝐴 𝑘+1 󵄩󵄩󵄩𝑋
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩Δ𝐵𝑘 󵄩󵄩󵄩𝑋 󵄩󵄩󵄩𝐴 𝑘+1 (𝑠 + Δ𝑠, 𝑡 + Δ𝑡)󵄩󵄩󵄩𝑋
󵄩 󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝐵𝑘 (𝑠, 𝑡)󵄩󵄩󵄩𝑋 󵄩󵄩󵄩Δ𝐴 𝑘+1 󵄩󵄩󵄩𝑋 .
(89)
󵄩󵄩
󵄩
󵄩󵄩Δ𝐵𝑘+1 󵄩󵄩󵄩𝑋
≤
󵄩
󵄩
󵄩󵄩 󵄩󵄩
󵄩󵄩𝐵𝑘 󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘 ,𝛿,𝛾 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾 (|Δ𝑠|𝛿 + (Δ𝑡)𝛾 )
(𝑡 − 𝑎)𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘 +𝛾 (𝑡 + Δ𝑡 − 𝑎)𝛼𝑘+1
󵄩
󵄩
󵄩󵄩 󵄩󵄩
󵄩󵄩𝐵𝑘 󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘 ,𝛿,𝛾 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾 (|Δ𝑠|𝛿𝑘+1 +(Δ𝑡)𝛾 )
+
(𝑡 − 𝑎)𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘 (𝑡 − 𝑎)𝛼𝑘+1 +𝛾
≤
󵄩
󵄩 󵄩
󵄩
2𝑘−1 𝐶𝑘 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾 (|Δ𝑠|𝛿 + (Δ𝑡)𝛾 )
(𝑡 − 𝑎)𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1 +𝛾
+
Then,
= 𝐵𝑘 (𝑠, 𝑡) 𝐴 𝑘+1 (𝑠, 𝑡) .
(𝑡 − 𝑎)𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1
By (89) and (83), we get
Now, let 𝐴 𝑖 ∈ 𝐻𝛼𝑖 ,𝛿𝑖 ,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋), (𝑖 = 1, 2, . . . , 𝑘 +
1) and 𝛿 = min{𝛿1 , 𝛿2 , . . . , 𝛿𝑘+1 }. Then, we must show that
𝐵𝑘+1 ∈ 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1 ,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and
𝐵𝑘+1 (𝑠, 𝑡) = 𝐴 1 (𝑠, 𝑡) 𝐴 2 (𝑠, 𝑡) ⋅ ⋅ ⋅ 𝐴 𝑘 (𝑠, 𝑡) 𝐴 𝑘+1 (𝑠, 𝑡)
󵄩 󵄩
󵄩
󵄩
2𝑘−1 𝐶𝑘 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾
(85)
󵄩 󵄩
󵄩
󵄩
2𝑘−1 𝐶𝑘 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾 (|Δ𝑠|𝛿𝑘+1 +(Δ𝑡)𝛾 )
(𝑡 − 𝑎)𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1 +𝛾
,
(90)
where 𝛿 = min{𝛿1 , 𝛿2 , . . . , 𝛿𝑘 }. Thus, we conclude that
󵄩
󵄩󵄩
󵄩󵄩Δ𝐵𝑘+1 󵄩󵄩󵄩𝑋
(1) If 𝑏 − 𝑎 ≤ 1, then we have from 0 ≤ |Δ𝑠| < 𝑏 − 𝑎 ≤ 1
that
𝛿𝑘+1
|Δ𝑠|
min{𝛿1 ,𝛿2 ,...,𝛿𝑘 }
, |Δ𝑠|
𝛿
≤ |Δ𝑠| ;
𝛿 = min {𝛿1 , 𝛿2 , . . . , 𝛿𝑘+1 } .
(86)
≤
󵄩 󵄩
󵄩
󵄩
2𝑘 𝐶𝑘+1 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾 (|Δ𝑠|𝛿 + (Δ𝑡)𝛾 )
(𝑡 − 𝑎)𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1 +𝛾
from (86) and (90), where 𝛿 = min{𝛿1 , 𝛿2 , . . . , 𝛿𝑘+1 }.
(91)
8
Abstract and Applied Analysis
(2) If 𝑏 − 𝑎 > 1, then 1 < 𝐷𝑘+1 , and since 𝐶𝑘+1 = 𝐶𝑘 𝐷𝑘+1
for all 𝑘 ∈ N, we have 1 ≤ 𝐶𝑘 < 𝐶𝑘+1 . That is, (88) is obtained
from (87). Besides,
(i) if 0 ≤ |Δ𝑠| < 1, then (91) is derived from (86) and
(90).
(ii) If 1 ≤ |Δ𝑠| < 𝑏 − 𝑎, since 0 < |Δ𝑠|/(𝑏 − 𝑎) < 1, we get
|Δ𝑠|𝛿𝑘+1 ≤ (𝑏 − 𝑎)𝛿𝑘+1 −𝛿 |Δ𝑠|𝛿 ,
min{𝛿1 ,𝛿2 ,...,𝛿𝑘 }
|Δ𝑠|
≤ (𝑏 − 𝑎)
min{𝛿1 ,𝛿2 ,...,𝛿𝑘 }−𝛿
𝛿
|Δ𝑠| ,
(92)
is bounded with
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑇2 󵄩󵄩 ≤ 𝑀 (𝛼2 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ,
that is, 𝑇2 ∈ 𝐵(𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏), 𝑋), 𝐻𝛼1 ,𝛾1 ((𝑎, 𝑏), 𝑋)).
(iii) Suppose that 𝛼1 + 𝛼2 + 𝛾1 and 𝛼1 + 𝛿1 < 1. Then, the
operator
𝑇3 : 𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏) , 𝑋) 󳨀→ 𝐻𝛼1 ,𝛿1 ((𝑎, 𝑏) , 𝑋)
𝑏
𝑎
𝐷𝑘+1 = max {(𝑏 − 𝑎)min{𝛿1 ,𝛿2 ,...,𝛿𝑘 }−𝛿 , (𝑏 − 𝑎)𝛿𝑘+1 −𝛿 } .
󵄩󵄩
󵄩
󵄩󵄩𝐵𝑘+1 󵄩󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1 ,𝛿,𝛾
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩
≤ 2𝑘 𝐶𝑘+1 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑘+1 󵄩󵄩󵄩𝛼𝑘+1 ,𝛿𝑘+1 ,𝛾 .
(94)
𝑠 ∈ (𝑎, 𝑏)
(102)
is bounded with
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑇3 󵄩󵄩 ≤ 𝑀 (𝛼1 , 𝛼2 , 𝛿1 , 𝛾1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ,
𝑀 (𝑐) =
Theorem 19. Let 𝑋 be a real or complex Banach algebra with
the norm ‖ ⋅ ‖𝑋 and 𝐴 ∈ 𝐻𝛼1 ,𝛿1 ,𝛾1 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋).
(i) Then, the integral operator
𝑇1 : 𝐶0,𝛼2 ((𝑎, 𝑏) , 𝑋) 󳨀→ 𝐶0,𝛿1 ((𝑎, 𝑏) , 𝑋)
(95)
defined by
𝑏
𝑎
(𝑏 − 𝑎)1−𝑐
,
1−𝑐
(104)
𝑀 (𝑑, 𝑒) = max {𝑀 (𝑑) , 𝑀 (𝑑 + 𝑒)} ,
2.2. Some Bounded Linear Integral Operators on the Spaces
𝐻𝛼,𝛿,𝛾 ((𝑎,𝑏) × (𝑎,𝑏), 𝑋) and 𝐻𝛼,𝛿 ((𝑎,𝑏) × (𝑎,𝑏), 𝑋). Hereafter,
by 𝐶𝑛 , we mean the constant defined by (79), and, by
the integral of the Banach valued functions, we mean the
Bochner integral unless stated otherwise.
𝑓 ∈ 𝐶0,𝛼2 ((𝑎, 𝑏) , 𝑋) ,
𝑠 ∈ (𝑎, 𝑏)
(96)
is bounded with
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑇1 󵄩󵄩 ≤ 𝑀 (𝛼1 , 𝛾1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ,
(97)
that is, 𝑇1 ∈ 𝐵(𝐶0,𝛼2 ((𝑎, 𝑏), 𝑋), 𝐶0,𝛿1 ((𝑎, 𝑏), 𝑋)).
(ii) The operator
𝑀 (𝑥, 𝑦, 𝑧, 𝑡)
= max {(𝑏 − 𝑎)𝑥 𝑀 (𝑥 + 𝑦) , (𝑏 − 𝑎)𝑥+𝑧 𝑀 (𝑥 + 𝑦 + 𝑡)}
(106)
for all the real numbers 𝑐, 𝑑, 𝑒, 𝑥, 𝑦, 𝑧, 𝑡 such that 𝑐 ≠ 1.
Proof. (i) We have
󵄩 󵄩
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩0,𝛼2
󵄩
󵄩󵄩
,
󵄩󵄩𝐴 (𝑠, 𝑡) 𝑓 (𝑡)󵄩󵄩󵄩𝑋 ≤
(𝑡 − 𝑎)𝛼1
󵄩 󵄩
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩0,𝛼2 (Δ𝑠)𝛿1
󵄩󵄩
󵄩󵄩
+
Δ𝑠,
𝑡)
−
𝐴
𝑡)]
𝑓
≤
(𝑠,
(𝑡)󵄩󵄩𝑋
󵄩󵄩[𝐴 (𝑠
(𝑡 − 𝑎)𝛼1 +𝛾1
(107)
for all 𝐴 ∈ 𝐻𝛼1 ,𝛿1 ,𝛾1 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋), 𝑓 ∈ 𝐶0,𝛼2 ((𝑎, 𝑏), 𝑋) and
𝑠, 𝑡, 𝑠 + Δ𝑠 ∈ (𝑎, 𝑏) with Δ𝑠 ≥ 0. By (107), it is obtained that
𝑏
𝑑𝑡
󵄩
󵄩 󵄩
󵄩󵄩
󵄩󵄩(𝑇1 𝑓) (𝑠)󵄩󵄩󵄩𝑋 ≤ ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩0,𝛼2 ∫
𝛼1
𝑎 (𝑡 − 𝑎)
󵄩 󵄩
= 𝑀 (𝛼1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩0,𝛼2 ,
𝑏
󵄩 󵄩
≤ ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩0,𝛼2 (Δ𝑠)𝛿1 ∫
(98)
𝑎
defined by
(𝑇2 𝑓) (𝑠) = ∫ 𝐴 (𝑡, 𝑠) 𝑓 (𝑡) 𝑑𝑡;
𝑎
(105)
󵄩󵄩
󵄩
󵄩󵄩Δ (𝑇1 𝑓)󵄩󵄩󵄩𝑋
𝑇2 : 𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏) , 𝑋) 󳨀→ 𝐻𝛼1 ,𝛾1 ((𝑎, 𝑏) , 𝑋)
𝑏
(103)
that is, 𝑇3 ∈ 𝐵(𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏), 𝑋), 𝐻𝛼1 ,𝛿1 ((𝑎, 𝑏), 𝑋)), where the
constants 𝑀(𝑐), 𝑀(𝑑, 𝑒), and 𝑀(𝑥, 𝑦, 𝑧, 𝑡) are given by
So, Lemma 18 holds for all 𝑛 ∈ N.
(𝑇1 𝑓) (𝑠) = ∫ 𝐴 (𝑠, 𝑡) 𝑓 (𝑡) 𝑑𝑡;
𝑓 ∈ 𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏) , 𝑋) ,
(93)
Since 1 < 𝐷𝑘+1 , (91) is derived from (90). By (1) and
(2), we show that (88) and (91) hold. Therefore, 𝐵𝑘+1 ∈
𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑘+1 ,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and
(101)
defined by
(𝑇3 𝑓) (𝑠) = ∫ 𝐴 (𝑠, 𝑡) 𝑓 (𝑡) 𝑑𝑡;
by taking |Δ𝑠|/(𝑏 − 𝑎) instead of |Δ𝑠| in (86). Let
(100)
󵄩 󵄩
= 𝑀 (𝛼1 + 𝛾1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩0,𝛼2 (Δ𝑠)𝛿1
𝑓 ∈ 𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏) , 𝑋) ,
𝑠 ∈ (𝑎, 𝑏)
(99)
𝑑𝑡
(𝑡 − 𝑎)𝛼1 +𝛾1
(108)
such that
Δ (𝑇1 𝑓) = (𝑇1 𝑓) (𝑠 + Δ𝑠) − (𝑇1 𝑓) (𝑠) .
(109)
Abstract and Applied Analysis
9
Thus, it is concluded from (108) that 𝑇1 𝑓 ∈ 𝐶0,𝛿1 ((𝑎, 𝑏), 𝑋)
and
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑇1 𝑓󵄩󵄩󵄩𝛿1 ≤ 𝑀 (𝛼1 , 𝛾1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩0,𝛼2 .
(110)
Besides, the norm of 𝑇1 holds the inequality
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑇1 󵄩󵄩 =
󵄩
󵄩
sup 󵄩󵄩󵄩𝑇1 𝑓󵄩󵄩󵄩𝛿1 ≤ 𝑀 (𝛼1 , 𝛾1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1
‖𝑓‖0,𝛼2 =1
(111)
from (110). Also, since 𝑇1 is linear, it is an element of the space
𝐵(𝐶0,𝛼2 ((𝑎, 𝑏), 𝑋), 𝐶0,𝛿1 ((𝑎, 𝑏), 𝑋)).
On the other hand:
(ii) The inequalities
󵄩 󵄩
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2
󵄩
󵄩
󵄩
󵄩󵄩
,
󵄩󵄩𝐴 (𝑡, 𝑠) 𝑓 (𝑡)󵄩󵄩󵄩𝑋 ≤ ‖𝐴 (𝑡, 𝑠)‖𝑋 󵄩󵄩󵄩𝑓 (𝑡)󵄩󵄩󵄩𝑋 ≤
(𝑠 − 𝑎)𝛼1 (𝑡 − 𝑎)𝛼2
(112)
󵄩
󵄩󵄩
󵄩󵄩[𝐴 (𝑡, 𝑠 + Δ𝑠) − 𝐴 (𝑡, 𝑠)] 𝑓 (𝑡)󵄩󵄩󵄩𝑋
󵄩
󵄩
≤ ‖𝐴 (𝑡, 𝑠 + Δ𝑠) − 𝐴 (𝑡, 𝑠)‖𝑋 󵄩󵄩󵄩𝑓 (𝑡)󵄩󵄩󵄩𝑋
󵄩 󵄩
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 (Δ𝑠)𝛾1
≤
(𝑡 − 𝑎)𝛼2 (𝑠 − 𝑎)𝛼1 +𝛾1
for all 𝑓 ∈ 𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏), 𝑋), 𝑠, 𝑡, 𝑠 + Δ𝑠 ∈ (𝑎, 𝑏) with Δ𝑠 ≥ 0.
By (118),
󵄩
󵄩󵄩
󵄩󵄩(𝑇3 𝑓) (𝑠)󵄩󵄩󵄩𝑋
𝑏
󵄩
󵄩
≤ ∫ 󵄩󵄩󵄩𝐴 (𝑠, 𝑡) 𝑓 (𝑡)󵄩󵄩󵄩𝑋 𝑑𝑡
𝑎
󵄩
󵄩󵄩
󵄩󵄩(𝑇2 𝑓) (𝑠)󵄩󵄩󵄩𝑋
(114)
=
𝑏
𝑏
󵄩 󵄩
≤ ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 (Δ𝑠)𝛿1 ∫
𝑎
𝑑𝑡
∫
𝛼2
𝑎 (𝑡 − 𝑎)
(115)
,
where Δ(𝑇2 𝑓) = (𝑇2 𝑓)(𝑠 + Δ𝑠) − (𝑇2 𝑓)(𝑠). By usage of (114)
and (115), we get that 𝑇2 𝑓 ∈ 𝐻𝛼1 ,𝛾1 ((𝑎, 𝑏), 𝑋) and
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑇2 𝑓󵄩󵄩󵄩𝛼1 ,𝛾1 ≤ 𝑀 (𝛼2 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 .
󵄩 󵄩
(𝑏 − 𝑎)𝛼1 +𝛿1 𝑀 (𝛼1 + 𝛼2 + 𝛾1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 (Δ𝑠)𝛿1
(𝑠 − 𝑎)𝛼1 +𝛿1
󵄩
󵄩
sup 󵄩󵄩󵄩𝑇2 𝑓󵄩󵄩󵄩𝛼1 ,𝛾1 ≤ 𝑀 (𝛼2 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1
‖𝑓‖𝛼2 ,𝛿2 =1
.
(121)
Thus, we have by (119) and (121) that 𝑇3 𝑓 ∈ 𝐻𝛼1 ,𝛿1 ((𝑎, 𝑏), 𝑋)
and
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑇3 𝑓󵄩󵄩󵄩𝛼1 ,𝛿1 ≤ 𝑀 (𝛼1 , 𝛼2 , 𝛿1 , 𝛾1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 . (122)
So, by (122), 𝑇3 is a bounded linear operator and also
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑇3 󵄩󵄩 =
󵄩
󵄩
sup 󵄩󵄩󵄩𝑇3 𝑓󵄩󵄩󵄩𝛼1 ,𝛿1 ≤ 𝑀 (𝛼1 , 𝛼2 , 𝛿1 , 𝛾1 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 .
‖𝑓‖𝛼
2 ,𝛿2
=1
(123)
(116)
Therefore, 𝑇3 ∈ 𝐵(𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏), 𝑋), 𝐻𝛼1 ,𝛿1 ((𝑎, 𝑏), 𝑋)).
Thus, 𝑇2 is bounded with
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑇2 󵄩󵄩 =
𝑑𝑡
(𝑡 − 𝑎)𝛼1 +𝛼2 +𝛾1
𝑏
󵄩 󵄩
𝑀 (𝛼2 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 (Δ𝑠)𝛾1
(𝑠 − 𝑎)𝛼1 +𝛾1
(120)
󵄩
󵄩
≤ ∫ 󵄩󵄩󵄩[𝐴 (𝑠 + Δ𝑠, 𝑡) − 𝐴 (𝑠, 𝑡)] 𝑓 (𝑡)󵄩󵄩󵄩𝑋 𝑑𝑡
𝑎
≤
(𝑠 − 𝑎)𝛼1 +𝛾1
Δ (𝑇3 𝑓) = (𝑇3 𝑓) (𝑠 + Δ𝑠) − (𝑇3 𝑓) (𝑠) ,
and by taking
󵄩
󵄩󵄩
󵄩󵄩Δ (𝑇3 𝑓)󵄩󵄩󵄩𝑋
󵄩󵄩
󵄩
󵄩󵄩Δ (𝑇2 𝑓)󵄩󵄩󵄩𝑋
≤
(119)
󵄩 󵄩
≤ ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 ∫
and by (113),
󵄩 󵄩
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 (Δ𝑠)𝛾1
𝑏
𝑑𝑡
𝛼 +𝛼
𝑎 (𝑡 − 𝑎) 1 2
󵄩 󵄩
(𝑏 − 𝑎)𝛼1 𝑀 (𝛼1 + 𝛼2 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2
≤
,
(𝑠 − 𝑎)𝛼1
(113)
hold for all 𝐴 ∈ 𝐻𝛼1 ,𝛿1 ,𝛾1 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋), 𝑓 ∈ 𝐻𝛼2 ,𝛿2
((𝑎, 𝑏), 𝑋), 𝑠, 𝑡, 𝑠 + Δ𝑠 ∈ (𝑎, 𝑏) with Δ𝑠 ≥ 0.
So, by (112),
󵄩 󵄩
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 𝑏
𝑑𝑡
≤
∫
𝛼1
𝛼2
(𝑠 − 𝑎)
𝑎 (𝑡 − 𝑎)
󵄩 󵄩
𝑀 (𝛼2 ) ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2
=
(𝑠 − 𝑎)𝛼1
from (116). Furthermore 𝑇2 is linear, and, hence, 𝑇2 ∈
𝐵(𝐻𝛼2 ,𝛿2 ((𝑎, 𝑏), 𝑋), 𝐻𝛼1 ,𝛾1 ((𝑎, 𝑏), 𝑋)).
(iii) It is clear that
󵄩 󵄩
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩
,
󵄩󵄩𝐴 (𝑠, 𝑡) 𝑓 (𝑡)󵄩󵄩𝑋 ≤ ‖𝐴 (𝑠, 𝑡)‖𝑋 󵄩󵄩𝑓 (𝑡)󵄩󵄩𝑋 ≤
(𝑡 − 𝑎)𝛼1 +𝛼2
󵄩
󵄩󵄩
󵄩󵄩[𝐴 (𝑠 + Δ𝑠, 𝑡) − 𝐴 (𝑠, 𝑡)] 𝑓 (𝑡)󵄩󵄩󵄩𝑋
󵄩
󵄩
≤ ‖𝐴 (𝑠 + Δ𝑠, 𝑡) − 𝐴 (𝑠, 𝑡)‖𝑋 󵄩󵄩󵄩𝑓 (𝑡)󵄩󵄩󵄩𝑋
󵄩 󵄩
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝛼2 ,𝛿2 (Δ𝑠)𝛿1
≤
(𝑡 − 𝑎)𝛼1 +𝛼2 +𝛾1
(118)
(117)
Hereafter, by 𝑀(𝑐), 𝑀(𝑑, 𝑒), and 𝑀(𝑥, 𝑦, 𝑧, 𝑡) for all the
real numbers 𝑐, 𝑑, 𝑒, 𝑥, 𝑦, 𝑧, 𝑡 with 𝑐 ≠ 1, we denote the constants given by (104), (105), and (106), respectively.
10
Abstract and Applied Analysis
Theorem 20. Let 𝑋 be a real or complex Banach algebra with
the norm ‖ ⋅ ‖𝑋 and 𝐴 ∈ 𝐻𝛼1 ,𝛿1 ,𝛾1 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) such that
𝛾2 < 𝛿1 . Then, 𝑇 given by
(𝑇𝐵) (𝑠1 , 𝑠2 )
𝑏
= ∫ 𝐴 (𝑠1 , 𝑡) 𝐵 (𝑡, 𝑠2 ) 𝑑𝑡;
𝑎
𝐵 ∈ 𝐻𝛼2 ,𝛿2 ,𝛾2 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋) ,
(124)
for all Δ𝑠2 ≥ 0 and 𝑡, 𝑠1 , 𝑠2 , 𝑠1 + Δ𝑠1 , 𝑠2 + Δ𝑠2 ∈ (𝑎, 𝑏). Since
Δ𝐷𝑡 = [(Δ 1,𝑡 𝐴) 𝐵 (𝑡, 𝑠2 + Δ𝑠2 ) + 𝐴 (𝑠1 , 𝑡) Δ 𝑡,2 𝐵] ,
we have
󵄨󵄨 󵄨󵄨𝛿1
󵄨󵄨Δ𝑠1 󵄨󵄨
󵄩
󵄩󵄩
𝛾
󵄩󵄩Δ𝐷𝑡 󵄩󵄩󵄩𝑋 ≤ (𝑏 − 𝑎) 2 𝑁1
𝛼 +𝛾
𝛼1 +𝛾1
(𝑠2 − 𝑎) 2 2
(𝑡 − 𝑎)
𝛾2
+ 𝑁1
𝑠1 , 𝑠2 ∈ (𝑎, 𝑏)
is a linear operator from 𝐻𝛼2 ,𝛿2 ,𝛾2 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) into
𝐻𝛼2 ,𝛿1 ,𝛾2 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋), and, also, 𝑇 is bounded with
‖𝑇‖ ≤ 𝑁‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ,
(125)
(Δ𝑠2 )
𝑁 = max {𝑀 (𝛼1 ) , max {(𝑏 − 𝑎) 𝑀 (𝛼1 + 𝛾1 ) , 𝑀 (𝛼1 )}} .
(126)
𝑡 ∈ (𝑎, 𝑏)
‖Δ (𝑇𝐵)‖𝑋 ≤
𝛾
󵄨 󵄨𝛿
𝑁2 ‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ‖𝐵‖𝛼2 ,𝛿2 ,𝛾2 (󵄨󵄨󵄨Δ𝑠1 󵄨󵄨󵄨 1 + (Δ𝑠2 ) 2 )
𝛼2 +𝛾2
(𝑠2 − 𝑎)
≤
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ‖𝐵‖𝛼2 ,𝛿2 ,𝛾2
𝛼2
(𝑡 − 𝑎)𝛼1 (𝑠2 − 𝑎)
;
𝑠1 , 𝑠2 , 𝑡 ∈ (𝑎, 𝑏) ,
(128)
𝑏
Lebesgue integral ∫𝑎 ‖𝐶𝑠1 ,𝑠2 (𝑡)‖𝑋 𝑑𝑡 exists, and, so, Bochner
integral in (124) exists by Theorem 8. Now, by taking 𝑁1 =
‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ‖𝐵‖𝛼2 ,𝛿2 ,𝛾2 , we get
󵄩󵄩
󵄩
󵄩󵄩(𝑇𝐵) (𝑠1 , 𝑠2 )󵄩󵄩󵄩𝑋
=
𝑏
𝑁1
𝑑𝑡
∫
𝛼2
𝛼1
(𝑠2 − 𝑎)
𝑎 (𝑡 − 𝑎)
(129)
where 𝑁2 = max{(𝑏 − 𝑎) 𝑀(𝛼1 + 𝛾1 ), 𝑀(𝛼1 )}. Since
max{𝑀(𝛼1 ), 𝑁2 } = 𝑁, it is found by (129) and (135) that
𝑁‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ‖𝐵‖𝛼2 ,𝛿2 ,𝛾2
󵄩󵄩
󵄩
,
󵄩󵄩(𝑇𝐵) (𝑠1 , 𝑠2 )󵄩󵄩󵄩𝑋 ≤
𝛼
(𝑠2 − 𝑎) 2
.
(136)
Hence, by (136), 𝑇𝐵 ∈ 𝐻𝛼2 ,𝛿1 ,𝛾2 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) and
‖𝑇𝐵‖𝛼2 ,𝛿1 ,𝛾2 ≤ 𝑁‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ‖𝐵‖𝛼2 ,𝛿2 ,𝛾2 .
(137)
That is, 𝑇 is bounded by (137) and
‖𝑇‖ =
sup
‖𝐵‖𝛼2 ,𝛿2 ,𝛾2 =1
‖𝑇𝐵‖𝛼2 ,𝛿1 ,𝛾2 ≤ 𝑁‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 .
(138)
Clearly, 𝑇 is linear. So, 𝑇 ∈ 𝐵(𝐻𝛼2 ,𝛿2 ,𝛾2 ((𝑎, 𝑏), 𝑋), 𝐻𝛼2 ,𝛿1 ,𝛾2
((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋)).
Theorem 21. Let 𝑋 be a real or complex Banach algebra and
define the function 𝐴 by
𝐴 𝑖 ∈ 𝐻𝛼𝑖 ,𝛿𝑖 ,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋)
from (128). Let define
𝛿 = min {𝛿1 , 𝛿2 , . . . , 𝛿𝑛 } ,
𝐷𝑡 (𝑠1 , 𝑠2 ) = 𝐴 (𝑠1 , 𝑡) 𝐵 (𝑡, 𝑠2 ) ,
𝑠, 𝑡 ∈ (𝑎, 𝑏) (139)
∀𝑖 ∈ {1, 2, . . . , 𝑛} ,
𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 + 𝛾 < 1
with 𝑛 ∈ N.
(140)
(130)
(i) Then, the linear Fredholm integral equation of the form
Δ 1,𝑡 𝐴 = 𝐴 (𝑠1 + Δ𝑠1 , 𝑡) − 𝐴 (𝑠1 , 𝑡) ,
Δ (𝑇𝐵) = (𝑇𝐵) (𝑠1 + Δ𝑠1 , 𝑠2 + Δ𝑠2 ) − (𝑇𝐵) (𝑠1 , 𝑠2 ) ,
𝛼2 +𝛾2
(𝑠2 − 𝑎)
such that
𝛼2
(𝑠2 − 𝑎)
Δ 𝑡,2 𝐵 = 𝐵 (𝑡, 𝑠2 + Δ𝑠2 ) − 𝐵 (𝑡, 𝑠2 ) ,
‖Δ (𝑇𝐵)‖𝑋 ≤
𝛾
󵄨 󵄨𝛿
𝑁‖𝐴‖𝛼1 ,𝛿1 ,𝛾1 ‖𝐵‖𝛼2 ,𝛿2 ,𝛾2 (󵄨󵄨󵄨Δ𝑠1 󵄨󵄨󵄨 1 + (Δ𝑠2 ) 2 )
𝐴 (𝑠, 𝑡) = 𝐴 1 (𝑠, 𝑡) 𝐴 2 (𝑠, 𝑡) ⋅ ⋅ ⋅ 𝐴 𝑛 (𝑠, 𝑡) ;
𝑀 (𝛼1 ) 𝑁1
Δ𝐷𝑡 = 𝐷𝑡 (𝑠1 + Δ𝑠1 , 𝑠2 + Δ𝑠2 ) − 𝐷𝑡 (𝑠1 , 𝑠2 ) ,
(135)
2.3. The Solutions of the Linear Fredholm Integral Equations in
the Spaces 𝐻𝛼,𝛿,𝛾 ((𝑎,𝑏) × (𝑎,𝑏), 𝑋) and 𝐻𝛼,𝛿 ((𝑎,𝑏) × (𝑎,𝑏), 𝑋).
We have the following.
𝑏
󵄩
󵄩
≤ ∫ 󵄩󵄩󵄩󵄩𝐶𝑠1 ,𝑠2 (𝑡)󵄩󵄩󵄩󵄩𝑋 𝑑𝑡
𝑎
≤
,
𝛾2
(127)
is continuous for each 𝑠1 , 𝑠2 ∈ (𝑎, 𝑏). Therefore, the function
𝐶𝑠1 ,𝑠2 : (𝑎, 𝑏) → 𝑋 is strongly measurable by Theorem 4,
and, hence, the function ‖𝐶𝑠1 ,𝑠2 ‖𝑋 : (𝑎, 𝑏) → R is Lebesgue
measurable from Theorem 5. Since
󵄩󵄩󵄩𝐶 (𝑡)󵄩󵄩󵄩 = 󵄩󵄩󵄩𝐴 (𝑠 , 𝑡) 𝐵 (𝑡, 𝑠 )󵄩󵄩󵄩
󵄩󵄩 𝑠1 ,𝑠2 󵄩󵄩𝑋 󵄩
1
2 󵄩𝑋
.
𝑏
Proof. The function 𝐶𝑠1 ,𝑠2 given by
𝐶𝑠1 ,𝑠2 (𝑡) = 𝐴 (𝑠1 , 𝑡) 𝐵 (𝑡, 𝑠2 ) ,
(134)
So, we obtain ‖Δ(𝑇𝐵)‖𝑋 ≤ ∫𝑎 ‖Δ𝐷𝑡 ‖𝑋 𝑑𝑡 which yields
where
𝛾2
𝛼2 +𝛾2
(𝑡 − 𝑎)𝛼1 (𝑠2 − 𝑎)
(133)
𝑏
(131)
(132)
𝑓 (𝑠) = 𝜙 (𝑠) + 𝜆 ∫ 𝐴 (𝑠, 𝑡) 𝑓 (𝑡) 𝑑𝑡
𝑎
has a unique solution 𝑓 in the space 𝐶0,𝛿 ((𝑎, 𝑏), 𝑋) with
(141)
Abstract and Applied Analysis
11
1
󵄩󵄩 󵄩󵄩
󵄩󵄩 󵄩󵄩
󵄩𝜙󵄩 ,
󵄩󵄩𝑓󵄩󵄩𝛿 ≤
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
1 − |𝜆| 2𝑛−1 𝐶𝑛 𝑀 (𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 , 𝛾) 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑛 󵄩󵄩󵄩𝛼𝑛 ,𝛿𝑛 ,𝛾 󵄩 󵄩𝛿
(142)
where 𝜙 ∈ 𝐶0,𝛿 ((𝑎, 𝑏), 𝑋) and 𝜆 is a real or complex parameter
satisfying the inequality:
|𝜆| <
2𝑛−1 𝐶𝑛 𝑀 (𝛼1
1
.
󵄩 󵄩
󵄩󵄩 󵄩󵄩
󵄩 󵄩
+ 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 , 𝛾) 󵄩󵄩𝐴 1 󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑛 󵄩󵄩󵄩𝛼𝑛 ,𝛿𝑛 ,𝛾
(ii) The solution of the equation
𝑏
𝑓 (𝑠) = 𝜙 (𝑠) + 𝜆 ∫ 𝐴 (𝑡, 𝑠) 𝑓 (𝑡) 𝑑𝑡
𝑎
Proof. (ii) Equation (144) may be rewritten as
𝑏
𝑓 (𝑠) − 𝜆 ∫ 𝐴 (𝑡, 𝑠) 𝑓 (𝑡) 𝑑𝑡 = 𝜙 (𝑠)
(144)
𝑎
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑓󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾
(145)
where 𝜙 ∈ 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ((𝑎, 𝑏), 𝑋), 𝜆 is a real or complex
parameter satisfying the inequality:
|𝜆| <
𝑓 ∈ 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ((𝑎, 𝑏) , 𝑋) ,
(154)
‖𝐴‖𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿,𝛾
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
≤ 2𝑛−1 𝐶𝑛 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑛 󵄩󵄩󵄩𝛼𝑛 ,𝛿𝑛 ,𝛾 .
(148)
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
× 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑛 󵄩󵄩󵄩𝛼𝑛 ,𝛿𝑛 ,𝛾 .
󳨀→ 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ((𝑎, 𝑏) , 𝑋)
(150)
(157)
Since ‖𝜆𝑇‖ < 1 by (146), the inverse operator
(𝐼 − 𝜆𝑇)−1 : 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ((𝑎, 𝑏) , 𝑋)
and 𝜆 is a real or complex parameter satisfying
(156)
from (117), and so
‖𝜆𝑇‖ ≤ |𝜆| 2𝑛−1 𝐶𝑛 𝑀 (𝛼1 + ⋅ ⋅ ⋅ + 𝛼𝑛 )
(149)
(155)
Also, we derive by Theorem 19 that 𝑇 ∈ 𝐵(𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾
((𝑎, 𝑏), 𝑋), 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ((𝑎, 𝑏), 𝑋)) and
‖𝑇‖ ≤ ‖𝐴‖𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿,𝛾 𝑀 (𝛼1 + ⋅ ⋅ ⋅ + 𝛼𝑛 )
where 𝜙 ∈ 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿 ((𝑎, 𝑏), 𝑋),
1
.
2𝑛−1 𝐶𝑛 𝐾
(153)
and 𝐼 is an identity operator on 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ((𝑎, 𝑏), 𝑋).
By Lemma 18, we have that 𝐴 ∈ 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿,𝛾 ((𝑎, 𝑏) ×
(𝑎, 𝑏), 𝑋) and
Then, (141) has a unique solution 𝑓 in the space
𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿 ((𝑎, 𝑏), 𝑋) such that
|𝜆| <
(152)
where 𝑇 is defined by
(146)
2 (𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 ) + 𝛾 and 𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 + 𝛿 < 1.
(147)
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
× 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑛 󵄩󵄩󵄩𝛼𝑛 ,𝛿𝑛 ,𝛾 ,
(𝐼 − 𝜆𝑇) 𝑓 = 𝜙,
𝑎
(iii) Suppose that
𝐾 = 𝑀 (𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 , 𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 , 𝛿, 𝛾)
𝑠 ∈ (𝑎, 𝑏)
which yields
(𝑇𝑓) (𝑠) = ∫ 𝐴 (𝑡, 𝑠) 𝑓 (𝑡) 𝑑𝑡,
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
× 󵄩󵄩󵄩𝐴 1 󵄩󵄩󵄩𝛼1 ,𝛿1 ,𝛾 󵄩󵄩󵄩𝐴 2 󵄩󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩󵄩𝐴 𝑛 󵄩󵄩󵄩𝛼𝑛 ,𝛿𝑛 ,𝛾 .
1
󵄩󵄩 󵄩󵄩
󵄩󵄩 󵄩󵄩
,
󵄩󵄩𝑓󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿 ≤
󵄩𝜙󵄩
1 − |𝜆| 2𝑛−1 𝐶𝑛 𝐾 󵄩 󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛿
((𝐼 − 𝜆𝑇) 𝑓) (𝑠) = 𝜙 (𝑠) ,
𝑏
1
,
2𝑛−1 𝐶𝑛 𝐿
𝐿 = 𝑀 (𝛼1 + 𝛼2 + ⋅ ⋅ ⋅ + 𝛼𝑛 )
(151)
or
uniquely exists in the space 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ((𝑎, 𝑏), 𝑋). Besides,
1
󵄩󵄩 󵄩󵄩
≤
,
󵄩𝜙󵄩
1 − |𝜆| 2𝑛−1 𝐶𝑛 𝐿 󵄩 󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾
(143)
exists and the norm of 𝐼 − 𝜆𝑇 satisfies the inequality:
(158)
12
Abstract and Applied Analysis
1
󵄩󵄩
󵄩
󵄩󵄩(𝐼 − 𝜆𝑇)−1 󵄩󵄩󵄩 ≤
󵄩󵄩 󵄩󵄩
󵄩󵄩 󵄩󵄩
󵄩
󵄩 1 − |𝜆| 2𝑛−1 𝐶 𝑀 (𝛼 + 𝛼 + ⋅ ⋅ ⋅ + 𝛼 ) 󵄩󵄩󵄩𝐴 󵄩󵄩󵄩
𝑛
1
2
𝑛 󵄩 1 󵄩𝛼1 ,𝛿1 ,𝛾 󵄩
󵄩𝐴 2 󵄩󵄩𝛼2 ,𝛿2 ,𝛾 ⋅ ⋅ ⋅ 󵄩󵄩𝐴 𝑛 󵄩󵄩𝛼𝑛 ,𝛿𝑛 ,𝛾
from Theorem 10. So, (144) has a unique solution 𝑓 in the
space 𝐻𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ((𝑎, 𝑏), 𝑋) such that
𝑓 = (𝐼 − 𝜆𝑇)−1 𝜙,
󵄩
󵄩󵄩 󵄩󵄩
−1 󵄩
󵄩󵄩𝑓󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 = 󵄩󵄩󵄩󵄩(𝐼 − 𝜆𝑇) 𝜙󵄩󵄩󵄩󵄩𝛼 +𝛼 +⋅⋅⋅+𝛼 ,𝛾
1
2
𝑛
󵄩󵄩
󵄩
󵄩
−1 󵄩
≤ 󵄩󵄩󵄩(𝐼 − 𝜆𝑇) 󵄩󵄩󵄩󵄩 󵄩󵄩󵄩𝜙󵄩󵄩󵄩𝛼1 +𝛼2 +⋅⋅⋅+𝛼𝑛 ,𝛾 ,
By Theorem 20, we have that 𝑇 is in the space
𝐵(𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋), 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋)), and
‖𝜆𝑇‖ ≤ |𝜆| 𝑁‖𝐴‖𝛼,𝛿,𝛾 .
(160)
Since ‖𝜆𝑇‖ < 1 by (164), the inverse operator
󳨀→ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋)
𝑏
𝑓 (𝑠1 , 𝑠2 ) = 𝜙 (𝑠1 , 𝑠2 ) + 𝜆 ∫ 𝐴 (𝑠1 , 𝑡) 𝑓 (𝑡, 𝑠2 ) 𝑑𝑡
(161)
𝑎
has a unique solution 𝑓 in 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) with
1
󵄩󵄩
󵄩
󵄩󵄩(𝐼 − 𝜆𝑇)−1 󵄩󵄩󵄩 ≤
󵄩
󵄩 1 − ‖𝜆𝑇‖
(162)
𝑁 = max {𝑀 (𝛼) , max {(𝑏 − 𝑎)𝛾 𝑀 (𝛼 + 𝛾) , 𝑀 (𝛼)}} , (163)
and 𝜆 is a real or complex parameter fulfilling
1
.
𝑁‖𝐴‖𝛼,𝛿,𝛾
(164)
Proof. Equation (161) can be expressed as
𝑏
𝑓 (𝑠1 , 𝑠2 ) − 𝜆 ∫ 𝐴 (𝑠1 , 𝑡) 𝑓 (𝑡, 𝑠2 ) 𝑑𝑡 = 𝜙 (𝑠1 , 𝑠2 )
𝑎
≤
1
1 − |𝜆| 𝑁‖𝐴‖𝛼,𝛿,𝛾
from Theorem 10. So, (161) has a unique solution 𝑓 in the
space 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋) such that
(165)
󵄩
󵄩󵄩 󵄩󵄩
−1 󵄩
󵄩󵄩𝑓󵄩󵄩𝛼,𝛿,𝛾 = 󵄩󵄩󵄩󵄩(𝐼 − 𝜆𝑇) 𝜙󵄩󵄩󵄩󵄩𝛼,𝛿,𝛾
󵄩󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩󵄩(𝐼 − 𝜆𝑇)−1 󵄩󵄩󵄩󵄩 󵄩󵄩󵄩𝜙󵄩󵄩󵄩𝛼,𝛿,𝛾
or
𝑠1 , 𝑠2 ∈ (𝑎, 𝑏)
(166)
which implies
(𝐼 − 𝜆𝑇) 𝑓 = 𝜙,
(167)
where 𝑇 is defined by
2.4. The Bounded Singular Integral Operators on the Space
𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋).
Lemma 23. Let 𝐴 ∈ 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋), 𝑠 ∈ 𝐼𝑝 = (𝑎, 𝑎 + 𝑝(𝑏 −
𝑎)/(𝑝 + 2)), and 0 ≤ Δ𝑠 < (𝑠 − 𝑎)/𝑝 with 𝑝 ∈ (1, ∞). Then the
improper integral given by
Γ (𝐴) (𝑠) = 𝑝.𝑣. ∫
𝑏
𝐴 (𝑡)
𝑑𝑡,
𝑡−𝑠
(𝑇𝑓) (𝑠1 , 𝑠2 ) = ∫ 𝐴 (𝑠1 , 𝑡) 𝑓 (𝑡, 𝑠2 ) 𝑑𝑡,
𝑎
(168)
𝑓 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋) ,
and 𝐼 is an identity operator on 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋).
𝑠 ∈ 𝐼𝑝
(173)
exists, and there exists the nonnegative constant 𝑀(𝛼, 𝛿, 𝑝)
depending only on 𝛼, 𝛿, and 𝑝 such that
‖Γ (𝐴) (𝑠)‖𝑋 ≤
𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿
,
(𝑠 − 𝑎)𝛼
‖Γ (𝐴) (𝑠 + Δ𝑠) − Γ (𝐴) (𝑠)‖𝑋 ≤
𝑏
(172)
which completes the proof.
𝑎
((𝐼 − 𝜆𝑇) 𝑓) (𝑠1 , 𝑠2 ) = 𝜙 (𝑠1 , 𝑠2 ) ;
(171)
𝑓 = (𝐼 − 𝜆𝑇)−1 𝜙,
where 𝐴, 𝜙 ∈ 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏), 𝑋), 𝑁 is the constant given
in Theorem 20, that is,
|𝜆| <
(170)
exists and the norm of 𝐼 − 𝜆𝑇 satisfies the inequality:
Theorem 22. Let 𝑋 be a real or complex Banach algebra. Then
Fredholm integral equation of the form
1
󵄩󵄩 󵄩󵄩
,
󵄩𝜙󵄩
1 − |𝜆| 𝑁‖𝐴‖𝛼,𝛿,𝛾 󵄩 󵄩𝛼,𝛿,𝛾
(169)
(𝐼 − 𝜆𝑇)−1 : 𝐻𝛼,𝛿,𝛾 ((𝑎, 𝑏) × (𝑎, 𝑏) , 𝑋)
and this also completes the proof of the second part.
The proof of the first and third parts of Theorem can
be completed by the similar way to that of the second part,
using Lemma 18, Theorem 19, (111), Lemma 18, Theorem 19,
and (123), respectively.
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑓󵄩󵄩𝛼,𝛿,𝛾 ≤
(159)
𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
(𝑠 − 𝑎)𝛼+𝛿
(174)
hold for all 𝑠, 𝑠 + Δ𝑠 ∈ (𝑎, 𝑏), where ‖𝐴‖𝛼,𝛿 is defined by (73).
Hereafter, by 𝐼𝑝 , we mean the interval 𝐼𝑝 = (𝑎, 𝑎 + 𝑝(𝑏 −
𝑎)/(𝑝 + 2)) with 1 < 𝑝 < ∞.
Abstract and Applied Analysis
13
Since 𝑡 ∈ (𝑠 − (𝑠 − 𝑎)/𝑝, 𝑠), we obtain 𝑠 − 𝑡 > 0, 𝑡 − 𝑎 >
(𝑝 − 1)(𝑠 − 𝑎)/𝑝 > 0 and 1/(𝑡 − 𝑎) < 𝑝/(𝑝 − 1)(𝑠 − 𝑎). So, we
have by (28) that
Proof. We can write
Γ (𝐴) (𝑠) = ∫
𝑠−((𝑠−𝑎)/𝑝)
𝑎
+∫
ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡
𝑠+((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
+∫
𝑏
𝑠+((𝑠−𝑎)/𝑝)
󵄨
󵄨
󵄨
󵄨󵄨
󵄨󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴 (𝑡) − 𝐴 (𝑠)‖𝑋 = 󵄨󵄨󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴 (𝑠) − 𝐴 (𝑡)‖𝑋
ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡
≤
ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡
for all 𝑠 ∈ 𝐼𝑝 . Here, ℎ𝑥 is defined for each 𝑥 ∈ (𝑎, 𝑏) by
ℎ𝑥 (𝑡) = 1/(𝑡 − 𝑥), 𝑡 ∈ (𝑎, 𝑏) and 𝑡 ≠ 𝑥. Since the functions
𝐴 ∈ 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋) and ℎ𝑥 are continuous for each 𝑥 ∈ (𝑎, 𝑏),
the function 𝐵𝑥 defined by 𝐵𝑥 (𝑡) = ℎ𝑥 (𝑡)𝐴(𝑡) is continuous,
and, so, it is strongly measurable by Theorem 4, where 𝑡 ∈
(𝑎, 𝑏) and 𝑡 ≠ 𝑥.
Since
󵄨
󵄨󵄨
󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴‖𝛼,𝛿
󵄩
󵄨
󵄨
󵄩󵄩
(176)
󵄩󵄩ℎ𝑠 (𝑡) 𝐴 (𝑡)󵄩󵄩󵄩𝑋 ≤ 󵄨󵄨󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴 (𝑡)‖𝑋 ≤ 󵄨
(𝑡 − 𝑎)𝛼
and 𝑠 − 𝑡 > (𝑠 − 𝑎)/𝑝 > 0 for all 𝑡 ∈ (𝑎, 𝑠 − (𝑠 − 𝑎)/𝑝), we get
1/(𝑠 − 𝑡) < 𝑝/(𝑠 − 𝑎). So,
∫
𝑠−((𝑠−𝑎)/𝑝)
𝑎
𝑠−((𝑠−𝑎)/𝑝)
𝑝
𝑑𝑡
𝑑𝑡
≤
∫
𝛼
(𝑡 − 𝑎) |𝑡 − 𝑠| 𝑠 − 𝑎 𝑎
(𝑡 − 𝑎)𝛼
=
= −ℎ𝑠 (𝑡) ‖𝐴 (𝑡 + 𝑠 − 𝑡) − 𝐴 (𝑡)‖𝑋
(175)
≤
𝑝𝛼+𝛿 ‖𝐴‖𝛼,𝛿 (𝑠 − 𝑡)𝛿−1
(𝑝 − 1)
𝛼+𝛿
(𝑠 − 𝑎)𝛼+𝛿
.
By (180) and
∫
𝑠
(𝑠−𝑎)/𝑝
(𝑠 − 𝑡)𝛿−1 𝑑𝑡 = ∫
𝑠−((𝑠−𝑎)/𝑝)
0
𝑢𝛿−1 𝑑𝑢 =
(𝑠 − 𝑎)𝛿
, (181)
𝛿𝑝𝛿
we have
𝑝𝛼 ‖𝐴‖𝛼,𝛿
𝑠
󵄨
󵄨󵄨
󵄨󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴 (𝑡) − 𝐴 (𝑠)‖𝑋 𝑑𝑡 ≤
𝑠−((𝑠−𝑎)/𝑝)
∫
(𝑝 − 1)
𝛼+𝛿
.
𝛿(𝑠 − 𝑎)𝛼
(182)
Similarly, since 𝑡 − 𝑠 > 0 for all 𝑡 ∈ (𝑠, 𝑠 + (𝑠 − 𝑎)/𝑝), we get
󵄨󵄨
󵄨
󵄨󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴 (𝑡) − 𝐴 (𝑠)‖𝑋
= ℎ𝑠 (𝑡) ‖𝐴 (𝑠 + 𝑡 − 𝑠) − 𝐴 (𝑠)‖𝑋
1−𝛼
‖𝐴‖𝛼,𝛿 (𝑡 − 𝑠)𝛿−1
≤
(177)
(183)
(𝑠 − 𝑎)𝛼+𝛿
from (28). Also, by
By (176),
󵄩󵄩
󵄩󵄩 𝑠−((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
󵄩󵄩
󵄩󵄩
󵄩
󵄩󵄩
󵄩󵄩 ≤ ∫
󵄩󵄩∫
𝐴
𝑑𝑡
ℎ
(𝑡)
(𝑡)
󵄩󵄩ℎ𝑠 (𝑡) 𝐴 (𝑡)󵄩󵄩󵄩𝑋 𝑑𝑡
𝑠
󵄩󵄩
󵄩󵄩 𝑎
𝑎
󵄩𝑋
󵄩
On the other hand, since (𝑠 − 𝑎)/𝑝
𝑠+(𝑠−𝑎)/𝑝
>
(178)
0, we get
∫𝑠−(𝑠−𝑎)/𝑝 𝑑𝑡/(𝑡 − 𝑠) = 0. Thus,
𝑠+((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
=∫
𝑠
𝑠−((𝑠−𝑎)/𝑝)
0
∫
𝑠+((𝑠−𝑎)/𝑝)
𝑠
𝑠+(𝑠−𝑎)/𝑝
𝑠−((𝑠−𝑎)/𝑝)
󵄨
󵄨󵄨
󵄨󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴 (𝑡) − 𝐴 (𝑠)‖𝑋 𝑑𝑡
𝑠+((𝑠−𝑎)/𝑝)
𝑠
(179)
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡.
(𝑠 − 𝑎)𝛿
(184)
𝛿𝑝𝛿
Thus, by taking 𝐼 = ∫𝑠−(𝑠−𝑎)/𝑝 ℎ𝑠 (𝑡)𝐴(𝑡)𝑑𝑡,
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡
𝑢𝛿−1 𝑑𝑢 =
‖𝐴‖𝛼,𝛿
󵄨
󵄨󵄨
. (185)
󵄨󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴 (𝑡) − 𝐴 (𝑠)‖𝑋 𝑑𝑡 ≤ 𝛿
𝛿𝑝 (𝑠 − 𝑎)𝛼
+∫
𝑠+((𝑠−𝑎)/𝑝)
𝑠
𝑠
(𝑠−𝑎)/𝑝
(𝑡 − 𝑠)𝛿−1 𝑑𝑡 = ∫
𝑠
ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡
𝑠+((𝑠−𝑎)/𝑝)
+∫
𝑠+((𝑠−𝑎)/𝑝)
‖𝐼‖ ≤ ∫
𝑠−((𝑠−𝑎)/𝑝)
=∫
∫
and (183), we derive
1−𝛼
𝑝𝛼 (𝑝 − 1) ‖𝐴‖𝛼,𝛿
.
≤
(1 − 𝛼) (𝑠 − 𝑎)𝛼
∫
(𝑡 − 𝑎)𝛼+𝛿
(180)
(𝑝−1)(𝑠−𝑎)/𝑝
𝑝
𝑢−𝛼 𝑑𝑢
∫
𝑠−𝑎 0
𝑝𝛼 (𝑝 − 1)
=
.
(1 − 𝛼) (𝑠 − 𝑎)𝛼
‖𝐴‖𝛼,𝛿 (𝑠 − 𝑡)𝛿−1
≤(
󵄨
󵄨󵄨
󵄨󵄨ℎ𝑠 (𝑡)󵄨󵄨󵄨 ‖𝐴 (𝑡) − 𝐴 (𝑠)‖𝑋 𝑑𝑡
𝑝𝛼
𝛼+𝛿
(𝑝 − 1)
𝛿
+
(186)
‖𝐴‖𝛼,𝛿
1
)
𝛿𝑝𝛿 (𝑠 − 𝑎)𝛼
from (182) and (185).
From the inequality 𝑡 − 𝑎 > 𝑠 + ((𝑠 − 𝑎)/𝑝) − 𝑎 = (𝑝 +
1)(𝑠−𝑎)/𝑝 which is satisfied for all 𝑡 ∈ (𝑠+(𝑠−𝑎)/𝑝, 𝑏), we get
14
Abstract and Applied Analysis
𝑡−𝑠 = 𝑡−𝑎−(𝑠−𝑎) > 𝑡−𝑎−𝑝(𝑡−𝑎)/(𝑝+1) = (𝑡−𝑎)/(𝑝+1) > 0,
and, thus, 1/(𝑡 − 𝑠) < (𝑝 + 1)/(𝑡 − 𝑎). Hence,
(𝑝 + 1) ‖𝐴‖𝛼,𝛿
‖𝐴‖𝛼,𝛿
󵄩
󵄩󵄩
.
≤
󵄩󵄩ℎ𝑠 (𝑡) 𝐴 (𝑡)󵄩󵄩󵄩𝑋 ≤
𝛼
(𝑡 − 𝑎) |𝑡 − 𝑠|
(𝑡 − 𝑎)𝛼+1
we can write
Γ (𝐴) (𝑠 + Δ𝑠) − Γ (𝐴) (𝑠)
(187)
𝑏
𝑎
𝑎
𝑠−((𝑠−𝑎)/𝑝)
From
∫
𝑏
= ∫ ℎ𝑠+Δ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡 − ∫ ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡
= Δ𝑠 ∫
𝑎
𝛼
𝑏
𝑠+((𝑠−𝑎)/𝑝)
𝑝
𝑑𝑡
=
𝛼
(𝑡 − 𝑎)𝛼+1 𝛼(𝑝 + 1) (𝑠 − 𝑎)𝛼
+∫
𝑝
1
𝛼
𝛼 ≤
𝛼(𝑏 − 𝑎)
𝛼(𝑝 + 1) (𝑠 − 𝑎)𝛼
(188)
−
and (187), it is obtained that
󵄩󵄩 𝑏
󵄩󵄩
𝑏
󵄩󵄩
󵄩󵄩
󵄩
󵄩󵄩
󵄩󵄩∫
󵄩󵄩 ≤ ∫
ℎ
𝐴
𝑑𝑡
(𝑡)
(𝑡)
󵄩󵄩ℎ𝑠 (𝑡) 𝐴 (𝑡)󵄩󵄩󵄩𝑋 𝑑𝑡
󵄩󵄩 𝑠+((𝑠−𝑎)/𝑝) 𝑠
󵄩󵄩
𝑠+((𝑠−𝑎)/𝑝)
󵄩
󵄩𝑋
≤
(𝑝 + 1) 𝑝𝛼 ‖𝐴‖𝛼,𝛿
𝛼
𝛼(𝑝 + 1) (𝑠 − 𝑎)𝛼
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
𝛼
𝑏
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
Thus,
󵄩󵄩
󵄩
󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) 𝐴 (𝑡)󵄩󵄩󵄩𝑋
≤
≤
By (178), (186), and (189), we have
∫
(𝑠 − 𝑎)2 (𝑡 − 𝑎)𝛼
,
1−𝛼
=
(𝑝 − 1)
(1 −
𝛼) 𝑝1−𝛼 (𝑠
− 𝑎)
𝛼−1
,
(195)
Δ𝑠 = (Δ𝑠)𝛿 (Δ𝑠)1−𝛿 ≤ (Δ𝑠)𝛿 (𝑠 − 𝑎)1−𝛿
and (194), we get
󵄩󵄩
󵄩󵄩
𝑠−((𝑠−𝑎)/𝑝)
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩Δ𝑠 ∫
ℎ
𝐴
𝑑𝑡
ℎ
(𝑡)
(𝑡)
(𝑡)
𝑠+Δ𝑠
𝑠
󵄩󵄩
󵄩󵄩
𝑎
󵄩𝑋
󵄩
1−𝛼
1−𝛼
(191)
≤
𝑝1+𝛼 (𝑝 − 1)
‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
(1 − 𝛼) (𝑠 − 𝑎)𝛼+𝛿
(196)
.
Since 𝑡 − 𝑎 > 𝑡 − 𝑠 > 𝑡 − 𝑠 − Δ𝑠 > 0 for all 𝑡 ∈ (𝑠 + Δ𝑠 + (𝑠 −
𝑎)/𝑝, 𝑏), the inequality 1/(𝑡 − 𝑎) < 1/(𝑡 − 𝑠) < 1/(𝑡 − 𝑠 − Δ𝑠)
holds, and, so,
1
(𝑡 − 𝑠 − Δ𝑠) (𝑡 − 𝑠)
1
1
1
(
−
),
Δ𝑠 𝑡 − 𝑠 − Δ𝑠 𝑡 − 𝑠
𝑠−𝑎
0 < Δ𝑠 <
,
𝑝
𝑝2 ‖𝐴‖𝛼,𝛿
(194)
((𝑝−1)(𝑠−𝑎))/𝑝
𝑑𝑡
𝑢−𝛼 𝑑𝑢
𝛼 =∫
(𝑡 − 𝑎)
0
(190)
Furthermore, since
=
𝑠−((𝑠−𝑎)/𝑝)
𝑎
such that
(𝑝 + 1) 𝑝𝛼
1
+
+
+
𝛼.
𝛼+𝛿
𝛿
𝛼(𝑝 + 1)
(𝑝 − 1) 𝛿 𝛿𝑝
‖𝐴‖𝛼,𝛿
(𝑡 − 𝑎) |𝑡 − 𝑠 − Δ𝑠| |𝑡 − 𝑠|
𝛼
since |𝑡 − 𝑠 − Δ𝑠| > |𝑡 − 𝑠| > (𝑠 − 𝑎)/𝑝 > 0 and 1/|𝑡 − 𝑠 − Δ𝑠| <
1/|𝑡 − 𝑠| < 𝑝/(𝑠 − 𝑎) for all 𝑡 ∈ (𝑎, 𝑠 − (𝑠 − 𝑎)/𝑝). Besides, by
𝑀1 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿
≤
(𝑠 − 𝑎)𝛼
𝑝𝛼
ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡.
(193)
(189)
𝑝𝛼 (𝑝 − 1)
𝑀1 (𝛼, 𝛿, 𝑝) =
1−𝛼
[ℎ𝑠+Δ𝑠 (𝑡) 𝐴 (𝑡) − ℎ𝑠 (𝑡) 𝐴 (𝑡)] 𝑑𝑡
+ Δ𝑠 ∫
.
󵄩󵄩 𝑠−((𝑠−𝑎)/𝑝)
󵄩󵄩󵄩
󵄩
ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡󵄩󵄩󵄩󵄩
‖Γ (𝐴) (𝑠)‖𝑋 ≤ 󵄩󵄩󵄩󵄩∫
󵄩󵄩𝑋
󵄩󵄩 𝑎
󵄩󵄩 𝑠+((𝑠−𝑎)/𝑝)
󵄩󵄩
󵄩
󵄩
ℎ𝑠 (𝑡) 𝐴 (𝑡) d𝑡󵄩󵄩󵄩󵄩
+ 󵄩󵄩󵄩󵄩∫
󵄩󵄩 𝑠−((𝑠−𝑎)/𝑝)
󵄩󵄩𝑋
󵄩󵄩 𝑏
󵄩󵄩
󵄩
󵄩
ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡󵄩󵄩󵄩󵄩
+ 󵄩󵄩󵄩󵄩∫
󵄩󵄩 𝑠+((𝑠−𝑎)/𝑝)
󵄩󵄩𝑋
ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) 𝐴 (𝑡) 𝑑𝑡
󵄩
󵄩󵄩
󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) 𝐴 (𝑡)󵄩󵄩󵄩𝑋
(192)
≤
≤
‖𝐴‖𝛼,𝛿
(𝑡 − 𝑎) (𝑡 − 𝑠 − Δ𝑠) (𝑡 − 𝑠)
𝛼
‖𝐴‖𝛼,𝛿
(𝑡 − 𝑠 − Δ𝑠)2+𝛼
.
(197)
Abstract and Applied Analysis
15
𝑠+Δ𝑠+(𝑠−𝑎)/𝑝
𝑢+(𝑠−𝑎)/𝑝
since ∫𝑠+Δ𝑠−(𝑠−𝑎)/𝑝 𝑑𝑡/(𝑡 − 𝑠 − Δ𝑠) = ∫𝑢−(𝑠−𝑎)/𝑝 𝑑𝑡/(𝑡 − 𝑢) = 0.
Therefore, we get by (195) that
By
𝑏
𝑑𝑡
2+𝛼
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝) (𝑡 − 𝑠 − Δ𝑠)
∫
=∫
𝑏−𝑠−Δ𝑠
𝑢−2−𝛼 𝑑𝑢
(𝑠−𝑎)/𝑝
=
≤
󵄨󵄨 𝑠+Δs+((𝑠−𝑎)/𝑝)
󵄨󵄨
𝑠+Δ𝑠−((𝑠−𝑎)/𝑝)
󵄨󵄨
󵄨󵄨
󵄨
󵄨󵄨
󵄨󵄨∫
󵄨󵄨 ≤ ∫
𝑑𝑡
ℎ
(𝑡)
󵄨󵄨ℎ𝑠+Δ𝑠 (𝑡)󵄨󵄨󵄨 𝑑𝑡
𝑠+Δ𝑠
󵄨󵄨 𝑠−((𝑠−𝑎)/𝑝)
󵄨󵄨
𝑠−
(𝑠−𝑎)/𝑝
(
)
󵄨
󵄨
1+𝛼
𝑝
1
−
1+𝛼
(1 + 𝛼) (𝑠 − 𝑎)
(1 + 𝛼) (𝑏 − 𝑠 − Δ𝑠)1+𝛼
𝑠+Δ𝑠−((𝑠−𝑎)/𝑝)
(198)
=∫
𝑠−((𝑠−𝑎)/𝑝)
≤
1+𝛼
𝑝
(1 + 𝛼) (𝑠 − 𝑎)1+𝛼
and (195), we obtain
󵄩󵄩
󵄩󵄩
𝑏
󵄩󵄩
󵄩󵄩
󵄩󵄩Δ𝑠 ∫
󵄩󵄩
ℎ
ℎ
𝐴
𝑑𝑡
(𝑡)
(𝑡)
(𝑡)
𝑠+Δ𝑠
𝑠
󵄩󵄩
󵄩󵄩
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
󵄩
󵄩𝑋
𝑝1+𝛼 ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
≤
󵄩󵄩
󵄩󵄩
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
󵄩󵄩
󵄩󵄩
󵄩󵄩𝐴 (𝑠 + Δ𝑠) ∫
󵄩󵄩
ℎ
𝑑𝑡
(𝑡)
𝑠+Δ𝑠
󵄩󵄩
󵄩󵄩
𝑠−((𝑠−𝑎)/𝑝)
󵄩
󵄩𝑋
(199)
≤
from (197). Then,
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
−∫
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
ℎ𝑠 (𝑡) [𝐴 (t) − 𝐴 (𝑠)] 𝑑𝑡
𝑠−((𝑠−𝑎)/𝑝)
+ 𝐴 (𝑠 + Δ𝑠) ∫
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
− 𝐴 (𝑠) ∫
𝑠−((𝑠−𝑎)/𝑝)
(200)
∫
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
ℎ𝑠+Δ𝑠 (𝑡) 𝑑𝑡
=∫
𝑠−((𝑠−𝑎)/𝑝)
ℎ𝑠 (𝑡) 𝑑𝑡.
=∫
ℎ𝑠+Δ𝑠 (𝑡) 𝑑𝑡
=∫
𝑠−((𝑠−𝑎)/𝑝)
𝑠+Δ𝑠−((𝑠−𝑎)/𝑝)
𝑠+Δ𝑠−((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
(201)
ℎ𝑠+Δ𝑠 (𝑡) 𝑑𝑡
ℎ𝑠+Δ𝑠 (𝑡) 𝑑𝑡,
(𝑠 − 𝑎)𝛼+𝛿
(203)
.
ℎ𝑠 (𝑡) 𝑑𝑡 + ∫
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑠+((𝑠−𝑎)/𝑝)
ℎ𝑠 (𝑡) 𝑑𝑡
(204)
ℎ𝑠 (𝑡) 𝑑𝑡
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑝
𝑝Δ𝑠
𝑝(Δ𝑠)𝛿
𝑑𝑡 =
≤
∫
𝑠 − 𝑎 𝑠+((𝑠−𝑎)/𝑝)
𝑠 − 𝑎 (𝑠 − 𝑎)𝛿
from (195). Hence,
ℎ𝑠+Δ𝑠 (𝑡) 𝑑𝑡
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
+∫
=∫
𝑠+((𝑠−𝑎)/𝑝)
𝑠+((𝑠−𝑎)/𝑝)
𝑠+Δ𝑠−((𝑠−𝑎)/𝑝)
≤
𝑝‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
ℎ𝑠 (𝑡) 𝑑𝑡
𝑠−((𝑠−𝑎)/𝑝)
≤
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
(𝑠 + Δ𝑠 − 𝑎)𝛼 (𝑠 − 𝑎)𝛿
𝑠−((𝑠−𝑎)/𝑝)
Also, we have
∫
𝑝‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
Since ∫𝑠−((𝑠−𝑎)/𝑝) 𝑑𝑡/(𝑡 − 𝑠) = 0 and 𝑡 − 𝑠 > (𝑠 − 𝑎)/𝑝 > 0 which
yields 1/(𝑡 − 𝑠) < 𝑝/(𝑠 − 𝑎) for all 𝑡 ∈ (𝑠 + (𝑠 − 𝑎)/𝑝, 𝑠 + Δ𝑠 +
(𝑠 − 𝑎)/𝑝), we obtain
ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)] 𝑑𝑡
=∫
(202)
𝑠+((𝑠−𝑎)/𝑝)
[ℎ𝑠+Δ𝑠 (𝑡) 𝐴 (𝑡) − ℎ𝑠 (𝑡) 𝐴 (𝑡)] 𝑑𝑡
𝑠−((𝑠−𝑎)/𝑝)
𝑠+Δ𝑠−((𝑠−𝑎)/𝑝)
𝑝Δ𝑠
𝑝
𝑝(Δ𝑠)𝛿
𝑑𝑡 ≤
,
≤
∫
𝑠 − 𝑎 𝑠−((𝑠−𝑎)/𝑝)
𝑠 − 𝑎 (𝑠 − 𝑎)𝛿
since 𝑠+Δ𝑠−𝑡 > (𝑠−𝑎)/𝑝 > 0 which implies that 1/(𝑠+Δ𝑠−𝑡) <
𝑝/(𝑠 − 𝑎) for all 𝑡 ∈ (𝑠 − (𝑠 − 𝑎)/𝑝, 𝑠 + Δ𝑠 − (𝑠 − 𝑎)/𝑝). So,
(1 + 𝛼) (𝑠 − 𝑎)𝛼+𝛿
∫
−ℎ𝑠+Δ𝑠 (𝑡) 𝑑𝑡
󵄩󵄩
󵄩󵄩
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
󵄩󵄩
󵄩󵄩
󵄩󵄩−𝐴 (𝑠) ∫
󵄩󵄩
𝑑𝑡
ℎ
(𝑡)
𝑠
󵄩󵄩
󵄩󵄩
𝑠−((𝑠−𝑎)/𝑝)
󵄩
󵄩𝑋
= ‖𝐴 (𝑠)‖𝑋 ∫
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
ℎ𝑠 (𝑡) 𝑑𝑡 ≤
𝑝‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
(𝑠 − 𝑎)𝛼+𝛿
.
(205)
16
Abstract and Applied Analysis
By (192),
∫
Since
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)] 𝑑𝑡
𝑠−((𝑠−𝑎)/𝑝)
−∫
𝑠−(Δ𝑠/2)
= Δ𝑠 ∫
𝑠−((𝑠−𝑎)/𝑝)
𝑠−(Δ𝑠/2)
𝑠−((𝑠−𝑎)/𝑝)
+∫
𝑠+(3Δ𝑠/2)
𝑠−(Δ𝑠/2)
−∫
𝑠+(3Δ𝑠/2)
𝑠−(Δ𝑠/2)
+ Δ𝑠 ∫
=∫
𝑠−(Δ𝑠/2)
+∫
ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡
ℎ𝑠 (𝑡) [𝐴 (𝑠 + Δ𝑠) − 𝐴 (𝑠)] 𝑑𝑡.
󵄩󵄩
󵄩
󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)]󵄩󵄩󵄩𝑋
󵄩
󵄩
= 󵄩󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) [𝐴 (𝑡 + 𝑠 − 𝑡) − 𝐴 (𝑡)]󵄩󵄩󵄩𝑋
(𝑡 − 𝑎)
≤
𝛼+𝛿
𝑝
‖𝐴‖𝛼,𝛿 (𝑠 − 𝑡)
𝛼+𝛿
(𝑝 − 1)
.
𝑠−(Δ𝑠/2)
𝑠−((𝑠−𝑎)/𝑝)
(𝑠 − 𝑡)𝛿−2 𝑑𝑡 = ∫
(𝑠−𝑎)/𝑝
Δ𝑠/2
𝑠+Δ𝑠
𝑠−(Δ𝑠/2)
𝑢𝛿−2 𝑑𝑢
󵄩󵄩
󵄩󵄩
𝑠−(Δ𝑠/2)
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩Δ𝑠 ∫
ℎ
−
𝐴
𝑑𝑡
ℎ
(𝑡)
(𝑡)
(𝑡)
(𝑠)]
[𝐴
𝑠+Δ𝑠
𝑠
󵄩󵄩
󵄩󵄩
𝑠−((𝑠−𝑎)/𝑝)
󵄩𝑋
󵄩
𝛼+𝛿
(1 − 𝛿) 2𝛿−1 (𝑠 − 𝑎)𝛼+𝛿
(𝑠 − 𝑎)𝛼+𝛿
(𝑠 + Δ𝑠 − 𝑡)𝛿−1 𝑑𝑡 = ∫
3Δ𝑠/2
0
≤
≤
(2𝑝)
.
𝑢𝛿−1 𝑑𝑢 =
3𝛿 (Δ𝑠)𝛿
𝛿2𝛿
𝛼+𝛿 𝛿
3 ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
𝛼+𝛿
(2𝑝 − 1)
𝛿2𝛿 (𝑠 − 𝑎)𝛼+𝛿
‖𝐴‖𝛼,𝛿 (𝑡 − 𝑠 − Δ𝑠)𝛿−1
𝑠+(3Δ𝑠/2)
𝑠+Δ𝑠
(𝑝 − 1)
𝛼+𝛿
and (211),
󵄩󵄩
󵄩󵄩 𝑠+Δ𝑠
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩∫
𝑑𝑡
−
𝐴
+
Δ𝑠)]
ℎ
(𝑡)
(𝑠
(𝑡)
[𝐴
𝑠+Δ𝑠
󵄩󵄩
󵄩󵄩 𝑠−(Δ𝑠/2)
󵄩𝑋
󵄩
∫
and (207),
≤
(2𝑝 − 1)
(213)
.
(𝑠 + Δ𝑠 − 𝑎)𝛼+𝛿
‖𝐴‖𝛼,𝛿 (𝑡 − 𝑠 − Δ𝑠)𝛿−1
≤
(𝑠 − 𝑎)𝛼+𝛿
.
(214)
By
(Δ𝑠)𝛿−1
,
(1 − 𝛿) 2𝛿−1
𝑝𝛼+𝛿 ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
‖𝐴‖𝛼,𝛿 (𝑠 + Δ𝑠 − 𝑡)𝛿−1
(212)
(Δ𝑠)𝛿−1
(𝑠 − 𝑎)𝛿−1
(208)
=
−
𝛿−1
(1 − 𝛿) 2
(1 − 𝛿) 𝑝𝛿−1
≤
𝛼+𝛿
Since 𝑡 − 𝑠 − Δ𝑠 > 0 for all 𝑡 ∈ (𝑠 + Δ𝑠, 𝑠 + (3Δ𝑠/2)), we
derive
󵄩
󵄩󵄩
󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)]󵄩󵄩󵄩𝑋
󵄩
󵄩
= 󵄩󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑠 + Δ𝑠 + 𝑡 − 𝑠 − Δ𝑠) − 𝐴 (𝑠 + Δ𝑠)]󵄩󵄩󵄩𝑋
By
∫
∫
(207)
𝛿−2
(𝑠 − 𝑎)𝛼+𝛿
(2𝑝)
By
Since 𝑠 − 𝑡 + Δ𝑠 > 𝑠 − 𝑡 > 0 and 𝑡 − 𝑎 > (𝑝 − 1)(𝑠 − 𝑎)/𝑝 > 0
which imply that 1/(𝑠 − 𝑡 + Δ𝑠) < 1/(𝑠 − 𝑡) and 1/(𝑡 − 𝑎) <
𝑝/(𝑝 − 1)(𝑠 − 𝑎) for all 𝑡 ∈ (𝑠 − ((𝑠 − 𝑎)/𝑝), 𝑠 − (Δ𝑠/2)), it is
found that
‖𝐴‖𝛼,𝛿 (𝑠 − 𝑡)
(211)
(𝑡 − 𝑎)𝛼+𝛿
≤
(206)
𝛼+𝛿
‖𝐴‖𝛼,𝛿 (𝑠 + Δ𝑠 − 𝑡)𝛿−1
≤
ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)] 𝑑𝑡
𝛿−2
ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)] 𝑑𝑡
= ℎ𝑠+Δ𝑠 (𝑡) ‖𝐴 (𝑡 + 𝑠 + Δ𝑠 − 𝑡) − 𝐴 (𝑡)‖𝑋
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡
𝑠+(3Δ𝑠/2)
𝑠+(3Δ𝑠/2)
(210)
and 𝑠 + Δ𝑠 − 𝑡 > 0, 𝑡 − 𝑎 > 𝑠 − 𝑎 − (Δ𝑠/2) > 𝑠 − 𝑎 − ((𝑠 − 𝑎)/2𝑝) =
(2𝑝 − 1)(𝑠 − 𝑎)/2𝑝 which yields 1/(𝑡 − 𝑎) < 2𝑝/(2𝑝 − 1)(𝑠 − 𝑎)
for all 𝑡 ∈ (𝑠 − (Δ𝑠/2), 𝑠 + Δ𝑠), we get
󵄩󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)]󵄩󵄩󵄩
󵄩𝑋
󵄩
ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)] 𝑑𝑡
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)] 𝑑𝑡
𝑠+Δ𝑠
ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑠 + Δ𝑠) − 𝐴 (𝑠)] 𝑑𝑡
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
≤
ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)] 𝑑𝑡
𝑠−(Δ𝑠/2)
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡
𝑠+(3Δ𝑠/2)
−∫
𝑠+(3Δ𝑠/2)
𝑠+Δ𝑠
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
𝑠−((𝑠−𝑎)/𝑝)
−∫
∫
.
(209)
(𝑡 − 𝑠 − Δ𝑠)𝛿−1 𝑑𝑡 = ∫
Δ𝑠/2
0
𝑢𝛿−1 𝑑𝑢 =
(Δ𝑠)𝛿
𝛿2𝛿
and (214),
󵄩󵄩
󵄩󵄩 𝑠+(3Δ𝑠/2)
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩∫
−
𝐴
+
Δ𝑠)]
𝑑𝑡
ℎ
(𝑡)
(𝑡)
(𝑠
[𝐴
𝑠+Δ𝑠
󵄩󵄩
󵄩󵄩 𝑠+Δ𝑠
󵄩𝑋
󵄩
≤
‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
𝛿2𝛿 (𝑠 − 𝑎)𝛼+𝛿
.
(215)
(216)
Abstract and Applied Analysis
17
So, we derive
equations (219) and (220), it is obtained that
󵄩󵄩
󵄩󵄩 𝑠+(3Δ𝑠/2)
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩∫
𝑑𝑡
−
𝐴
+
Δ𝑠)]
ℎ
(𝑡)
(𝑠
(𝑡)
[𝐴
󵄩󵄩
󵄩󵄩 𝑠−(Δ𝑠/2) 𝑠+Δ𝑠
󵄩𝑋
󵄩
𝛼+𝛿 𝛿
‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
1
≤(
+
)
𝛼+𝛿
𝛿
(𝑠 − 𝑎)𝛼+𝛿
(2𝑝 − 1) 𝛿2𝛿 𝛿2
(2𝑝)
3
󵄩󵄩
󵄩󵄩 𝑠
󵄩
󵄩󵄩
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡󵄩󵄩󵄩
󵄩󵄩∫
󵄩󵄩𝑋
󵄩󵄩 𝑠−(Δ𝑠/2)
(217)
≤
𝑠+(3Δ𝑠/2)
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡
𝑠−(Δ𝑠/2)
=∫
𝑠
𝑠−(Δ𝑠/2)
+∫
𝛼+𝛿
(2𝑝 − 1)
‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
𝛼+𝛿
𝛿2𝛿 (𝑠 − 𝑎)𝛼+𝛿
,
(222)
󵄩󵄩
󵄩󵄩 𝑠+(3Δ𝑠/2)
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩∫
−
𝐴
𝑑𝑡
ℎ
(𝑡)
(𝑡)
(𝑠)]
[𝐴
𝑠
󵄩󵄩
󵄩󵄩 𝑠
󵄩𝑋
󵄩
from (213) and (216).
∫
(2𝑝)
≤
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡
𝑠+(3Δ𝑠/2)
𝑠
3𝛿 ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
𝛿2𝛿 (𝑠 − 𝑎)𝛼+𝛿
.
(218)
By (222),
ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)] 𝑑𝑡.
󵄩󵄩
󵄩󵄩 𝑠+(3Δ𝑠/2)
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩∫
−
𝐴
𝑑𝑡
ℎ
(𝑡)
(𝑡)
(𝑠)]
[𝐴
󵄩󵄩
󵄩󵄩 𝑠−(Δ𝑠/2) 𝑠
󵄩𝑋
󵄩
Since 𝑡 − 𝑠 < 0 and 𝑡 − 𝑎 > (2𝑝 − 1)(𝑠 − 𝑎)/2𝑝, which implies
1/(𝑡 − 𝑎) < 2𝑝/(2𝑝 − 1)(𝑠 − 𝑎) for all 𝑡 ∈ (𝑠 − (Δ𝑠/2), 𝑠),
󵄩
󵄩󵄩
󵄩󵄩ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)]󵄩󵄩󵄩𝑋
𝛼+𝛿
𝛿
3𝛿 ‖𝐴‖𝛼,𝛿 (Δ𝑠)
≤(
+
)
.
𝛼+𝛿
𝛿
(𝑠 − 𝑎)𝛼+𝛿
(2𝑝 − 1) 𝛿2𝛿 𝛿2
(2𝑝)
= −ℎ𝑠 (t) ‖𝐴 (𝑠) − 𝐴 (𝑡)‖𝑋
(223)
= −ℎ𝑠 (𝑡) ‖𝐴 (𝑡 + 𝑠 − 𝑡) − 𝐴 (𝑡)‖𝑋
≤
≤
‖𝐴‖𝛼,𝛿 (𝑠 − 𝑡)𝛿−1
(219)
(𝑡 − 𝑎)𝛼+𝛿
(2𝑝)
𝛼+𝛿
‖𝐴‖𝛼,𝛿 (𝑠 − 𝑡)𝛿−1
𝛼+𝛿
(2𝑝 − 1)
(𝑠 − 𝑎)𝛼+𝛿
.
Since 𝑡 − 𝑠 > 𝑡 − 𝑠 − Δ𝑠 > 0 and 𝑠 + Δ𝑠 − 𝑎 > 𝑠 − 𝑎 > 0 which
yield 1/(𝑡 − 𝑠) < 1/(𝑡 − 𝑠 − Δ𝑠) and 1/(𝑠 + Δ𝑠 − 𝑎) < 1/(𝑠 − 𝑎),
󵄩
󵄩󵄩
󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠 + Δ𝑠)]󵄩󵄩󵄩𝑋
󵄩
󵄩
= 󵄩󵄩󵄩ℎ𝑠+Δ𝑠 (𝑡) ℎ𝑠 (𝑡) [𝐴 (𝑠 + Δ𝑠 + 𝑡 − 𝑠 − Δ𝑠) − 𝐴 (𝑠 + Δ𝑠)]󵄩󵄩󵄩𝑋
Since 𝑡 − 𝑠 > 0 for all 𝑡 ∈ (𝑠, 𝑠 + (3Δ𝑠/2)),
≤
󵄩
󵄩󵄩
󵄩󵄩ℎ𝑠 (𝑡) [𝐴 (𝑡) − 𝐴 (𝑠)]󵄩󵄩󵄩𝑋 = ℎ𝑠 (𝑡) ‖𝐴 (𝑠 + 𝑡 − 𝑠) − 𝐴 (𝑠)‖𝑋
‖𝐴‖𝛼,𝛿 (𝑡 − 𝑠)
≤
‖𝐴‖𝛼,𝛿 (𝑡 − 𝑠 − Δ𝑠)𝛿−2
(𝑠 + Δ𝑠 − 𝑎)𝛼+𝛿
≤
‖𝐴‖𝛼,𝛿 (𝑡 − 𝑠 − Δ𝑠)𝛿−2
(𝑠 − 𝑎)𝛼+𝛿
.
(224)
𝛿−1
(𝑠 − 𝑎)𝛼+𝛿
.
(220)
By
By
∫
∫
𝑠
𝑠−(Δ𝑠/2)
Δ𝑠/2
0
∫
𝑠+(3Δ𝑠/2)
𝑠
𝑠+(3Δ𝑠/2)
(𝑠 − 𝑡)𝛿−1 𝑑𝑡
=∫
𝑢𝛿−1 𝑑𝑢 =
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
(Δ𝑠)𝛿
,
𝛿2𝛿
(𝑡 − 𝑠)𝛿−1 𝑑𝑡 = ∫
3Δ𝑠/2
0
(𝑡 − 𝑠 − Δ𝑠)𝛿−2 𝑑𝑡 = ∫
Δ𝑠/2
𝑢𝛿−1 𝑑𝑢 =
𝛿
3 (Δ𝑠)
,
𝛿2𝛿
𝑢𝛿−2 𝑑𝑢
=
(Δ𝑠)𝛿−1
(𝑠 − 𝑎)𝛿−1
−
(1 − 𝛿) 2𝛿−1 (1 − 𝛿) 𝑝𝛿−1
≤
(Δ𝑠)𝛿−1
(1 − 𝛿) 2𝛿−1
(221)
𝛿
(𝑠−𝑎)/𝑝
(225)
18
Abstract and Applied Analysis
and (224),
By (190) and (229),
󵄩󵄩
󵄩󵄩
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
󵄩󵄩
󵄩󵄩
󵄩󵄩Δ𝑠 ∫
󵄩󵄩
ℎ
−
𝐴
+
Δ𝑠)]
𝑑𝑡
ℎ
(𝑡)
(𝑡)
(𝑡)
(𝑠
[𝐴
𝑠+Δ𝑠
𝑠
󵄩󵄩
󵄩󵄩
𝑠+(3Δ𝑠/2)
󵄩
󵄩𝑋
≤
−∫
‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
(1 − 𝛿) 2𝛿−1 (𝑠 − 𝑎)
𝑠−(Δ𝑠/2)
𝑠−((𝑠−𝑎)/𝑝)
−∫
𝑠+(3Δ𝑠/2)
−3Δ𝑠/2
−((𝑠−𝑎)/𝑝)−Δ𝑠
= [𝐴 (𝑠)−𝐴 (𝑠+Δ𝑠)] (∫
Δ𝑠+((𝑠−𝑎)/𝑝)
𝑑𝑢 Δ𝑠+((𝑠−𝑎)/𝑝) 𝑑𝑢
−∫
)
𝑢
𝑢
3Δ𝑠/2
= 0.
(226)
Thus, by taking
𝑀2 (𝛼, 𝛿, 𝑝) = 2𝑝 +
+
+
𝑝
(𝑝 − 1)
(2𝑝)
(1 − 𝛿) 2𝛿−1
𝛼+𝛿 𝛿
3
𝛼+𝛿
(2𝑝 − 1)
(2𝑝)
𝛿2𝛿
+
𝛼+𝛿
𝛼+𝛿
(2𝑝 − 1)
𝛿2𝛿
+
which terminates the proof.
Hereafter, we assume unless stated otherwise that
𝑀(𝛼, 𝛿, 𝑝) is defined by (232).
Theorem 24. The operator 𝑇 defined by
𝑇 (𝐴) (𝑠) = 𝑝.𝑣. ∫
𝑎
3𝛿
𝛿2𝛿
𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿
,
(𝑠 − 𝑎)𝛼
(234)
𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
(𝑠 − 𝑎)𝛼+𝛿
1
1
≤
,
𝑠 + Δ𝑠 − 𝑎 𝑠 − 𝑎
(228)
1
(Δ𝑠)𝛿
(𝑠 − 𝑎)𝛼+𝛿
from (203), (205), (209), (217), (223), (226). By (196), (199),
and (228), we find
‖Γ (𝐴) (𝑠 + Δ𝑠) − Γ (𝐴) (𝑠)‖𝑋
(235)
≤
such that
1−𝛼
[(1 + 𝛼) (𝑝 − 1)
(1 − 𝛼) (1 + 𝛼)
+ 1 − 𝛼]
≤
.
(230)
(𝑠 − 𝑎)𝛿
‖Δ𝑇 (𝐴)‖𝑋 ≤ ‖𝑇 (𝐴) (𝑠 + Δ𝑠)‖𝑋 + ‖𝑇 (𝐴) (𝑠)‖𝑋
(229)
(𝑠 − 𝑎)𝛼+𝛿
≤
(236)
𝑝𝛿
that
≤
(𝑀2 (𝛼, 𝛿, 𝑝) + 𝑀3 (𝛼, 𝑝)) ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
𝑝
(233)
hold for all 𝐴 ∈ 𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋), 𝑠, 𝑠 + Δ𝑠 ∈ 𝐼𝑝 . Furthermore, if
Δ𝑠 ≥ (𝑠 − 𝑎)/𝑝 > 0, we obtain by (234) and
𝑀2 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
𝑀3 (𝛼, 𝑝) =
𝐴 ∈ 𝐻𝛼,𝛿 ((𝑎, 𝑏) , 𝑋) , 𝑠 ∈ 𝐼𝑝
‖𝑇 (𝐴) (𝑠 + Δ𝑠) − 𝑇 (𝐴) (𝑠)‖𝑋 ≤
󵄩󵄩 𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
󵄩󵄩
󵄩󵄩
󵄩󵄩
󵄩󵄩∫
󵄩󵄩
[ℎ
𝐴
−
ℎ
𝐴
𝑑𝑡
(𝑡)
(𝑡)
(𝑡)
(𝑡)]
𝑠+Δ𝑠
𝑠
󵄩󵄩 𝑠−((𝑠−𝑎)/𝑝)
󵄩󵄩
󵄩
󵄩𝑋
1+𝛼
𝐴 (𝑡)
𝑑𝑡;
𝑡−𝑠
‖𝑇 (𝐴) (𝑠)‖𝑋 ≤
(227)
we obtain
≤
𝑏
Proof. If 0 ≤ Δ𝑠 < (𝑠 − 𝑎)/𝑝, we get by Lemma 23 that
1
𝛿2𝛿
1
+
,
(1 − 𝛿) 2𝛿−1
≤
(231)
is the singular integral operator in the space 𝐵(𝐻𝛼,𝛿
((𝑎, 𝑏), 𝑋), 𝐻𝛼,𝛿 (𝐼𝑝 , 𝑋)), and ‖𝑇‖ satisfies the inequality
‖𝑇‖ ≤ 2𝑝𝛿 𝑀(𝛼, 𝛿, 𝑝).
𝛼+𝛿
𝛼+𝛿
(𝑠 − 𝑎)𝛼+𝛿
𝑀 (𝛼, 𝛿, 𝑝) = max {𝑀1 (𝛼, 𝛿, 𝑝) , 𝑀2 (𝛼, 𝛿, 𝑝) + 𝑀3 (𝛼, 𝑝)}
(232)
𝑑𝑢 Δ𝑠+((𝑠−𝑎)/𝑝) 𝑑𝑢
−∫
)
𝑢
𝑢
3Δ𝑠/2
3Δ𝑠/2
𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
with
ℎ𝑠 (𝑡) [𝐴 (𝑠 + Δ𝑠) − 𝐴 (𝑠)] 𝑑𝑡
= [𝐴 (𝑠)−𝐴 (𝑠+Δ𝑠)] (∫
𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿
,
(𝑠 − 𝑎)𝛼
‖Γ (𝐴) (𝑠 + Δ𝑠) − Γ (𝐴) (𝑠)‖𝑋 ≤
,
𝛼+𝛿
ℎ𝑠+Δ𝑠 (𝑡) [𝐴 (𝑠 + Δ𝑠) − 𝐴 (𝑠)] 𝑑𝑡
𝑠+Δ𝑠+((𝑠−𝑎)/𝑝)
‖Γ (𝐴) (𝑠)‖𝑋 ≤
𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿
+
(𝑠 + Δ𝑠 − 𝑎)𝛼
(𝑠 − 𝑎)𝛼
2𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
(𝑠 − 𝑎)𝛼 (Δ𝑠)𝛿
2𝑝𝛿 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 (Δ𝑠)𝛿
(𝑠 − 𝑎)𝛼+𝛿
,
(237)
Abstract and Applied Analysis
19
which holds for all 𝑠, 𝑠 + Δ𝑠 ∈ 𝐼𝑝 , where Δ𝑇(𝐴) = 𝑇(𝐴)(𝑠 +
Δ𝑠) − 𝑇(𝐴)(𝑠). Thus, by (235) and (237),
𝛿
𝛿
‖𝑇 (𝐴) (𝑠 + Δ𝑠) − 𝑇 (𝐴) (𝑠)‖𝑋 ≤
2𝑝 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 (Δ𝑠)
(𝑠 − 𝑎)𝛼+𝛿
2𝑝𝛿 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿
.
(𝑠 − 𝑎)𝛼
Acknowledgment
(240)
The authors would like to thank the referee for his/her careful
reading and helpful comments that have improved the quality
of the paper.
References
𝛿
‖𝑇‖ = sup ‖𝑇 (𝐴)‖𝛼,𝛿 ≤ 2𝑝 𝑀 (𝛼, 𝛿, 𝑝) .
(241)
‖𝐴‖𝛼,𝛿 =1
Also, 𝑇 is linear. So, 𝑇 ∈ 𝐵(𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋), 𝐻𝛼,𝛿 (𝐼𝑝 , 𝑋)). This
also concludes the proof.
Theorem 25. The operator 𝑇 defined by
𝑏
𝑇 (𝐴) (𝑠) = 𝑝.𝑣. ∫ 𝐾 (𝑠, 𝑡) 𝐴 (𝑡) 𝑑𝑡,
𝑎
(242)
𝐴 ∈ 𝐻𝛼,𝛿 ((𝑎, 𝑏) , 𝑋) , 𝑠 ∈ (𝑎, 𝑏)
is the singular integral operator
𝐵(𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋), 𝐿 𝑞 ((𝑎, 𝑏), 𝑋)) such that
in
‖𝑇‖ ≤ 2𝑝𝛿 𝑀 (𝛼, 𝛿, 𝑝) [𝑀 (𝛼𝑞)]
1/𝑞
the
,
space
(243)
where 1 ≤ 𝑞 < ∞, 𝛼 < 1/𝑞 and 𝐾 : (𝑎, 𝑏) × (𝑎, 𝑏) → R is
given by
𝑠 ∈ 𝐼𝑝 , 𝑡 ∈ (𝑎, 𝑏) , 𝑠 ≠ 𝑡
(244)
𝑠 ∉ 𝐼𝑝 , 𝑡 ∈ (𝑎, 𝑏) .
Proof. Equation (242) may be rewritten as
𝑏
{𝑝.𝑣. ∫ 𝐴 (𝑡) 𝑑𝑡,
𝑇 (𝐴) (𝑠) = {
𝑎 𝑡−𝑠
{0,
𝑠 ∈ 𝐼𝑝
(245)
𝑠 ∉ 𝐼𝑝 .
From (239) and (245),
‖𝑇 (𝐴) (𝑠)‖𝑋 ≤
2𝑝𝛿 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿
(𝑠 − 𝑎)𝛼
(246)
for all 𝑠 ∈ (𝑎, 𝑏). Thus, we have by (246) that
𝑏
𝑞
(248)
(239)
from (238) and (239), we get
1
{
,
𝐾 (𝑠, 𝑡) = { 𝑡 − 𝑠
{0,
1/𝑞
holds. From (248), we obtain ‖𝑇‖ ≤ 2𝑝𝛿 𝑀(𝛼, 𝛿, 𝑝)[𝑀(𝛼𝑞)]1/𝑞
and since 𝑇 is linear, 𝑇 ∈ 𝐵(𝐻𝛼,𝛿 ((𝑎, 𝑏), 𝑋), 𝐿 𝑞 ((𝑎, 𝑏), 𝑋)), and
this completes the proof of theorem.
Since 𝑇(𝐴) ∈ 𝐻𝛼,𝛿 (𝐼𝑝 , 𝑋) and
‖𝑇 (𝐴)‖𝛼,𝛿 ≤ 2𝑝𝛿 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿
‖𝑇 (𝐴)‖𝐿 𝑞 ((𝑎,𝑏),𝑋) ≤ 2𝑝𝛿 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 [𝑀 (𝛼𝑞)]
(238)
and by (234),
‖𝑇 (𝐴) (𝑠)‖𝑋 ≤
So, it is obvious by (247) that the inequality
𝑞
∫ ‖𝑇 (𝐴) (𝑠)‖𝑋 𝑑𝑠 ≤ (2𝑝𝛿 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 ) ∫
𝑎
𝑏
𝑎
𝑑𝑠
(𝑠 − 𝑎)𝛼𝑞
𝑞
= (2𝑝𝛿 𝑀 (𝛼, 𝛿, 𝑝) ‖𝐴‖𝛼,𝛿 ) 𝑀 (𝛼𝑞) .
(247)
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