# Some Bounded Linear Integral Operators and Linear Fredholm Integral Equations in the SpacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b),X)

Authors: İsmet Özdemir; Ali M. Akhmedov; Ö. Faruk Temizer
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The spacesHα,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 the form f(s)=ϕ(s)+λ∫abA(s,t)f(t)dt,f(s)=ϕ(s)+λ∫abA(t,s)f(t)dt and f(s,t)=ϕ(s,t)+λ∫abA(s,t)f(t,s)dt are investigated in these spaces by analytical methods.


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## Body

## 1. Preliminaries, Background, and Notation

The approximate solution of the singular integral equation(1)u(s)-1πp.v.∫-11‍k(s,t)u(t)t-sdt=f(t)
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:(2)λu(t)-p.v.∫ab‍|t-s|p-1k(t,s)u(s)ds=g(t),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 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:
(3)k(x,y)=g(x,y)(x-y)α+h(x,y),0<α≤1,
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:
(4)k(x,y)=1(x-y)α,0<α≤1,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: (5)μ(x)ϕ(x)+λ(x)p.v.∫-11‍ϕ(y)(y-x)αdy=f(x),|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(6)A:L2(0,T)→L2(0,T),(Af)(x)=∫0T‍k(x-y)f(y)dy
was studied in [9].An integral equation of the form:(7)f(s)=ϕ(s)+λ∫ab‍A(s,t)f(t)dt
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]).
LetX 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=∑i=1n‍xiχEi, where
(8)χEi(w)={1,w∈Ei0,w∉Ei.Definition 2 (see [11, page 88]).
The Bochner integral of a simple functionf given by Definition 1 with respect to μ on E∈Σ is defined by
(9)∫E‍fdμ=∑i=1n‍xiμ(Ei∩E).Definition 3 (see [12, page 201]).
LetX 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
(10)limn∥fn(t)-g(t)∥X=0;t∈S-N,μ(N)=0.Theorem 4 (see [11, page 88]).
All continuous functionsf:S→X are strongly measurable.Theorem 5 (see [13, page 336]).
LetX 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
(11)h:S→ℝ,h(t)=∥g(t)∥X
is Σ-measurable.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,
(12)limn∫E‍fndμ=∫E‍gdμ.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 ∫E‍gdμ exists, then
(13)∥∫E‍gdμ∥X≤∫E‍∥g∥Xdμ.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 ∫E‍gdμ exists if and only if g is strongly measurable and ∫E‍∥g∥Xdμ<∞.Theorem 9 (see [14, page 82]).
LetX 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:
(14)∥A(s,t)∥X≤M,∥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:
(15)(A+B)(s,t)=A(s,t)+B(s,t),(λA)(s,t)=λA(s,t),λ∈K,(K=ℝorK=ℂ),
and C0,α,δ((a,b)×(a,b),X) is a Banach space with the norm
(16)∥A∥0,α,δ=max{sups,t∈(a,b)∥A(s,t)∥X,mA},
where
(17)mA=sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0∥A(s+Δs,t+Δt)-A(s,t)∥X|Δs|α+|Δt|δ.
Again, letX be a real or complex Banach space with the norm ∥·∥X. Let a,b,α be the real numbers with -∞<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
(18)∥A(s)∥X≤M,∥A(s+Δs)-A(s)∥X≤M(Δs)α
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:
(19)(A+B)(s)=A(s)+B(s),(λA)(s)=λA(s),λ∈K,(K=ℝorK=ℂ),
and C0,α((a,b),X) is a Banach space with the norm:
(20)∥A∥0,α=max{{sups+Δs∈(a,b)Δs≠0∥A(s+Δs)-A(s)∥X(Δs)α}sups∈(a,b)∥A(s)∥X,sups+Δs∈(a,b)Δs≠0∥A(s+Δs)-A(s)∥X(Δs)α}.C
0
,
α
,
δ
(
(
a
,
b
)
×
(
a
,
b
)
,
X
) and C0,α((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
(21)k(x,y)=O(|x-y|-mh(|x-y|))
and
(22)k(x′,y)-k(x′′,y)=O(h(py(x′,x′′))e(|x′-x′′|)pym(x′,x′′)e(py(x′-x′′)))
was introduced in [18], where functions h,e:(0,∞)→(0,∞) are increasing,
(23)py(x′,x′′)=min{|y-x′|,|y-x′′|}
and m is a natural number.Theorem 10 (see [14, page 16]).
LetT: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:
(24)(I-T)-1=∑k=0∞‍Tk
which satisfies
(25)∥(I-T)-1∥≤11-∥T∥.
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).


## 2. The Main Results

### 2.1. The SpacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b),X)

In this section, we determine essentially the spacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b),X).Theorem 11.
LetX 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
(26)A:(a,b)×(a,b)→X,(s,t)→A(s,t)
satisfying the inequalities
(27)∥A(s,t)∥X≤M(t-a)α,(28)∥ΔA∥X≤M(|Δs|δ+(Δt)γ)(t-a)α+γ,ΔA=A(s+Δs,t+Δt)-A(s,t)
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
(29)(A+B)(s,t)=A(s,t)+B(s,t),(λA)(s,t)=λA(s,t),λ∈K,(K=ℝorK=ℂ),
and Hα,δ,γ((a,b)×(a,b),X) is a Banach space with the norm
(30)∥A∥α,δ,γ=max{pA,kA},
where pA and kA are defined by
(31)pA=sups,t∈(a,b)(t-a)α∥A(s,t)∥X,(32)kA=sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥ΔA∥X|Δs|δ+(Δt)γ.Proof.
LetX 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
(33)Hα,δ,γ((a,b)×(a,b),X)={A∣A:(a,b)×(a,b)→X,Asatisifes(27)and(28)}⊂𝒜,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).(N2)
IfA∈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
(34)∥A∥α,δ,γ=max{pA,kA}=0.
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)
Letλ∈K,(K=ℝorK=ℂ). Since
(35)p(λA)=sups,t∈(a,b)(t-a)α∥(λA)(s,t)∥X=|λ|pA,k(λA)=sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥Δ(λA)∥X|Δs|δ+(Δt)γ=|λ|kAby (31) and (32),
(36)∥λA∥α,δ,γ=max{p(λA),k(λA)}=|λ|max{pA,kA}=|λ|∥A∥α,δ,γ.(N4)
IfA,B∈Hα,δ,γ((a,b)×(a,b),X),
(37)p(A+B)=sups,t∈(a,b)(t-a)α∥(A+B)(s,t)∥X≤sups,t∈(a,b)(t-a)α∥A(s,t)∥X+sups,t∈(a,b)(t-a)α∥B(s,t)∥X=pA+pB
by (31) and
(38)k(A+B)=sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥Δ(A+B)∥X|Δs|δ+(Δt)γ≤sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥ΔA∥X|Δs|δ+(Δt)γ+sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥ΔB∥X|Δs|δ+(Δt)γ=kA+kB
by (32). Hence,
(39)∥A+B∥α,δ,γ=max{p(A+B),k(A+B)}≤max{pA+pB,kA+kB}.From the inequality
(40)max{pA+pB,kA+kB}≤max{pA,kA}+max{pB,kB}
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 spaceHα,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 (An) in Hα,δ,γ((a,b)×(a,b),X), and we show that (An) converges to a function A∈Hα,δ,γ((a,b)×(a,b),X). Since (An) is Cauchy, for every ε>0, there exists n0(ε)∈ℕ such that
(41)∥An-Am∥α,δ,γ<ε,(m,n>n0(ε)).
So from (41),
(42)sups,t∈(a,b)(t-a)α∥An(s,t)-Am(s,t)∥X<ε,sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥Δ(An-Am)∥X|Δs|δ+(Δt)γ<ε,
where
(43)Δ(An-Am)=(An-Am)(s+Δs,t+Δt)-(An-Am)(s,t).
By (42), while Δs≠0 or Δt≠0, we have
(44)∥(t-a)αAn(s,t)-(t-a)αAm(s,t)∥X<ε,(45)(t-a)α+γ∥Δ(An-Am)∥X|Δs|δ+(Δt)γ<ε,(m,n>n0(ε))
for all s,t,s+Δs,t+Δt∈(a,b),(Δt≥0). We see by (44) that ((t-a)αAn(s,t)) is Cauchy in X. Since X is complete, there exists a unique x∈X such that
(46)limn→∞(t-a)αAn(s,t)=xorlimn→∞An(s,t)=1(t-a)αx.
The limit x∈X depends on the choice of s,t∈(a,b). This defines a function:
(47)A:(a,b)×(a,b)→X,(s,t)→A(s,t),
where
(48)limn→∞An(s,t)=A(s,t).
Now, we want to show that limn→∞∥An-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
(49)(t-a)α∥An(s,t)-A(s,t)∥X=(t-a)α∥An(s,t)-limm→∞Am(s,t)∥X=limm→∞(t-a)α∥An(s,t)-Am(s,t)∥X<ε,(50)(t-a)α+γ∥Δ(An-A)∥X|Δs|δ+(Δt)γ=(t-a)α+γ∥ΔAn-limm→∞ΔAm∥X|Δs|δ+(Δt)γ=limm→∞(t-a)α+γ∥Δ(An-Am)∥X|Δs|δ+(Δt)γ<ε
for all s,t,s+Δs,t+Δt∈(a,b) with Δt≥0 such that Δs≠0 or Δt≠0. Since ∥An-A∥α,δ,γ<ε for all n>n0(ε) by (49) and (50), we derive limn→∞∥An-A∥α,δ,γ=0. Furthermore, since (An) is bounded, there exists the nonnegative constant C such that ∥An∥α,δ,γ≤C which yields
(51)sups,t∈(a,b)(t-a)α∥An(s,t)∥X≤C,sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥ΔAn∥X|Δs|δ+(Δt)γ≤C
for all n∈ℕ. Thus, it is obtained by (51) that
(52)∥An(s,t)∥X≤C(t-a)α,(53)∥An(s+Δs,t+Δt)-An(s,t)∥X≤C(|Δs|δ+(Δt)γ)(t-a)α+γ
for all s,t,s+Δs,t+Δt∈(a,b) with Δt≥0 and n∈ℕ. By (52) and (49),
(54)∥A(s,t)∥X=∥An(s,t)+A(s,t)-An(s,t)∥X≤∥An(s,t)∥X+∥An(s,t)-A(s,t)∥X≤C+ε(t-a)α.
By (53) and (50),
(55)∥ΔA∥X=∥ΔA+ΔAn-ΔAn∥X≤∥ΔAn∥X+∥Δ(An-A)∥X≤(C+ε)(|Δs|δ+(Δt)γ)(t-a)α+γ
for all ε>0 and n>n0(ε). Therefore, we obtain from (54) and (55) that there exists the nonnegative constant C such that
(56)∥A(s,t)∥X≤C(t-a)α,∥A(s+Δs,t+Δt)-A(s,t)∥X≤C(|Δs|δ+(Δt)γ)(t-a)α+γ.
Hence, A∈Hα,δ,γ((a,b)×(a,b),X). This step completes the proof.Theorem 12.
The inclusion(57)C0,δ,γ((a,b)×(a,b),X)⊂Hα,δ,γ((a,b)×(a,b),X)
and the inequality
(58)∥A∥α,δ,γ≤max{∥A∥0,δ,γ(b-a)α,∥A∥0,δ,γ(b-a)α+γ}
hold, where ∥A∥0,δ,γ is defined by (16).Proof.
IfA∈C0,δ,γ((a,b)×(a,b),X), then by (16) and (17) there exists the constant ∥A∥0,δ,γ satisfying the inequalities:
(59)∥A(s,t)∥X≤∥A∥0,δ,γ(b-a)α(t-a)α,∥A(s+Δs,t+Δt)-A(s,t)∥X≤∥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)M1=max{∥A∥0,δ,γ(b-a)α,∥A∥0,δ,γ(b-a)α+γ},
it is obtained by (59) that A∈Hα,δ,γ((a,b)×(a,b),X) and ∥A∥α,δ,γ≤M1. That is,
(61)C0,δ,γ((a,b)×(a,b),X)⊂Hα,δ,γ((a,b)×(a,b),X),
as desired.Example 13.
Let us takeℝ instead of X and define the function A as
(62)A:(a,b)×(a,b)→ℝ,A(s,t)=s+t.
Then, we have
(63)∥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)|≤|Δs|+|Δt|=|Δs|δ|Δs|1-δ+|Δt|γ|Δt|1-γ≤max{(b-a)1-δ,(b-a)1-γ}(|Δs|δ+|Δt|γ)=M2(|Δs|δ+|Δt|γ)
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
(64)∥A∥α,δ,γ≤max{M(b-a)α,M(b-a)α+γ}.Theorem 14.
The functionA∈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
(65)lim(s,t)→(s0,t0)d((s,t),(s0,t0))=lim(s,t)→(s0,t0)(s-s0)2+(t-t0)2=0.
Besides, since
(66)0≤|s-s0|≤d((s,t),(s0,t0)),0≤|t-t0|≤d((s,t),(s0,t0)),
we have by (65) that s→s0 and t→t0. Thus, since the equalities:
(67)∥A(s,t)-A(s0,t0)∥X=∥A(s0+s-s0,t0+t-t0)-A(s0,t0)∥X,(68)∥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
hold, there exists the nonnegative constant M by (28) that the inequalities:
(69)0≤∥A(s,t)-A(s0,t0)∥X≤M(|s-s0|δ+(t-t0)γ)(t0-a)α+γ,(t≥t0)0≤∥A(s,t)-A(s0,t0)∥X≤M(|s0-s|δ+(t0-t)γ)(t-a)α+γ,(t<t0)
hold from (67) and (68), respectively. By (69), we have
(70)lim(s,t)→(s0,t0)∥A(s,t)-A(s0,t0)∥X=0.
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).Theorem 15.
LetX 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
(71)∥A(s)∥X≤M(s-a)α,∥A(s+Δs)-A(s)∥X≤M(Δs)δ(s-a)α+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
(72)(A+B)(s)=A(s)+B(s),(λA)(s)=λA(s),λ∈K,(K=ℝorK=ℂ),
and Hα,δ((a,b),X) is a Banach space with the norm:
(73)∥A∥α,δ=max{{sups,s+Δs∈(a,b)Δs≠0(s-a)α+δ∥ΔA∥X(Δs)δ}sups∈(a,b)(s-a)α∥A(s)∥X,sups,s+Δs∈(a,b)Δs≠0(s-a)α+δ∥ΔA∥X(Δs)δ},
where ΔA=A(s+Δs)-A(s).Theorem 16.
The functionA∈Hα,δ((a,b),X) is continuous with respect to the Euclidean metric on (a,b)⊂ℝ.Theorem 17.
The inclusion(74)C0,δ((a,b),X)⊂Hα,δ((a,b),X)
and the inequality
(75)∥A∥α,δ≤max{∥A∥0,δ(b-a)α,∥A∥0,δ(b-a)α+δ}
hold, where ∥A∥0,δ is the norm in the space C0,δ((a,b),X).Since the proofs of Theorems15–17 are completely similar to that of Theorems 11–14, we leave them to the reader.Lemma 18.
LetX be a real or complex Banach algebra with the norm ∥·∥X and define Bn by
(76)Bn(s,t)=A1(s,t)A2(s,t)⋯An(s,t);s,t∈(a,b),
where
(77)Ai∈Hαi,δi,γ((a,b)×(a,b),X),(i=1,2,…,n),α1+α2+⋯+αn+γ<1forn≥1.
Then, if
(78)δ=min{δ1,δ2,…,δn},(n≥1),D1=1,Dn=max{(b-a)min{δ1,δ2,…,δn-1}-δ,(b-a)δn-δ},(n≥2),(79)Cn=D1D2⋯Dn,
then Bn∈Hα1+α2+⋯+αn,δ,γ((a,b)×(a,b),X) and
(80)∥Bn∥α1+α2+⋯+αn,δ,γ≤2n-1Cn∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.Proof.
We use the induction method. IfA1∈Hα1,δ1,γ((a,b)×(a,b),X), then
(81)δ=min{δ1}=δ1,B1(s,t)=A1(s,t),B1∈Hα1,δ1,γ((a,b)×(a,b),X),∥B1∥α1,δ1,γ=C1∥A1∥α1,δ1,γ.
Thus, Lemma is true for n=1.
Assume that if(82)Ai∈Hαi,δi,γ((a,b)×(a,b),X),(i=1,2,…,k),δ=min{δ1,δ2,…,δk},
then Bk∈Hα1+α2+⋯+αk,δ,γ((a,b)×(a,b),X) and
(83)∥Bk∥α1+α2+⋯+αk,δ,γ≤2k-1Ck∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥Ak∥αk,δk,γ.
Now, letAi∈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(84)∥Bk+1∥α1+α2+⋯+αk+1,δ,γ≤2kCk+1∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥Ak+1∥αk+1,δk+1,γ.
Then,(85)Bk+1(s,t)=A1(s,t)A2(s,t)⋯Ak(s,t)Ak+1(s,t)=Bk(s,t)Ak+1(s,t).(1) Ifb-a≤1, then we have from 0≤|Δs|<b-a≤1 that
(86)|Δs|δk+1,|Δs|min{δ1,δ2,…,δk}≤|Δs|δ;δ=min{δ1,δ2,…,δk+1}.
Furthermore,
(87)∥Bk+1(s,t)∥X=∥Bk(s,t)Ak+1(s,t)∥X≤∥Bk(s,t)∥X∥Ak+1(s,t)∥X≤∥Bk∥α1+α2+⋯+αk,δ,γ∥Ak+1∥αk+1,δk+1,γ(t-a)α1+α2+⋯+αk(t-a)αk+1≤2k-1Ck∥A1∥α1,δ1,γ⋯∥Ak+1∥αk+1,δk+1,γ(t-a)α1+α2+⋯+αk+1
for all s,t∈(a,b). Since Dk=1 for all k∈ℕ, Ck=1. So, we have
(88)∥Bk+1(s,t)∥X≤2kCk+1∥A1∥α1,δ1,γ⋯∥Ak+1∥αk+1,δk+1,γ(t-a)α1+α2+⋯+αk+1
from (87). Additionally,
(89)∥ΔBk+1∥X=∥Bk(s+Δs,t+Δt)Ak+1(s+Δs,t+Δt)-Bk(s,t)Ak+1(s,t)∥X=∥(ΔBk)Ak+1(s+Δs,t+Δt)+Bk(s,t)ΔAk+1∥X≤∥ΔBk∥X∥Ak+1(s+Δs,t+Δt)∥X+∥Bk(s,t)∥X∥ΔAk+1∥X.
By (89) and (83), we get
(90)∥ΔBk+1∥X≤∥Bk∥α1+α2+⋯+αk,δ,γ∥Ak+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∥Ak+1∥αk+1,δk+1,γ(|Δs|δk+1+(Δt)γ)(t-a)α1+α2+⋯+αk(t-a)αk+1+γ≤2k-1Ck∥A1∥α1,δ1,γ⋯∥Ak+1∥αk+1,δk+1,γ(|Δs|δ+(Δt)γ)(t-a)α1+α2+⋯+αk+1+γ+2k-1Ck∥A1∥α1,δ1,γ⋯∥Ak+1∥αk+1,δk+1,γ(|Δs|δk+1+(Δt)γ)(t-a)α1+α2+⋯+αk+1+γ,
where δ=min{δ1,δ2,…,δk}. Thus, we conclude that
(91)∥ΔBk+1∥X≤2kCk+1∥A1∥α1,δ1,γ⋯∥Ak+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}.(2) Ifb-a>1, then 1<Dk+1, and since Ck+1=CkDk+1 for all k∈ℕ, we have 1≤Ck<Ck+1. That is, (88) is obtained from (87). Besides,(i)
if0≤|Δs|<1, then (91) is derived from (86) and (90).(ii)
If1≤|Δs|<b-a, since 0<|Δs|/(b-a)<1, we get
(92)|Δs|δk+1≤(b-a)δk+1-δ|Δs|δ,|Δs|min{δ1,δ2,…,δk}≤(b-a)min{δ1,δ2,…,δk}-δ|Δs|δ,by taking |Δs|/(b-a) instead of |Δs| in (86). Let
(93)Dk+1=max{(b-a)min{δ1,δ2,…,δk}-δ,(b-a)δk+1-δ}.
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
(94)∥Bk+1∥α1+α2+⋯+αk+1,δ,γ≤2kCk+1∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥Ak+1∥αk+1,δk+1,γ.
So, Lemma 18 holds for all n∈ℕ.


### 2.2. Some Bounded Linear Integral Operators on the SpacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b)×(a,b),X)

Hereafter, byCn, we mean the constant defined by (79), and, by the integral of the Banach valued functions, we mean the Bochner integral unless stated otherwise.Theorem 19.
LetX 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(95)T1:C0,α2((a,b),X)→C0,δ1((a,b),X)
defined by
(96)(T1f)(s)=∫ab‍A(s,t)f(t)dt;f∈C0,α2((a,b),X),s∈(a,b)
is bounded with
(97)∥T1∥≤M(α1,γ1)∥A∥α1,δ1,γ1,
that is, T1∈B(C0,α2((a,b),X),C0,δ1((a,b),X)).
(ii) The operator(98)T2:Hα2,δ2((a,b),X)→Hα1,γ1((a,b),X)
defined by
(99)(T2f)(s)=∫ab‍A(t,s)f(t)dt;f∈Hα2,δ2((a,b),X),s∈(a,b)
is bounded with
(100)∥T2∥≤M(α2)∥A∥α1,δ1,γ1,
that is, T2∈B(Hα2,δ2((a,b),X),Hα1,γ1((a,b),X)).
(iii) Suppose thatα1+α2+γ1 and  α1+δ1<1. Then, the operator
(101)T3:Hα2,δ2((a,b),X)→Hα1,δ1((a,b),X)
defined by
(102)(T3f)(s)=∫ab‍A(s,t)f(t)dt;f∈Hα2,δ2((a,b),X),s∈(a,b)
is bounded with
(103)∥T3∥≤M(α1,α2,δ1,γ1)∥A∥α1,δ1,γ1,
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
(104)M(c)=(b-a)1-c1-c,(105)M(d,e)=max{M(d),M(d+e)},(106)M(x,y,z,t)=max{(b-a)xM(x+y),(b-a)x+zM(x+y+t)}
for all the real numbers c,d,e,x,y,z,t such that c≠1.Proof.
(i) We have(107)∥A(s,t)f(t)∥X≤∥A∥α1,δ1,γ1∥f∥0,α2(t-a)α1,∥[A(s+Δs,t)-A(s,t)]f(t)∥X≤∥A∥α1,δ1,γ1∥f∥0,α2(Δs)δ1(t-a)α1+γ1
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
(108)∥(T1f)(s)∥X≤∥A∥α1,δ1,γ1∥f∥0,α2∫ab‍dt(t-a)α1=M(α1)∥A∥α1,δ1,γ1∥f∥0,α2,∥Δ(T1f)∥X≤∥A∥α1,δ1,γ1∥f∥0,α2(Δs)δ1∫ab‍dt(t-a)α1+γ1=M(α1+γ1)∥A∥α1,δ1,γ1∥f∥0,α2(Δs)δ1
such that
(109)Δ(T1f)=(T1f)(s+Δs)-(T1f)(s).
Thus, it is concluded from (108) that T1f∈C0,δ1((a,b),X) and
(110)∥T1f∥δ1≤M(α1,γ1)∥A∥α1,δ1,γ1∥f∥0,α2.
Besides, the norm of T1 holds the inequality
(111)∥T1∥=sup∥f∥0,α2=1∥T1f∥δ1≤M(α1,γ1)∥A∥α1,δ1,γ1
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(112)∥A(t,s)f(t)∥X≤∥A(t,s)∥X∥f(t)∥X≤∥A∥α1,δ1,γ1∥f∥α2,δ2(s-a)α1(t-a)α2,(113)∥[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
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),
(114)∥(T2f)(s)∥X≤∥A∥α1,δ1,γ1∥f∥α2,δ2(s-a)α1∫ab‍dt(t-a)α2=M(α2)∥A∥α1,δ1,γ1∥f∥α2,δ2(s-a)α1
and by (113),
(115)∥Δ(T2f)∥X≤∥A∥α1,δ1,γ1∥f∥α2,δ2(Δs)γ1(s-a)α1+γ1∫ab‍dt(t-a)α2=M(α2)∥A∥α1,δ1,γ1∥f∥α2,δ2(Δs)γ1(s-a)α1+γ1,
where Δ(T2f)=(T2f)(s+Δs)-(T2f)(s). By usage of (114) and (115), we get that T2f∈Hα1,γ1((a,b),X) and
(116)∥T2f∥α1,γ1≤M(α2)∥A∥α1,δ1,γ1∥f∥α2,δ2.
Thus, T2 is bounded with
(117)∥T2∥=sup∥f∥α2,δ2=1∥T2f∥α1,γ1≤M(α2)∥A∥α1,δ1,γ1
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(118)∥A(s,t)f(t)∥X≤∥A(s,t)∥X∥f(t)∥X≤∥A∥α1,δ1,γ1∥f∥α2,δ2(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
for all f∈Hα2,δ2((a,b),X),s,t,s+Δs∈(a,b)  with  Δs≥0.
By (118),
(119)∥(T3f)(s)∥X≤∫ab‍∥A(s,t)f(t)∥Xdt≤∥A∥α1,δ1,γ1∥f∥α2,δ2∫ab‍dt(t-a)α1+α2≤(b-a)α1M(α1+α2)∥A∥α1,δ1,γ1∥f∥α2,δ2(s-a)α1,
and by taking
(120)Δ(T3f)=(T3f)(s+Δs)-(T3f)(s),(121)∥Δ(T3f)∥X≤∫ab‍∥[A(s+Δs,t)-A(s,t)]f(t)∥Xdt≤∥A∥α1,δ1,γ1∥f∥α2,δ2(Δs)δ1∫ab‍dt(t-a)α1+α2+γ1≤(b-a)α1+δ1M(α1+α2+γ1)∥A∥α1,δ1,γ1∥f∥α2,δ2(Δs)δ1(s-a)α1+δ1.
Thus, we have by (119) and (121) that T3f∈Hα1,δ1((a,b),X) and
(122)∥T3f∥α1,δ1≤M(α1,α2,δ1,γ1)∥A∥α1,δ1,γ1∥f∥α2,δ2.
So, by (122), T3 is a bounded linear operator and also
(123)∥T3∥=sup∥f∥α2,δ2=1∥T3f∥α1,δ1≤M(α1,α2,δ1,γ1)∥A∥α1,δ1,γ1.
Therefore, T3∈B(Hα2,δ2((a,b),X),Hα1,δ1((a,b),X)).Hereafter, byM(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.Theorem 20.
LetX 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
(124)(TB)(s1,s2)=∫ab‍A(s1,t)B(t,s2)dt;B∈Hα2,δ2,γ2((a,b)×(a,b),X),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
(125)∥T∥≤N∥A∥α1,δ1,γ1,
where
(126)N=max{M(α1),max{(b-a)γ2M(α1+γ1),M(α1)}}.Proof.
The functionCs1,s2 given by
(127)Cs1,s2(t)=A(s1,t)B(t,s2),t∈(a,b)
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
(128)∥Cs1,s2(t)∥X=∥A(s1,t)B(t,s2)∥X≤∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2(t-a)α1(s2-a)α2;s1,s2,t∈(a,b),
Lebesgue integral ∫ab‍∥Cs1,s2(t)∥Xdt 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
(129)∥(TB)(s1,s2)∥X≤∫ab‍∥Cs1,s2(t)∥Xdt≤N1(s2-a)α2∫ab‍dt(t-a)α1=M(α1)N1(s2-a)α2
from (128). Let define
(130)Dt(s1,s2)=A(s1,t)B(t,s2),ΔDt=Dt(s1+Δs1,s2+Δs2)-Dt(s1,s2),Δ1,tA=A(s1+Δs1,t)-A(s1,t),(131)Δt,2B=B(t,s2+Δs2)-B(t,s2),(132)Δ(TB)=(TB)(s1+Δs1,s2+Δs2)-(TB)(s1,s2),
for all Δs2≥0 and t,s1,s2,s1+Δs1,s2+Δs2∈(a,b). Since
(133)ΔDt=[(Δ1,tA)B(t,s2+Δs2)+A(s1,t)Δt,2B],
we have
(134)∥ΔDt∥X≤(b-a)γ2N1|Δs1|δ1(t-a)α1+γ1(s2-a)α2+γ2+N1(Δs2)γ2(t-a)α1(s2-a)α2+γ2.
So, we obtain ∥Δ(TB)∥X≤∫ab‍∥ΔDt∥Xdt which yields
(135)∥Δ(TB)∥X≤N2∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2(|Δs1|δ1+(Δs2)γ2)(s2-a)α2+γ2,
where N2=max{(b-a)γ2M(α1+γ1),M(α1)}. Since max{M(α1),N2}=N, it is found by (129) and (135) that
(136)∥(TB)(s1,s2)∥X≤N∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2(s2-a)α2,∥Δ(TB)∥X≤N∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2(|Δs1|δ1+(Δs2)γ2)(s2-a)α2+γ2.
Hence, by (136), TB∈Hα2,δ1,γ2((a,b)×(a,b),X) and
(137)∥TB∥α2,δ1,γ2≤N∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2.
That is, T is bounded by (137) and
(138)∥T∥=sup∥B∥α2,δ2,γ2=1∥TB∥α2,δ1,γ2≤N∥A∥α1,δ1,γ1.
Clearly, T is linear. So, T∈B(Hα2,δ2,γ2((a,b),X),Hα2,δ1,γ2((a,b)×(a,b),X)).


### 2.3.  The Solutions of the Linear Fredholm Integral Equations in the SpacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b)×(a,b),X)

We have the following.Theorem 21.
LetX be a real or complex Banach algebra and define the function A by
(139)A(s,t)=A1(s,t)A2(s,t)⋯An(s,t);s,t∈(a,b)
such that
(140)Ai∈Hαi,δi,γ((a,b)×(a,b),X)∀i∈{1,2,…,n},δ=min{δ1,δ2,…,δn},α1+α2+⋯+αn+γ<1withn∈ℕ.
(i) Then, the linear Fredholm integral equation of the form
(141)f(s)=ϕ(s)+λ∫ab‍A(s,t)f(t)dt
has a unique solution f in the space C0,δ((a,b),X) with(142)∥f∥δ≤11-|λ|2n-1CnM(α1+α2+⋯+αn,γ)∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ∥ϕ∥δ,where ϕ∈C0,δ((a,b),X) and λ is a real or complex parameter satisfying the inequality: (143)|λ|<12n-1CnM(α1+α2+⋯+αn,γ)∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.
(ii) The solution of the equation(144)f(s)=ϕ(s)+λ∫ab‍A(t,s)f(t)dt
uniquely exists in the space Hα1+α2+⋯+αn,γ((a,b),X). Besides,
(145)∥f∥α1+α2+⋯+αn,γ≤11-|λ|2n-1CnL∥ϕ∥α1+α2+⋯+αn,γ,
where ϕ∈Hα1+α2+⋯+αn,γ((a,b),X), λ is a real or complex parameter satisfying the inequality:
(146)|λ|<12n-1CnL,L=M(α1+α2+⋯+αn)×∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.
(iii) Suppose that(147)2(α1+α2+⋯+αn)+γandα1+α2+⋯+αn+δ<1.
Then, (141) has a unique solution f in the space Hα1+α2+⋯+αn,δ((a,b),X) such that
(148)∥f∥α1+α2+⋯+αn,δ≤11-|λ|2n-1CnK∥ϕ∥α1+α2+⋯+αn,δ,
where ϕ∈Hα1+α2+⋯+αn,δ((a,b),X),
(149)K=M(α1+α2+⋯+αn,α1+α2+⋯+αn,δ,γ)×∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ,
and λ is a real or complex parameter satisfying
(150)|λ|<12n-1CnK.Proof.
(ii) Equation (144) may be rewritten as
(151)f(s)-λ∫ab‍A(t,s)f(t)dt=ϕ(s)
or
(152)((I-λT)f)(s)=ϕ(s),s∈(a,b)
which yields
(153)(I-λT)f=ϕ,
where T is defined by
(154)(Tf)(s)=∫ab‍A(t,s)f(t)dt,f∈Hα1+α2+⋯+αn,γ((a,b),X),
and I is an identity operator on Hα1+α2+⋯+αn,γ((a,b),X).
By Lemma18, we have that A∈Hα1+α2+⋯+αn,δ,γ((a,b)×(a,b),X) and
(155)∥A∥α1+α2+⋯+αn,δ,γ≤2n-1Cn∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.
Also, we derive by Theorem 19 that T∈B(Hα1+α2+⋯+αn,γ((a,b),X),Hα1+α2+⋯+αn,γ((a,b),X)) and
(156)∥T∥≤∥A∥α1+α2+⋯+αn,δ,γM(α1+⋯+αn)
from (117), and so
(157)∥λT∥≤|λ|2n-1CnM(α1+⋯+αn)×∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.
Since ∥λT∥<1 by (146), the inverse operator
(158)(I-λT)-1:Hα1+α2+⋯+αn,γ((a,b),X)→Hα1+α2+⋯+αn,γ((a,b),X)
exists and the norm of I-λT satisfies the inequality: (159)∥(I-λT)-1∥≤11-|λ|2n-1CnM(α1+α2+⋯+αn)∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γfrom Theorem 10. So, (144) has a unique solution f in the space Hα1+α2+⋯+αn,γ((a,b),X) such that
(160)f=(I-λT)-1ϕ,∥f∥α1+α2+⋯+αn,γ=∥(I-λT)-1ϕ∥α1+α2+⋯+αn,γ≤∥(I-λT)-1∥∥ϕ∥α1+α2+⋯+αn,γ,
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 Lemma18, Theorem 19, (111), Lemma 18, Theorem 19, and (123), respectively.Theorem 22.
LetX be a real or complex Banach algebra. Then Fredholm integral equation of the form
(161)f(s1,s2)=ϕ(s1,s2)+λ∫ab‍A(s1,t)f(t,s2)dt
has a unique solution f in Hα,δ,γ((a,b)×(a,b),X) with
(162)∥f∥α,δ,γ≤11-|λ|N∥A∥α,δ,γ∥ϕ∥α,δ,γ,
where A,ϕ∈Hα,δ,γ((a,b)×(a,b),X), N is the constant given in Theorem 20, that is,
(163)N=max{M(α),max{(b-a)γM(α+γ),M(α)}},
and λ is a real or complex parameter fulfilling
(164)|λ|<1N∥A∥α,δ,γ.Proof.
Equation (161) can be expressed as
(165)f(s1,s2)-λ∫ab‍A(s1,t)f(t,s2)dt=ϕ(s1,s2)
or
(166)((I-λT)f)(s1,s2)=ϕ(s1,s2);s1,s2∈(a,b)
which implies
(167)(I-λT)f=ϕ,
where T is defined by
(168)(Tf)(s1,s2)=∫ab‍A(s1,t)f(t,s2)dt,f∈Hα,δ,γ((a,b)×(a,b),X),
and I is an identity operator on Hα,δ,γ((a,b)×(a,b),X).
By Theorem20, 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
(169)∥λT∥≤|λ|N∥A∥α,δ,γ.
Since ∥λT∥<1 by (164), the inverse operator
(170)(I-λT)-1:Hα,δ,γ((a,b)×(a,b),X)→Hα,δ,γ((a,b)×(a,b),X)
exists and the norm of I-λT satisfies the inequality:
(171)∥(I-λT)-1∥≤11-∥λT∥≤11-|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
(172)f=(I-λT)-1ϕ,∥f∥α,δ,γ=∥(I-λT)-1ϕ∥α,δ,γ≤∥(I-λT)-1∥∥ϕ∥α,δ,γ
which completes the proof.


### 2.4. The Bounded Singular Integral Operators on the SpaceHα,δ((a,b),X)

Lemma 23.
LetA∈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
(173)Γ(A)(s)=p.v.∫ab‍A(t)t-sdt,s∈Ip
exists, and there exists the nonnegative constant M(α,δ,p) depending only on α, δ, and p such that
(174)∥Γ(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(s-a)α,∥Γ(A)(s+Δs)-Γ(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
hold for all s,s+Δs∈(a,b), where ∥A∥α,δ is defined by (73).Hereafter, byIp, we mean the interval Ip=(a,a+p(b-a)/(p+2)) with 1<p<∞.Proof.
We can write(175)Γ(A)(s)=∫as-((s-a)/p)‍hs(t)A(t)dt+∫s-((s-a)/p)s+((s-a)/p)‍hs(t)A(t)dt+∫s+((s-a)/p)b‍hs(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(176)∥hs(t)A(t)∥X≤|hs(t)|∥A(t)∥X≤|hs(t)|∥A∥α,δ(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,
(177)∫as-((s-a)/p)‍dt(t-a)α|t-s|≤ps-a∫as-((s-a)/p)‍dt(t-a)α=ps-a∫0(p-1)(s-a)/p‍u-αdu=pα(p-1)1-α(1-α)(s-a)α.
By (176),
(178)∥∫as-((s-a)/p)‍hs(t)A(t)dt∥X≤∫as-((s-a)/p)‍∥hs(t)A(t)∥Xdt≤pα(p-1)1-α∥A∥α,δ(1-α)(s-a)α.
On the other hand, since(s-a)/p>0, we get ∫s-(s-a)/ps+(s-a)/p‍dt/(t-s)=0. Thus,
(179)∫s-((s-a)/p)s+((s-a)/p)‍hs(t)A(t)dt=∫s-((s-a)/p)s+((s-a)/p)‍hs(t)[A(t)-A(s)]dt=∫s-((s-a)/p)s‍hs(t)[A(t)-A(s)]dt+∫ss+((s-a)/p)‍hs(t)[A(t)-A(s)]dt.
Sincet∈(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
(180)|hs(t)|∥A(t)-A(s)∥X=|hs(t)|∥A(s)-A(t)∥X=-hs(t)∥A(t+s-t)-A(t)∥X≤∥A∥α,δ(s-t)δ-1(t-a)α+δ≤pα+δ∥A∥α,δ(s-t)δ-1(p-1)α+δ(s-a)α+δ.
By (180) and
(181)∫s-((s-a)/p)s‍(s-t)δ-1dt=∫0(s-a)/p‍uδ-1du=(s-a)δδpδ,
we have
(182)∫s-((s-a)/p)s‍|hs(t)|∥A(t)-A(s)∥Xdt≤pα∥A∥α,δ(p-1)α+δδ(s-a)α.
Similarly, since t-s>0 for all t∈(s,s+(s-a)/p), we get
(183)|hs(t)|∥A(t)-A(s)∥X=hs(t)∥A(s+t-s)-A(s)∥X≤∥A∥α,δ(t-s)δ-1(s-a)α+δ
from (28). Also, by
(184)∫ss+((s-a)/p)‍(t-s)δ-1dt=∫0(s-a)/p‍uδ-1du=(s-a)δδpδ
and (183), we derive
(185)∫ss+((s-a)/p)‍|hs(t)|∥A(t)-A(s)∥Xdt≤∥A∥α,δδpδ(s-a)α.
Thus, by taking I=∫s-(s-a)/ps+(s-a)/p‍hs(t)A(t)dt,
(186)∥I∥≤∫s-((s-a)/p)s‍|hs(t)|∥A(t)-A(s)∥Xdt+∫ss+((s-a)/p)‍|hs(t)|∥A(t)-A(s)∥Xdt≤(pα(p-1)α+δδ+1δpδ)∥A∥α,δ(s-a)α
from (182) and (185).
From the inequalityt-a>s+((s-a)/p)-a=(p+1)(s-a)/p which is satisfied for all t∈(s+(s-a)/p,b), we get 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,
(187)∥hs(t)A(t)∥X≤∥A∥α,δ(t-a)α|t-s|≤(p+1)∥A∥α,δ(t-a)α+1.
From
(188)∫s+((s-a)/p)b‍dt(t-a)α+1=pαα(p+1)α(s-a)α-1α(b-a)α≤pαα(p+1)α(s-a)α
and (187), it is obtained that
(189)∥∫s+((s-a)/p)b‍hs(t)A(t)dt∥X≤∫s+((s-a)/p)b‍∥hs(t)A(t)∥Xdt≤(p+1)pα∥A∥α,δα(p+1)α(s-a)α.
By (178), (186), and (189), we have
(190)∥Γ(A)(s)∥X≤∥∫as-((s-a)/p)‍hs(t)A(t)dt∥X+∥∫s-((s-a)/p)s+((s-a)/p)‍hs(t)A(t)dt∥X+∥∫s+((s-a)/p)b‍hs(t)A(t)dt∥X≤M1(α,δ,p)∥A∥α,δ(s-a)α
such that
(191)M1(α,δ,p)=pα(p-1)1-α1-α+pα(p-1)α+δδ+1δpδ+(p+1)pαα(p+1)α.
Furthermore, since(192)1(t-s-Δs)(t-s)=1Δs(1t-s-Δs-1t-s),0<Δs<s-ap,
we can write
(193)Γ(A)(s+Δs)-Γ(A)(s)=∫ab‍hs+Δs(t)A(t)dt-∫ab‍hs(t)A(t)dt=Δs∫as-((s-a)/p)‍hs+Δs(t)hs(t)A(t)dt+∫s-((s-a)/p)s+Δs+((s-a)/p)‍[hs+Δs(t)A(t)-hs(t)A(t)]dt+Δs∫s+Δs+((s-a)/p)b‍hs+Δs(t)hs(t)A(t)dt.
Thus,
(194)∥hs+Δs(t)hs(t)A(t)∥X≤∥A∥α,δ(t-a)α|t-s-Δs||t-s|≤p2∥A∥α,δ(s-a)2(t-a)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
(195)∫as-((s-a)/p)‍dt(t-a)α=∫0((p-1)(s-a))/p‍u-αdu=(p-1)1-α(1-α)p1-α(s-a)α-1,Δs=(Δs)δ(Δs)1-δ≤(Δs)δ(s-a)1-δ
and (194), we get
(196)∥Δs∫as-((s-a)/p)‍hs+Δs(t)hs(t)A(t)dt∥X≤p1+α(p-1)1-α∥A∥α,δ(Δs)δ(1-α)(s-a)α+δ.
Sincet-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,
(197)∥hs+Δs(t)hs(t)A(t)∥X≤∥A∥α,δ(t-a)α(t-s-Δs)(t-s)≤∥A∥α,δ(t-s-Δs)2+α.
By
(198)∫s+Δs+((s-a)/p)b‍dt(t-s-Δs)2+α=∫(s-a)/pb-s-Δs‍u-2-αdu=p1+α(1+α)(s-a)1+α-1(1+α)(b-s-Δs)1+α≤p1+α(1+α)(s-a)1+α
and (195), we obtain (199)∥Δs∫s+Δs+((s-a)/p)b‍hs+Δs(t)hs(t)A(t)dt∥X≤p1+α∥A∥α,δ(Δs)δ(1+α)(s-a)α+δ
from (197). Then,(200)∫s-((s-a)/p)s+Δs+((s-a)/p)‍[hs+Δs(t)A(t)-hs(t)A(t)]dt=∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)[A(t)-A(s+Δs)]dt-∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)[A(t)-A(s)]dt+A(s+Δs)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt-A(s)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt.
Also, we have
(201)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt=∫s-((s-a)/p)s+Δs-((s-a)/p)‍hs+Δs(t)dt+∫s+Δs-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt=∫s-((s-a)/p)s+Δs-((s-a)/p)‍hs+Δs(t)dt,
since ∫s+Δs-(s-a)/ps+Δs+(s-a)/p‍dt/(t-s-Δs)=∫u-(s-a)/pu+(s-a)/p‍dt/(t-u)=0. Therefore, we get by (195) that
(202)|∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt|≤∫s-((s-a)/p)s+Δs-((s-a)/p)‍|hs+Δs(t)|dt=∫s-((s-a)/p)s+Δs-((s-a)/p)‍-hs+Δs(t)dt≤ps-a∫s-((s-a)/p)s+Δs-((s-a)/p)‍dt≤pΔss-a≤p(Δs)δ(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,
(203)∥A(s+Δs)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt∥X≤p∥A∥α,δ(Δs)δ(s+Δs-a)α(s-a)δ≤p∥A∥α,δ(Δs)δ(s-a)α+δ.
Since ∫s-((s-a)/p)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
(204)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt=∫s-((s-a)/p)s+((s-a)/p)‍hs(t)dt+∫s+((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt=∫s+((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt≤ps-a∫s+((s-a)/p)s+Δs+((s-a)/p)‍dt=pΔss-a≤p(Δs)δ(s-a)δ
from (195). Hence,
(205)∥-A(s)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt∥X=∥A(s)∥X∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt≤p∥A∥α,δ(Δs)δ(s-a)α+δ.
By (192),
(206)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)[A(t)-A(s+Δs)]dt-∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)[A(t)-A(s)]dt=Δs∫s-((s-a)/p)s-(Δs/2)‍hs+Δs(t)hs(t)[A(t)-A(s)]dt-∫s-((s-a)/p)s-(Δs/2)‍hs+Δs(t)[A(s+Δs)-A(s)]dt+∫s-(Δs/2)s+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt-∫s-(Δs/2)s+(3Δs/2)‍hs(t)[A(t)-A(s)]dt+Δs∫s+(3Δs/2)s+Δs+((s-a)/p)‍hs+Δs(t)hs(t)[A(t)-A(s+Δs)]dt-∫s+(3Δs/2)s+Δs+((s-a)/p)‍hs(t)[A(s+Δs)-A(s)]dt.
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
(207)∥hs+Δs(t)hs(t)[A(t)-A(s)]∥X=∥hs+Δs(t)hs(t)[A(t+s-t)-A(t)]∥X≤∥A∥α,δ(s-t)δ-2(t-a)α+δ≤pα+δ∥A∥α,δ(s-t)δ-2(p-1)α+δ(s-a)α+δ.
By
(208)∫s-((s-a)/p)s-(Δs/2)‍(s-t)δ-2dt=∫Δs/2(s-a)/p‍uδ-2du=(Δs)δ-1(1-δ)2δ-1-(s-a)δ-1(1-δ)pδ-1≤(Δs)δ-1(1-δ)2δ-1,
and (207),
(209)∥Δs∫s-((s-a)/p)s-(Δs/2)‍hs+Δs(t)hs(t)[A(t)-A(s)]dt∥X≤pα+δ∥A∥α,δ(Δs)δ(p-1)α+δ(1-δ)2δ-1(s-a)α+δ.
Since
(210)∫s-(Δs/2)s+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt=∫s-(Δs/2)s+Δs‍hs+Δs(t)[A(t)-A(s+Δs)]dt+∫s+Δss+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt
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
(211)∥hs+Δs(t)[A(t)-A(s+Δs)]∥X=hs+Δs(t)∥A(t+s+Δs-t)-A(t)∥X≤∥A∥α,δ(s+Δs-t)δ-1(t-a)α+δ≤(2p)α+δ∥A∥α,δ(s+Δs-t)δ-1(2p-1)α+δ(s-a)α+δ.
By
(212)∫s-(Δs/2)s+Δs‍(s+Δs-t)δ-1dt=∫03Δs/2‍uδ-1du=3δ(Δs)δδ2δ
and (211),
(213)∥∫s-(Δs/2)s+Δs‍hs+Δs(t)[A(t)-A(s+Δs)]dt∥X≤(2p)α+δ3δ∥A∥α,δ(Δs)δ(2p-1)α+δδ2δ(s-a)α+δ.
Sincet-s-Δs>0 for all t∈(s+Δs,s+(3Δs/2)), we derive
(214)∥hs+Δs(t)[A(t)-A(s+Δs)]∥X=∥hs+Δs(t)[A(s+Δs+t-s-Δs)-A(s+Δs)]∥X≤∥A∥α,δ(t-s-Δs)δ-1(s+Δs-a)α+δ≤∥A∥α,δ(t-s-Δs)δ-1(s-a)α+δ.
By
(215)∫s+Δss+(3Δs/2)‍(t-s-Δs)δ-1dt=∫0Δs/2‍uδ-1du=(Δs)δδ2δ
and (214),
(216)∥∫s+Δss+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt∥X≤∥A∥α,δ(Δs)δδ2δ(s-a)α+δ.
So, we derive(217)∥∫s-(Δs/2)s+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt∥X≤((2p)α+δ3δ(2p-1)α+δδ2δ+1δ2δ)∥A∥α,δ(Δs)δ(s-a)α+δ
from (213) and (216).
(218)∫s-(Δs/2)s+(3Δs/2)‍hs(t)[A(t)-A(s)]dt=∫s-(Δs/2)s‍hs(t)[A(t)-A(s)]dt+∫ss+(3Δs/2)‍hs(t)[A(t)-A(s)]dt.
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),
(219)∥hs(t)[A(t)-A(s)]∥X=-hs(t)∥A(s)-A(t)∥X=-hs(t)∥A(t+s-t)-A(t)∥X≤∥A∥α,δ(s-t)δ-1(t-a)α+δ≤(2p)α+δ∥A∥α,δ(s-t)δ-1(2p-1)α+δ(s-a)α+δ.
Sincet-s>0 for all t∈(s,s+(3Δs/2)),
(220)∥hs(t)[A(t)-A(s)]∥X=hs(t)∥A(s+t-s)-A(s)∥X≤∥A∥α,δ(t-s)δ-1(s-a)α+δ.
By
(221)∫s-(Δs/2)s‍(s-t)δ-1dt=∫0Δs/2‍uδ-1du=(Δs)δδ2δ,∫ss+(3Δs/2)‍(t-s)δ-1dt=∫03Δs/2‍uδ-1du=3δ(Δs)δδ2δ,
equations (219) and (220), it is obtained that
(222)∥∫s-(Δs/2)s‍hs(t)[A(t)-A(s)]dt∥X≤(2p)α+δ∥A∥α,δ(Δs)δ(2p-1)α+δδ2δ(s-a)α+δ,∥∫ss+(3Δs/2)‍hs(t)[A(t)-A(s)]dt∥X≤3δ∥A∥α,δ(Δs)δδ2δ(s-a)α+δ.
By (222),
(223)∥∫s-(Δs/2)s+(3Δs/2)‍hs(t)[A(t)-A(s)]dt∥X≤((2p)α+δ(2p-1)α+δδ2δ+3δδ2δ)∥A∥α,δ(Δs)δ(s-a)α+δ.
Sincet-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),
(224)∥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≤∥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.
By
(225)∫s+(3Δs/2)s+Δs+((s-a)/p)‍(t-s-Δs)δ-2dt=∫Δs/2(s-a)/p‍uδ-2du=(Δs)δ-1(1-δ)2δ-1-(s-a)δ-1(1-δ)pδ-1≤(Δs)δ-1(1-δ)2δ-1
and (224),
(226)∥Δs∫s+(3Δs/2)s+Δs+((s-a)/p)‍hs+Δs(t)hs(t)[A(t)-A(s+Δs)]dt∥X≤∥A∥α,δ(Δs)δ(1-δ)2δ-1(s-a)α+δ,-∫s-((s-a)/p)s-(Δs/2)‍hs+Δs(t)[A(s+Δs)-A(s)]dt-∫s+(3Δs/2)s+Δs+((s-a)/p)‍hs(t)[A(s+Δs)-A(s)]dt=[A(s)-A(s+Δs)](∫-((s-a)/p)-Δs-3Δs/2‍duu-∫3Δs/2Δs+((s-a)/p)‍duu)=[A(s)-A(s+Δs)](∫3Δs/2Δs+((s-a)/p)‍duu-∫3Δs/2Δs+((s-a)/p)‍duu)=0.
Thus, by taking
(227)M2(α,δ,p)=2p+pα+δ(p-1)α+δ(1-δ)2δ-1+(2p)α+δ3δ(2p-1)α+δδ2δ+1δ2δ+(2p)α+δ(2p-1)α+δδ2δ+3δδ2δ+1(1-δ)2δ-1,
we obtain
(228)∥∫s-((s-a)/p)s+Δs+((s-a)/p)‍[hs+Δs(t)A(t)-hs(t)A(t)]dt∥X≤M2(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
from (203), (205), (209), (217), (223), (226). By (196), (199), and (228), we find
(229)∥Γ(A)(s+Δs)-Γ(A)(s)∥X≤(M2(α,δ,p)+M3(α,p))∥A∥α,δ(Δs)δ(s-a)α+δ
such that
(230)M3(α,p)=p1+α[(1+α)(p-1)1-α+1-α](1-α)(1+α).
By (190) and (229),
(231)∥Γ(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(s-a)α,∥Γ(A)(s+Δs)-Γ(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
with
(232)M(α,δ,p)=max{M1(α,δ,p),M2(α,δ,p)+M3(α,p)}
which terminates the proof.Hereafter, we assume unless stated otherwise thatM(α,δ,p) is defined by (232).Theorem 24.
The operatorT defined by
(233)T(A)(s)=p.v.∫ab‍A(t)t-sdt;A∈Hα,δ((a,b),X),s∈Ip
is the singular integral operator in the space B(Hα,δ((a,b),X),Hα,δ(Ip,X)), and ∥T∥ satisfies the inequality ∥T∥≤2pδM(α,δ,p).Proof.
If0≤Δs<(s-a)/p, we get by Lemma 23 that
(234)∥T(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(s-a)α,(235)∥T(A)(s+Δs)-T(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
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
(236)1s+Δs-a≤1s-a,1(Δs)δ≤pδ(s-a)δ
that
(237)∥ΔT(A)∥X≤∥T(A)(s+Δs)∥X+∥T(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(s+Δs-a)α+M(α,δ,p)∥A∥α,δ(s-a)α≤2M(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α(Δs)δ≤2pδM(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ,
which holds for all s,s+Δs∈Ip, where ΔT(A)=T(A)(s+Δs)-T(A)(s). Thus, by (235) and (237),
(238)∥T(A)(s+Δs)-T(A)(s)∥X≤2pδM(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
and by (234),
(239)∥T(A)(s)∥X≤2pδM(α,δ,p)∥A∥α,δ(s-a)α.
Since T(A)∈Hα,δ(Ip,X) and
(240)∥T(A)∥α,δ≤2pδM(α,δ,p)∥A∥α,δ
from (238) and (239), we get
(241)∥T∥=sup∥A∥α,δ=1∥T(A)∥α,δ≤2pδM(α,δ,p).
Also, T is linear. So, T∈B(Hα,δ((a,b),X),Hα,δ(Ip,X)). This also concludes the proof.Theorem 25.
The operatorT defined by
(242)T(A)(s)=p.v.∫ab‍K(s,t)A(t)dt,A∈Hα,δ((a,b),X),s∈(a,b)
is the singular integral operator in the space B(Hα,δ((a,b),X),Lq((a,b),X)) such that
(243)∥T∥≤2pδM(α,δ,p)[M(αq)]1/q,
where 1≤q<∞,α<1/q and K:(a,b)×(a,b)→ℝ is given by
(244)K(s,t)={1t-s,s∈Ip,t∈(a,b),s≠t0,s∉Ip,t∈(a,b).Proof.
Equation (242) may be rewritten as
(245)T(A)(s)={p.v.∫ab‍A(t)t-sdt,s∈Ip0,s∉Ip.
From (239) and (245),
(246)∥T(A)(s)∥X≤2pδM(α,δ,p)∥A∥α,δ(s-a)α
for all s∈(a,b). Thus, we have by (246) that
(247)∫ab‍∥T(A)(s)∥Xqds≤(2pδM(α,δ,p)∥A∥α,δ)q∫ab‍ds(s-a)αq=(2pδM(α,δ,p)∥A∥α,δ)qM(αq).
So, it is obvious by (247) that the inequality
(248)∥T(A)∥Lq((a,b),X)≤2pδM(α,δ,p)∥A∥α,δ[M(αq)]1/q
holds. From (248), we obtain ∥T∥≤2pδM(α,δ,p)[M(αq)]1/q and since T is linear, T∈B(Hα,δ((a,b),X),Lq((a,b),X)), and this completes the proof of theorem.


## 2.1. The SpacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b),X)

In this section, we determine essentially the spacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b),X).Theorem 11.
LetX 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
(26)A:(a,b)×(a,b)→X,(s,t)→A(s,t)
satisfying the inequalities
(27)∥A(s,t)∥X≤M(t-a)α,(28)∥ΔA∥X≤M(|Δs|δ+(Δt)γ)(t-a)α+γ,ΔA=A(s+Δs,t+Δt)-A(s,t)
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
(29)(A+B)(s,t)=A(s,t)+B(s,t),(λA)(s,t)=λA(s,t),λ∈K,(K=ℝorK=ℂ),
and Hα,δ,γ((a,b)×(a,b),X) is a Banach space with the norm
(30)∥A∥α,δ,γ=max{pA,kA},
where pA and kA are defined by
(31)pA=sups,t∈(a,b)(t-a)α∥A(s,t)∥X,(32)kA=sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥ΔA∥X|Δs|δ+(Δt)γ.Proof.
LetX 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
(33)Hα,δ,γ((a,b)×(a,b),X)={A∣A:(a,b)×(a,b)→X,Asatisifes(27)and(28)}⊂𝒜,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).(N2)
IfA∈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
(34)∥A∥α,δ,γ=max{pA,kA}=0.
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)
Letλ∈K,(K=ℝorK=ℂ). Since
(35)p(λA)=sups,t∈(a,b)(t-a)α∥(λA)(s,t)∥X=|λ|pA,k(λA)=sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥Δ(λA)∥X|Δs|δ+(Δt)γ=|λ|kAby (31) and (32),
(36)∥λA∥α,δ,γ=max{p(λA),k(λA)}=|λ|max{pA,kA}=|λ|∥A∥α,δ,γ.(N4)
IfA,B∈Hα,δ,γ((a,b)×(a,b),X),
(37)p(A+B)=sups,t∈(a,b)(t-a)α∥(A+B)(s,t)∥X≤sups,t∈(a,b)(t-a)α∥A(s,t)∥X+sups,t∈(a,b)(t-a)α∥B(s,t)∥X=pA+pB
by (31) and
(38)k(A+B)=sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥Δ(A+B)∥X|Δs|δ+(Δt)γ≤sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥ΔA∥X|Δs|δ+(Δt)γ+sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥ΔB∥X|Δs|δ+(Δt)γ=kA+kB
by (32). Hence,
(39)∥A+B∥α,δ,γ=max{p(A+B),k(A+B)}≤max{pA+pB,kA+kB}.From the inequality
(40)max{pA+pB,kA+kB}≤max{pA,kA}+max{pB,kB}
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 spaceHα,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 (An) in Hα,δ,γ((a,b)×(a,b),X), and we show that (An) converges to a function A∈Hα,δ,γ((a,b)×(a,b),X). Since (An) is Cauchy, for every ε>0, there exists n0(ε)∈ℕ such that
(41)∥An-Am∥α,δ,γ<ε,(m,n>n0(ε)).
So from (41),
(42)sups,t∈(a,b)(t-a)α∥An(s,t)-Am(s,t)∥X<ε,sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥Δ(An-Am)∥X|Δs|δ+(Δt)γ<ε,
where
(43)Δ(An-Am)=(An-Am)(s+Δs,t+Δt)-(An-Am)(s,t).
By (42), while Δs≠0 or Δt≠0, we have
(44)∥(t-a)αAn(s,t)-(t-a)αAm(s,t)∥X<ε,(45)(t-a)α+γ∥Δ(An-Am)∥X|Δs|δ+(Δt)γ<ε,(m,n>n0(ε))
for all s,t,s+Δs,t+Δt∈(a,b),(Δt≥0). We see by (44) that ((t-a)αAn(s,t)) is Cauchy in X. Since X is complete, there exists a unique x∈X such that
(46)limn→∞(t-a)αAn(s,t)=xorlimn→∞An(s,t)=1(t-a)αx.
The limit x∈X depends on the choice of s,t∈(a,b). This defines a function:
(47)A:(a,b)×(a,b)→X,(s,t)→A(s,t),
where
(48)limn→∞An(s,t)=A(s,t).
Now, we want to show that limn→∞∥An-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
(49)(t-a)α∥An(s,t)-A(s,t)∥X=(t-a)α∥An(s,t)-limm→∞Am(s,t)∥X=limm→∞(t-a)α∥An(s,t)-Am(s,t)∥X<ε,(50)(t-a)α+γ∥Δ(An-A)∥X|Δs|δ+(Δt)γ=(t-a)α+γ∥ΔAn-limm→∞ΔAm∥X|Δs|δ+(Δt)γ=limm→∞(t-a)α+γ∥Δ(An-Am)∥X|Δs|δ+(Δt)γ<ε
for all s,t,s+Δs,t+Δt∈(a,b) with Δt≥0 such that Δs≠0 or Δt≠0. Since ∥An-A∥α,δ,γ<ε for all n>n0(ε) by (49) and (50), we derive limn→∞∥An-A∥α,δ,γ=0. Furthermore, since (An) is bounded, there exists the nonnegative constant C such that ∥An∥α,δ,γ≤C which yields
(51)sups,t∈(a,b)(t-a)α∥An(s,t)∥X≤C,sups,t,s+Δs,t+Δt∈(a,b)Δs≠0∨Δt≠0(t-a)α+γ∥ΔAn∥X|Δs|δ+(Δt)γ≤C
for all n∈ℕ. Thus, it is obtained by (51) that
(52)∥An(s,t)∥X≤C(t-a)α,(53)∥An(s+Δs,t+Δt)-An(s,t)∥X≤C(|Δs|δ+(Δt)γ)(t-a)α+γ
for all s,t,s+Δs,t+Δt∈(a,b) with Δt≥0 and n∈ℕ. By (52) and (49),
(54)∥A(s,t)∥X=∥An(s,t)+A(s,t)-An(s,t)∥X≤∥An(s,t)∥X+∥An(s,t)-A(s,t)∥X≤C+ε(t-a)α.
By (53) and (50),
(55)∥ΔA∥X=∥ΔA+ΔAn-ΔAn∥X≤∥ΔAn∥X+∥Δ(An-A)∥X≤(C+ε)(|Δs|δ+(Δt)γ)(t-a)α+γ
for all ε>0 and n>n0(ε). Therefore, we obtain from (54) and (55) that there exists the nonnegative constant C such that
(56)∥A(s,t)∥X≤C(t-a)α,∥A(s+Δs,t+Δt)-A(s,t)∥X≤C(|Δs|δ+(Δt)γ)(t-a)α+γ.
Hence, A∈Hα,δ,γ((a,b)×(a,b),X). This step completes the proof.Theorem 12.
The inclusion(57)C0,δ,γ((a,b)×(a,b),X)⊂Hα,δ,γ((a,b)×(a,b),X)
and the inequality
(58)∥A∥α,δ,γ≤max{∥A∥0,δ,γ(b-a)α,∥A∥0,δ,γ(b-a)α+γ}
hold, where ∥A∥0,δ,γ is defined by (16).Proof.
IfA∈C0,δ,γ((a,b)×(a,b),X), then by (16) and (17) there exists the constant ∥A∥0,δ,γ satisfying the inequalities:
(59)∥A(s,t)∥X≤∥A∥0,δ,γ(b-a)α(t-a)α,∥A(s+Δs,t+Δt)-A(s,t)∥X≤∥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)M1=max{∥A∥0,δ,γ(b-a)α,∥A∥0,δ,γ(b-a)α+γ},
it is obtained by (59) that A∈Hα,δ,γ((a,b)×(a,b),X) and ∥A∥α,δ,γ≤M1. That is,
(61)C0,δ,γ((a,b)×(a,b),X)⊂Hα,δ,γ((a,b)×(a,b),X),
as desired.Example 13.
Let us takeℝ instead of X and define the function A as
(62)A:(a,b)×(a,b)→ℝ,A(s,t)=s+t.
Then, we have
(63)∥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)|≤|Δs|+|Δt|=|Δs|δ|Δs|1-δ+|Δt|γ|Δt|1-γ≤max{(b-a)1-δ,(b-a)1-γ}(|Δs|δ+|Δt|γ)=M2(|Δs|δ+|Δt|γ)
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
(64)∥A∥α,δ,γ≤max{M(b-a)α,M(b-a)α+γ}.Theorem 14.
The functionA∈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
(65)lim(s,t)→(s0,t0)d((s,t),(s0,t0))=lim(s,t)→(s0,t0)(s-s0)2+(t-t0)2=0.
Besides, since
(66)0≤|s-s0|≤d((s,t),(s0,t0)),0≤|t-t0|≤d((s,t),(s0,t0)),
we have by (65) that s→s0 and t→t0. Thus, since the equalities:
(67)∥A(s,t)-A(s0,t0)∥X=∥A(s0+s-s0,t0+t-t0)-A(s0,t0)∥X,(68)∥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
hold, there exists the nonnegative constant M by (28) that the inequalities:
(69)0≤∥A(s,t)-A(s0,t0)∥X≤M(|s-s0|δ+(t-t0)γ)(t0-a)α+γ,(t≥t0)0≤∥A(s,t)-A(s0,t0)∥X≤M(|s0-s|δ+(t0-t)γ)(t-a)α+γ,(t<t0)
hold from (67) and (68), respectively. By (69), we have
(70)lim(s,t)→(s0,t0)∥A(s,t)-A(s0,t0)∥X=0.
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).Theorem 15.
LetX 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
(71)∥A(s)∥X≤M(s-a)α,∥A(s+Δs)-A(s)∥X≤M(Δs)δ(s-a)α+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
(72)(A+B)(s)=A(s)+B(s),(λA)(s)=λA(s),λ∈K,(K=ℝorK=ℂ),
and Hα,δ((a,b),X) is a Banach space with the norm:
(73)∥A∥α,δ=max{{sups,s+Δs∈(a,b)Δs≠0(s-a)α+δ∥ΔA∥X(Δs)δ}sups∈(a,b)(s-a)α∥A(s)∥X,sups,s+Δs∈(a,b)Δs≠0(s-a)α+δ∥ΔA∥X(Δs)δ},
where ΔA=A(s+Δs)-A(s).Theorem 16.
The functionA∈Hα,δ((a,b),X) is continuous with respect to the Euclidean metric on (a,b)⊂ℝ.Theorem 17.
The inclusion(74)C0,δ((a,b),X)⊂Hα,δ((a,b),X)
and the inequality
(75)∥A∥α,δ≤max{∥A∥0,δ(b-a)α,∥A∥0,δ(b-a)α+δ}
hold, where ∥A∥0,δ is the norm in the space C0,δ((a,b),X).Since the proofs of Theorems15–17 are completely similar to that of Theorems 11–14, we leave them to the reader.Lemma 18.
LetX be a real or complex Banach algebra with the norm ∥·∥X and define Bn by
(76)Bn(s,t)=A1(s,t)A2(s,t)⋯An(s,t);s,t∈(a,b),
where
(77)Ai∈Hαi,δi,γ((a,b)×(a,b),X),(i=1,2,…,n),α1+α2+⋯+αn+γ<1forn≥1.
Then, if
(78)δ=min{δ1,δ2,…,δn},(n≥1),D1=1,Dn=max{(b-a)min{δ1,δ2,…,δn-1}-δ,(b-a)δn-δ},(n≥2),(79)Cn=D1D2⋯Dn,
then Bn∈Hα1+α2+⋯+αn,δ,γ((a,b)×(a,b),X) and
(80)∥Bn∥α1+α2+⋯+αn,δ,γ≤2n-1Cn∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.Proof.
We use the induction method. IfA1∈Hα1,δ1,γ((a,b)×(a,b),X), then
(81)δ=min{δ1}=δ1,B1(s,t)=A1(s,t),B1∈Hα1,δ1,γ((a,b)×(a,b),X),∥B1∥α1,δ1,γ=C1∥A1∥α1,δ1,γ.
Thus, Lemma is true for n=1.
Assume that if(82)Ai∈Hαi,δi,γ((a,b)×(a,b),X),(i=1,2,…,k),δ=min{δ1,δ2,…,δk},
then Bk∈Hα1+α2+⋯+αk,δ,γ((a,b)×(a,b),X) and
(83)∥Bk∥α1+α2+⋯+αk,δ,γ≤2k-1Ck∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥Ak∥αk,δk,γ.
Now, letAi∈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(84)∥Bk+1∥α1+α2+⋯+αk+1,δ,γ≤2kCk+1∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥Ak+1∥αk+1,δk+1,γ.
Then,(85)Bk+1(s,t)=A1(s,t)A2(s,t)⋯Ak(s,t)Ak+1(s,t)=Bk(s,t)Ak+1(s,t).(1) Ifb-a≤1, then we have from 0≤|Δs|<b-a≤1 that
(86)|Δs|δk+1,|Δs|min{δ1,δ2,…,δk}≤|Δs|δ;δ=min{δ1,δ2,…,δk+1}.
Furthermore,
(87)∥Bk+1(s,t)∥X=∥Bk(s,t)Ak+1(s,t)∥X≤∥Bk(s,t)∥X∥Ak+1(s,t)∥X≤∥Bk∥α1+α2+⋯+αk,δ,γ∥Ak+1∥αk+1,δk+1,γ(t-a)α1+α2+⋯+αk(t-a)αk+1≤2k-1Ck∥A1∥α1,δ1,γ⋯∥Ak+1∥αk+1,δk+1,γ(t-a)α1+α2+⋯+αk+1
for all s,t∈(a,b). Since Dk=1 for all k∈ℕ, Ck=1. So, we have
(88)∥Bk+1(s,t)∥X≤2kCk+1∥A1∥α1,δ1,γ⋯∥Ak+1∥αk+1,δk+1,γ(t-a)α1+α2+⋯+αk+1
from (87). Additionally,
(89)∥ΔBk+1∥X=∥Bk(s+Δs,t+Δt)Ak+1(s+Δs,t+Δt)-Bk(s,t)Ak+1(s,t)∥X=∥(ΔBk)Ak+1(s+Δs,t+Δt)+Bk(s,t)ΔAk+1∥X≤∥ΔBk∥X∥Ak+1(s+Δs,t+Δt)∥X+∥Bk(s,t)∥X∥ΔAk+1∥X.
By (89) and (83), we get
(90)∥ΔBk+1∥X≤∥Bk∥α1+α2+⋯+αk,δ,γ∥Ak+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∥Ak+1∥αk+1,δk+1,γ(|Δs|δk+1+(Δt)γ)(t-a)α1+α2+⋯+αk(t-a)αk+1+γ≤2k-1Ck∥A1∥α1,δ1,γ⋯∥Ak+1∥αk+1,δk+1,γ(|Δs|δ+(Δt)γ)(t-a)α1+α2+⋯+αk+1+γ+2k-1Ck∥A1∥α1,δ1,γ⋯∥Ak+1∥αk+1,δk+1,γ(|Δs|δk+1+(Δt)γ)(t-a)α1+α2+⋯+αk+1+γ,
where δ=min{δ1,δ2,…,δk}. Thus, we conclude that
(91)∥ΔBk+1∥X≤2kCk+1∥A1∥α1,δ1,γ⋯∥Ak+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}.(2) Ifb-a>1, then 1<Dk+1, and since Ck+1=CkDk+1 for all k∈ℕ, we have 1≤Ck<Ck+1. That is, (88) is obtained from (87). Besides,(i)
if0≤|Δs|<1, then (91) is derived from (86) and (90).(ii)
If1≤|Δs|<b-a, since 0<|Δs|/(b-a)<1, we get
(92)|Δs|δk+1≤(b-a)δk+1-δ|Δs|δ,|Δs|min{δ1,δ2,…,δk}≤(b-a)min{δ1,δ2,…,δk}-δ|Δs|δ,by taking |Δs|/(b-a) instead of |Δs| in (86). Let
(93)Dk+1=max{(b-a)min{δ1,δ2,…,δk}-δ,(b-a)δk+1-δ}.
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
(94)∥Bk+1∥α1+α2+⋯+αk+1,δ,γ≤2kCk+1∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥Ak+1∥αk+1,δk+1,γ.
So, Lemma 18 holds for all n∈ℕ.


## 2.2. Some Bounded Linear Integral Operators on the SpacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b)×(a,b),X)

Hereafter, byCn, we mean the constant defined by (79), and, by the integral of the Banach valued functions, we mean the Bochner integral unless stated otherwise.Theorem 19.
LetX 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(95)T1:C0,α2((a,b),X)→C0,δ1((a,b),X)
defined by
(96)(T1f)(s)=∫ab‍A(s,t)f(t)dt;f∈C0,α2((a,b),X),s∈(a,b)
is bounded with
(97)∥T1∥≤M(α1,γ1)∥A∥α1,δ1,γ1,
that is, T1∈B(C0,α2((a,b),X),C0,δ1((a,b),X)).
(ii) The operator(98)T2:Hα2,δ2((a,b),X)→Hα1,γ1((a,b),X)
defined by
(99)(T2f)(s)=∫ab‍A(t,s)f(t)dt;f∈Hα2,δ2((a,b),X),s∈(a,b)
is bounded with
(100)∥T2∥≤M(α2)∥A∥α1,δ1,γ1,
that is, T2∈B(Hα2,δ2((a,b),X),Hα1,γ1((a,b),X)).
(iii) Suppose thatα1+α2+γ1 and  α1+δ1<1. Then, the operator
(101)T3:Hα2,δ2((a,b),X)→Hα1,δ1((a,b),X)
defined by
(102)(T3f)(s)=∫ab‍A(s,t)f(t)dt;f∈Hα2,δ2((a,b),X),s∈(a,b)
is bounded with
(103)∥T3∥≤M(α1,α2,δ1,γ1)∥A∥α1,δ1,γ1,
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
(104)M(c)=(b-a)1-c1-c,(105)M(d,e)=max{M(d),M(d+e)},(106)M(x,y,z,t)=max{(b-a)xM(x+y),(b-a)x+zM(x+y+t)}
for all the real numbers c,d,e,x,y,z,t such that c≠1.Proof.
(i) We have(107)∥A(s,t)f(t)∥X≤∥A∥α1,δ1,γ1∥f∥0,α2(t-a)α1,∥[A(s+Δs,t)-A(s,t)]f(t)∥X≤∥A∥α1,δ1,γ1∥f∥0,α2(Δs)δ1(t-a)α1+γ1
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
(108)∥(T1f)(s)∥X≤∥A∥α1,δ1,γ1∥f∥0,α2∫ab‍dt(t-a)α1=M(α1)∥A∥α1,δ1,γ1∥f∥0,α2,∥Δ(T1f)∥X≤∥A∥α1,δ1,γ1∥f∥0,α2(Δs)δ1∫ab‍dt(t-a)α1+γ1=M(α1+γ1)∥A∥α1,δ1,γ1∥f∥0,α2(Δs)δ1
such that
(109)Δ(T1f)=(T1f)(s+Δs)-(T1f)(s).
Thus, it is concluded from (108) that T1f∈C0,δ1((a,b),X) and
(110)∥T1f∥δ1≤M(α1,γ1)∥A∥α1,δ1,γ1∥f∥0,α2.
Besides, the norm of T1 holds the inequality
(111)∥T1∥=sup∥f∥0,α2=1∥T1f∥δ1≤M(α1,γ1)∥A∥α1,δ1,γ1
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(112)∥A(t,s)f(t)∥X≤∥A(t,s)∥X∥f(t)∥X≤∥A∥α1,δ1,γ1∥f∥α2,δ2(s-a)α1(t-a)α2,(113)∥[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
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),
(114)∥(T2f)(s)∥X≤∥A∥α1,δ1,γ1∥f∥α2,δ2(s-a)α1∫ab‍dt(t-a)α2=M(α2)∥A∥α1,δ1,γ1∥f∥α2,δ2(s-a)α1
and by (113),
(115)∥Δ(T2f)∥X≤∥A∥α1,δ1,γ1∥f∥α2,δ2(Δs)γ1(s-a)α1+γ1∫ab‍dt(t-a)α2=M(α2)∥A∥α1,δ1,γ1∥f∥α2,δ2(Δs)γ1(s-a)α1+γ1,
where Δ(T2f)=(T2f)(s+Δs)-(T2f)(s). By usage of (114) and (115), we get that T2f∈Hα1,γ1((a,b),X) and
(116)∥T2f∥α1,γ1≤M(α2)∥A∥α1,δ1,γ1∥f∥α2,δ2.
Thus, T2 is bounded with
(117)∥T2∥=sup∥f∥α2,δ2=1∥T2f∥α1,γ1≤M(α2)∥A∥α1,δ1,γ1
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(118)∥A(s,t)f(t)∥X≤∥A(s,t)∥X∥f(t)∥X≤∥A∥α1,δ1,γ1∥f∥α2,δ2(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
for all f∈Hα2,δ2((a,b),X),s,t,s+Δs∈(a,b)  with  Δs≥0.
By (118),
(119)∥(T3f)(s)∥X≤∫ab‍∥A(s,t)f(t)∥Xdt≤∥A∥α1,δ1,γ1∥f∥α2,δ2∫ab‍dt(t-a)α1+α2≤(b-a)α1M(α1+α2)∥A∥α1,δ1,γ1∥f∥α2,δ2(s-a)α1,
and by taking
(120)Δ(T3f)=(T3f)(s+Δs)-(T3f)(s),(121)∥Δ(T3f)∥X≤∫ab‍∥[A(s+Δs,t)-A(s,t)]f(t)∥Xdt≤∥A∥α1,δ1,γ1∥f∥α2,δ2(Δs)δ1∫ab‍dt(t-a)α1+α2+γ1≤(b-a)α1+δ1M(α1+α2+γ1)∥A∥α1,δ1,γ1∥f∥α2,δ2(Δs)δ1(s-a)α1+δ1.
Thus, we have by (119) and (121) that T3f∈Hα1,δ1((a,b),X) and
(122)∥T3f∥α1,δ1≤M(α1,α2,δ1,γ1)∥A∥α1,δ1,γ1∥f∥α2,δ2.
So, by (122), T3 is a bounded linear operator and also
(123)∥T3∥=sup∥f∥α2,δ2=1∥T3f∥α1,δ1≤M(α1,α2,δ1,γ1)∥A∥α1,δ1,γ1.
Therefore, T3∈B(Hα2,δ2((a,b),X),Hα1,δ1((a,b),X)).Hereafter, byM(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.Theorem 20.
LetX 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
(124)(TB)(s1,s2)=∫ab‍A(s1,t)B(t,s2)dt;B∈Hα2,δ2,γ2((a,b)×(a,b),X),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
(125)∥T∥≤N∥A∥α1,δ1,γ1,
where
(126)N=max{M(α1),max{(b-a)γ2M(α1+γ1),M(α1)}}.Proof.
The functionCs1,s2 given by
(127)Cs1,s2(t)=A(s1,t)B(t,s2),t∈(a,b)
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
(128)∥Cs1,s2(t)∥X=∥A(s1,t)B(t,s2)∥X≤∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2(t-a)α1(s2-a)α2;s1,s2,t∈(a,b),
Lebesgue integral ∫ab‍∥Cs1,s2(t)∥Xdt 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
(129)∥(TB)(s1,s2)∥X≤∫ab‍∥Cs1,s2(t)∥Xdt≤N1(s2-a)α2∫ab‍dt(t-a)α1=M(α1)N1(s2-a)α2
from (128). Let define
(130)Dt(s1,s2)=A(s1,t)B(t,s2),ΔDt=Dt(s1+Δs1,s2+Δs2)-Dt(s1,s2),Δ1,tA=A(s1+Δs1,t)-A(s1,t),(131)Δt,2B=B(t,s2+Δs2)-B(t,s2),(132)Δ(TB)=(TB)(s1+Δs1,s2+Δs2)-(TB)(s1,s2),
for all Δs2≥0 and t,s1,s2,s1+Δs1,s2+Δs2∈(a,b). Since
(133)ΔDt=[(Δ1,tA)B(t,s2+Δs2)+A(s1,t)Δt,2B],
we have
(134)∥ΔDt∥X≤(b-a)γ2N1|Δs1|δ1(t-a)α1+γ1(s2-a)α2+γ2+N1(Δs2)γ2(t-a)α1(s2-a)α2+γ2.
So, we obtain ∥Δ(TB)∥X≤∫ab‍∥ΔDt∥Xdt which yields
(135)∥Δ(TB)∥X≤N2∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2(|Δs1|δ1+(Δs2)γ2)(s2-a)α2+γ2,
where N2=max{(b-a)γ2M(α1+γ1),M(α1)}. Since max{M(α1),N2}=N, it is found by (129) and (135) that
(136)∥(TB)(s1,s2)∥X≤N∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2(s2-a)α2,∥Δ(TB)∥X≤N∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2(|Δs1|δ1+(Δs2)γ2)(s2-a)α2+γ2.
Hence, by (136), TB∈Hα2,δ1,γ2((a,b)×(a,b),X) and
(137)∥TB∥α2,δ1,γ2≤N∥A∥α1,δ1,γ1∥B∥α2,δ2,γ2.
That is, T is bounded by (137) and
(138)∥T∥=sup∥B∥α2,δ2,γ2=1∥TB∥α2,δ1,γ2≤N∥A∥α1,δ1,γ1.
Clearly, T is linear. So, T∈B(Hα2,δ2,γ2((a,b),X),Hα2,δ1,γ2((a,b)×(a,b),X)).


## 2.3.  The Solutions of the Linear Fredholm Integral Equations in the SpacesHα,δ,γ((a,b)×(a,b),X) and Hα,δ((a,b)×(a,b),X)

We have the following.Theorem 21.
LetX be a real or complex Banach algebra and define the function A by
(139)A(s,t)=A1(s,t)A2(s,t)⋯An(s,t);s,t∈(a,b)
such that
(140)Ai∈Hαi,δi,γ((a,b)×(a,b),X)∀i∈{1,2,…,n},δ=min{δ1,δ2,…,δn},α1+α2+⋯+αn+γ<1withn∈ℕ.
(i) Then, the linear Fredholm integral equation of the form
(141)f(s)=ϕ(s)+λ∫ab‍A(s,t)f(t)dt
has a unique solution f in the space C0,δ((a,b),X) with(142)∥f∥δ≤11-|λ|2n-1CnM(α1+α2+⋯+αn,γ)∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ∥ϕ∥δ,where ϕ∈C0,δ((a,b),X) and λ is a real or complex parameter satisfying the inequality: (143)|λ|<12n-1CnM(α1+α2+⋯+αn,γ)∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.
(ii) The solution of the equation(144)f(s)=ϕ(s)+λ∫ab‍A(t,s)f(t)dt
uniquely exists in the space Hα1+α2+⋯+αn,γ((a,b),X). Besides,
(145)∥f∥α1+α2+⋯+αn,γ≤11-|λ|2n-1CnL∥ϕ∥α1+α2+⋯+αn,γ,
where ϕ∈Hα1+α2+⋯+αn,γ((a,b),X), λ is a real or complex parameter satisfying the inequality:
(146)|λ|<12n-1CnL,L=M(α1+α2+⋯+αn)×∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.
(iii) Suppose that(147)2(α1+α2+⋯+αn)+γandα1+α2+⋯+αn+δ<1.
Then, (141) has a unique solution f in the space Hα1+α2+⋯+αn,δ((a,b),X) such that
(148)∥f∥α1+α2+⋯+αn,δ≤11-|λ|2n-1CnK∥ϕ∥α1+α2+⋯+αn,δ,
where ϕ∈Hα1+α2+⋯+αn,δ((a,b),X),
(149)K=M(α1+α2+⋯+αn,α1+α2+⋯+αn,δ,γ)×∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ,
and λ is a real or complex parameter satisfying
(150)|λ|<12n-1CnK.Proof.
(ii) Equation (144) may be rewritten as
(151)f(s)-λ∫ab‍A(t,s)f(t)dt=ϕ(s)
or
(152)((I-λT)f)(s)=ϕ(s),s∈(a,b)
which yields
(153)(I-λT)f=ϕ,
where T is defined by
(154)(Tf)(s)=∫ab‍A(t,s)f(t)dt,f∈Hα1+α2+⋯+αn,γ((a,b),X),
and I is an identity operator on Hα1+α2+⋯+αn,γ((a,b),X).
By Lemma18, we have that A∈Hα1+α2+⋯+αn,δ,γ((a,b)×(a,b),X) and
(155)∥A∥α1+α2+⋯+αn,δ,γ≤2n-1Cn∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.
Also, we derive by Theorem 19 that T∈B(Hα1+α2+⋯+αn,γ((a,b),X),Hα1+α2+⋯+αn,γ((a,b),X)) and
(156)∥T∥≤∥A∥α1+α2+⋯+αn,δ,γM(α1+⋯+αn)
from (117), and so
(157)∥λT∥≤|λ|2n-1CnM(α1+⋯+αn)×∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γ.
Since ∥λT∥<1 by (146), the inverse operator
(158)(I-λT)-1:Hα1+α2+⋯+αn,γ((a,b),X)→Hα1+α2+⋯+αn,γ((a,b),X)
exists and the norm of I-λT satisfies the inequality: (159)∥(I-λT)-1∥≤11-|λ|2n-1CnM(α1+α2+⋯+αn)∥A1∥α1,δ1,γ∥A2∥α2,δ2,γ⋯∥An∥αn,δn,γfrom Theorem 10. So, (144) has a unique solution f in the space Hα1+α2+⋯+αn,γ((a,b),X) such that
(160)f=(I-λT)-1ϕ,∥f∥α1+α2+⋯+αn,γ=∥(I-λT)-1ϕ∥α1+α2+⋯+αn,γ≤∥(I-λT)-1∥∥ϕ∥α1+α2+⋯+αn,γ,
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 Lemma18, Theorem 19, (111), Lemma 18, Theorem 19, and (123), respectively.Theorem 22.
LetX be a real or complex Banach algebra. Then Fredholm integral equation of the form
(161)f(s1,s2)=ϕ(s1,s2)+λ∫ab‍A(s1,t)f(t,s2)dt
has a unique solution f in Hα,δ,γ((a,b)×(a,b),X) with
(162)∥f∥α,δ,γ≤11-|λ|N∥A∥α,δ,γ∥ϕ∥α,δ,γ,
where A,ϕ∈Hα,δ,γ((a,b)×(a,b),X), N is the constant given in Theorem 20, that is,
(163)N=max{M(α),max{(b-a)γM(α+γ),M(α)}},
and λ is a real or complex parameter fulfilling
(164)|λ|<1N∥A∥α,δ,γ.Proof.
Equation (161) can be expressed as
(165)f(s1,s2)-λ∫ab‍A(s1,t)f(t,s2)dt=ϕ(s1,s2)
or
(166)((I-λT)f)(s1,s2)=ϕ(s1,s2);s1,s2∈(a,b)
which implies
(167)(I-λT)f=ϕ,
where T is defined by
(168)(Tf)(s1,s2)=∫ab‍A(s1,t)f(t,s2)dt,f∈Hα,δ,γ((a,b)×(a,b),X),
and I is an identity operator on Hα,δ,γ((a,b)×(a,b),X).
By Theorem20, 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
(169)∥λT∥≤|λ|N∥A∥α,δ,γ.
Since ∥λT∥<1 by (164), the inverse operator
(170)(I-λT)-1:Hα,δ,γ((a,b)×(a,b),X)→Hα,δ,γ((a,b)×(a,b),X)
exists and the norm of I-λT satisfies the inequality:
(171)∥(I-λT)-1∥≤11-∥λT∥≤11-|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
(172)f=(I-λT)-1ϕ,∥f∥α,δ,γ=∥(I-λT)-1ϕ∥α,δ,γ≤∥(I-λT)-1∥∥ϕ∥α,δ,γ
which completes the proof.


## 2.4. The Bounded Singular Integral Operators on the SpaceHα,δ((a,b),X)

Lemma 23.
LetA∈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
(173)Γ(A)(s)=p.v.∫ab‍A(t)t-sdt,s∈Ip
exists, and there exists the nonnegative constant M(α,δ,p) depending only on α, δ, and p such that
(174)∥Γ(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(s-a)α,∥Γ(A)(s+Δs)-Γ(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
hold for all s,s+Δs∈(a,b), where ∥A∥α,δ is defined by (73).Hereafter, byIp, we mean the interval Ip=(a,a+p(b-a)/(p+2)) with 1<p<∞.Proof.
We can write(175)Γ(A)(s)=∫as-((s-a)/p)‍hs(t)A(t)dt+∫s-((s-a)/p)s+((s-a)/p)‍hs(t)A(t)dt+∫s+((s-a)/p)b‍hs(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(176)∥hs(t)A(t)∥X≤|hs(t)|∥A(t)∥X≤|hs(t)|∥A∥α,δ(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,
(177)∫as-((s-a)/p)‍dt(t-a)α|t-s|≤ps-a∫as-((s-a)/p)‍dt(t-a)α=ps-a∫0(p-1)(s-a)/p‍u-αdu=pα(p-1)1-α(1-α)(s-a)α.
By (176),
(178)∥∫as-((s-a)/p)‍hs(t)A(t)dt∥X≤∫as-((s-a)/p)‍∥hs(t)A(t)∥Xdt≤pα(p-1)1-α∥A∥α,δ(1-α)(s-a)α.
On the other hand, since(s-a)/p>0, we get ∫s-(s-a)/ps+(s-a)/p‍dt/(t-s)=0. Thus,
(179)∫s-((s-a)/p)s+((s-a)/p)‍hs(t)A(t)dt=∫s-((s-a)/p)s+((s-a)/p)‍hs(t)[A(t)-A(s)]dt=∫s-((s-a)/p)s‍hs(t)[A(t)-A(s)]dt+∫ss+((s-a)/p)‍hs(t)[A(t)-A(s)]dt.
Sincet∈(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
(180)|hs(t)|∥A(t)-A(s)∥X=|hs(t)|∥A(s)-A(t)∥X=-hs(t)∥A(t+s-t)-A(t)∥X≤∥A∥α,δ(s-t)δ-1(t-a)α+δ≤pα+δ∥A∥α,δ(s-t)δ-1(p-1)α+δ(s-a)α+δ.
By (180) and
(181)∫s-((s-a)/p)s‍(s-t)δ-1dt=∫0(s-a)/p‍uδ-1du=(s-a)δδpδ,
we have
(182)∫s-((s-a)/p)s‍|hs(t)|∥A(t)-A(s)∥Xdt≤pα∥A∥α,δ(p-1)α+δδ(s-a)α.
Similarly, since t-s>0 for all t∈(s,s+(s-a)/p), we get
(183)|hs(t)|∥A(t)-A(s)∥X=hs(t)∥A(s+t-s)-A(s)∥X≤∥A∥α,δ(t-s)δ-1(s-a)α+δ
from (28). Also, by
(184)∫ss+((s-a)/p)‍(t-s)δ-1dt=∫0(s-a)/p‍uδ-1du=(s-a)δδpδ
and (183), we derive
(185)∫ss+((s-a)/p)‍|hs(t)|∥A(t)-A(s)∥Xdt≤∥A∥α,δδpδ(s-a)α.
Thus, by taking I=∫s-(s-a)/ps+(s-a)/p‍hs(t)A(t)dt,
(186)∥I∥≤∫s-((s-a)/p)s‍|hs(t)|∥A(t)-A(s)∥Xdt+∫ss+((s-a)/p)‍|hs(t)|∥A(t)-A(s)∥Xdt≤(pα(p-1)α+δδ+1δpδ)∥A∥α,δ(s-a)α
from (182) and (185).
From the inequalityt-a>s+((s-a)/p)-a=(p+1)(s-a)/p which is satisfied for all t∈(s+(s-a)/p,b), we get 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,
(187)∥hs(t)A(t)∥X≤∥A∥α,δ(t-a)α|t-s|≤(p+1)∥A∥α,δ(t-a)α+1.
From
(188)∫s+((s-a)/p)b‍dt(t-a)α+1=pαα(p+1)α(s-a)α-1α(b-a)α≤pαα(p+1)α(s-a)α
and (187), it is obtained that
(189)∥∫s+((s-a)/p)b‍hs(t)A(t)dt∥X≤∫s+((s-a)/p)b‍∥hs(t)A(t)∥Xdt≤(p+1)pα∥A∥α,δα(p+1)α(s-a)α.
By (178), (186), and (189), we have
(190)∥Γ(A)(s)∥X≤∥∫as-((s-a)/p)‍hs(t)A(t)dt∥X+∥∫s-((s-a)/p)s+((s-a)/p)‍hs(t)A(t)dt∥X+∥∫s+((s-a)/p)b‍hs(t)A(t)dt∥X≤M1(α,δ,p)∥A∥α,δ(s-a)α
such that
(191)M1(α,δ,p)=pα(p-1)1-α1-α+pα(p-1)α+δδ+1δpδ+(p+1)pαα(p+1)α.
Furthermore, since(192)1(t-s-Δs)(t-s)=1Δs(1t-s-Δs-1t-s),0<Δs<s-ap,
we can write
(193)Γ(A)(s+Δs)-Γ(A)(s)=∫ab‍hs+Δs(t)A(t)dt-∫ab‍hs(t)A(t)dt=Δs∫as-((s-a)/p)‍hs+Δs(t)hs(t)A(t)dt+∫s-((s-a)/p)s+Δs+((s-a)/p)‍[hs+Δs(t)A(t)-hs(t)A(t)]dt+Δs∫s+Δs+((s-a)/p)b‍hs+Δs(t)hs(t)A(t)dt.
Thus,
(194)∥hs+Δs(t)hs(t)A(t)∥X≤∥A∥α,δ(t-a)α|t-s-Δs||t-s|≤p2∥A∥α,δ(s-a)2(t-a)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
(195)∫as-((s-a)/p)‍dt(t-a)α=∫0((p-1)(s-a))/p‍u-αdu=(p-1)1-α(1-α)p1-α(s-a)α-1,Δs=(Δs)δ(Δs)1-δ≤(Δs)δ(s-a)1-δ
and (194), we get
(196)∥Δs∫as-((s-a)/p)‍hs+Δs(t)hs(t)A(t)dt∥X≤p1+α(p-1)1-α∥A∥α,δ(Δs)δ(1-α)(s-a)α+δ.
Sincet-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,
(197)∥hs+Δs(t)hs(t)A(t)∥X≤∥A∥α,δ(t-a)α(t-s-Δs)(t-s)≤∥A∥α,δ(t-s-Δs)2+α.
By
(198)∫s+Δs+((s-a)/p)b‍dt(t-s-Δs)2+α=∫(s-a)/pb-s-Δs‍u-2-αdu=p1+α(1+α)(s-a)1+α-1(1+α)(b-s-Δs)1+α≤p1+α(1+α)(s-a)1+α
and (195), we obtain (199)∥Δs∫s+Δs+((s-a)/p)b‍hs+Δs(t)hs(t)A(t)dt∥X≤p1+α∥A∥α,δ(Δs)δ(1+α)(s-a)α+δ
from (197). Then,(200)∫s-((s-a)/p)s+Δs+((s-a)/p)‍[hs+Δs(t)A(t)-hs(t)A(t)]dt=∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)[A(t)-A(s+Δs)]dt-∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)[A(t)-A(s)]dt+A(s+Δs)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt-A(s)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt.
Also, we have
(201)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt=∫s-((s-a)/p)s+Δs-((s-a)/p)‍hs+Δs(t)dt+∫s+Δs-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt=∫s-((s-a)/p)s+Δs-((s-a)/p)‍hs+Δs(t)dt,
since ∫s+Δs-(s-a)/ps+Δs+(s-a)/p‍dt/(t-s-Δs)=∫u-(s-a)/pu+(s-a)/p‍dt/(t-u)=0. Therefore, we get by (195) that
(202)|∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt|≤∫s-((s-a)/p)s+Δs-((s-a)/p)‍|hs+Δs(t)|dt=∫s-((s-a)/p)s+Δs-((s-a)/p)‍-hs+Δs(t)dt≤ps-a∫s-((s-a)/p)s+Δs-((s-a)/p)‍dt≤pΔss-a≤p(Δs)δ(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,
(203)∥A(s+Δs)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)dt∥X≤p∥A∥α,δ(Δs)δ(s+Δs-a)α(s-a)δ≤p∥A∥α,δ(Δs)δ(s-a)α+δ.
Since ∫s-((s-a)/p)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
(204)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt=∫s-((s-a)/p)s+((s-a)/p)‍hs(t)dt+∫s+((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt=∫s+((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt≤ps-a∫s+((s-a)/p)s+Δs+((s-a)/p)‍dt=pΔss-a≤p(Δs)δ(s-a)δ
from (195). Hence,
(205)∥-A(s)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt∥X=∥A(s)∥X∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)dt≤p∥A∥α,δ(Δs)δ(s-a)α+δ.
By (192),
(206)∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs+Δs(t)[A(t)-A(s+Δs)]dt-∫s-((s-a)/p)s+Δs+((s-a)/p)‍hs(t)[A(t)-A(s)]dt=Δs∫s-((s-a)/p)s-(Δs/2)‍hs+Δs(t)hs(t)[A(t)-A(s)]dt-∫s-((s-a)/p)s-(Δs/2)‍hs+Δs(t)[A(s+Δs)-A(s)]dt+∫s-(Δs/2)s+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt-∫s-(Δs/2)s+(3Δs/2)‍hs(t)[A(t)-A(s)]dt+Δs∫s+(3Δs/2)s+Δs+((s-a)/p)‍hs+Δs(t)hs(t)[A(t)-A(s+Δs)]dt-∫s+(3Δs/2)s+Δs+((s-a)/p)‍hs(t)[A(s+Δs)-A(s)]dt.
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
(207)∥hs+Δs(t)hs(t)[A(t)-A(s)]∥X=∥hs+Δs(t)hs(t)[A(t+s-t)-A(t)]∥X≤∥A∥α,δ(s-t)δ-2(t-a)α+δ≤pα+δ∥A∥α,δ(s-t)δ-2(p-1)α+δ(s-a)α+δ.
By
(208)∫s-((s-a)/p)s-(Δs/2)‍(s-t)δ-2dt=∫Δs/2(s-a)/p‍uδ-2du=(Δs)δ-1(1-δ)2δ-1-(s-a)δ-1(1-δ)pδ-1≤(Δs)δ-1(1-δ)2δ-1,
and (207),
(209)∥Δs∫s-((s-a)/p)s-(Δs/2)‍hs+Δs(t)hs(t)[A(t)-A(s)]dt∥X≤pα+δ∥A∥α,δ(Δs)δ(p-1)α+δ(1-δ)2δ-1(s-a)α+δ.
Since
(210)∫s-(Δs/2)s+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt=∫s-(Δs/2)s+Δs‍hs+Δs(t)[A(t)-A(s+Δs)]dt+∫s+Δss+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt
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
(211)∥hs+Δs(t)[A(t)-A(s+Δs)]∥X=hs+Δs(t)∥A(t+s+Δs-t)-A(t)∥X≤∥A∥α,δ(s+Δs-t)δ-1(t-a)α+δ≤(2p)α+δ∥A∥α,δ(s+Δs-t)δ-1(2p-1)α+δ(s-a)α+δ.
By
(212)∫s-(Δs/2)s+Δs‍(s+Δs-t)δ-1dt=∫03Δs/2‍uδ-1du=3δ(Δs)δδ2δ
and (211),
(213)∥∫s-(Δs/2)s+Δs‍hs+Δs(t)[A(t)-A(s+Δs)]dt∥X≤(2p)α+δ3δ∥A∥α,δ(Δs)δ(2p-1)α+δδ2δ(s-a)α+δ.
Sincet-s-Δs>0 for all t∈(s+Δs,s+(3Δs/2)), we derive
(214)∥hs+Δs(t)[A(t)-A(s+Δs)]∥X=∥hs+Δs(t)[A(s+Δs+t-s-Δs)-A(s+Δs)]∥X≤∥A∥α,δ(t-s-Δs)δ-1(s+Δs-a)α+δ≤∥A∥α,δ(t-s-Δs)δ-1(s-a)α+δ.
By
(215)∫s+Δss+(3Δs/2)‍(t-s-Δs)δ-1dt=∫0Δs/2‍uδ-1du=(Δs)δδ2δ
and (214),
(216)∥∫s+Δss+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt∥X≤∥A∥α,δ(Δs)δδ2δ(s-a)α+δ.
So, we derive(217)∥∫s-(Δs/2)s+(3Δs/2)‍hs+Δs(t)[A(t)-A(s+Δs)]dt∥X≤((2p)α+δ3δ(2p-1)α+δδ2δ+1δ2δ)∥A∥α,δ(Δs)δ(s-a)α+δ
from (213) and (216).
(218)∫s-(Δs/2)s+(3Δs/2)‍hs(t)[A(t)-A(s)]dt=∫s-(Δs/2)s‍hs(t)[A(t)-A(s)]dt+∫ss+(3Δs/2)‍hs(t)[A(t)-A(s)]dt.
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),
(219)∥hs(t)[A(t)-A(s)]∥X=-hs(t)∥A(s)-A(t)∥X=-hs(t)∥A(t+s-t)-A(t)∥X≤∥A∥α,δ(s-t)δ-1(t-a)α+δ≤(2p)α+δ∥A∥α,δ(s-t)δ-1(2p-1)α+δ(s-a)α+δ.
Sincet-s>0 for all t∈(s,s+(3Δs/2)),
(220)∥hs(t)[A(t)-A(s)]∥X=hs(t)∥A(s+t-s)-A(s)∥X≤∥A∥α,δ(t-s)δ-1(s-a)α+δ.
By
(221)∫s-(Δs/2)s‍(s-t)δ-1dt=∫0Δs/2‍uδ-1du=(Δs)δδ2δ,∫ss+(3Δs/2)‍(t-s)δ-1dt=∫03Δs/2‍uδ-1du=3δ(Δs)δδ2δ,
equations (219) and (220), it is obtained that
(222)∥∫s-(Δs/2)s‍hs(t)[A(t)-A(s)]dt∥X≤(2p)α+δ∥A∥α,δ(Δs)δ(2p-1)α+δδ2δ(s-a)α+δ,∥∫ss+(3Δs/2)‍hs(t)[A(t)-A(s)]dt∥X≤3δ∥A∥α,δ(Δs)δδ2δ(s-a)α+δ.
By (222),
(223)∥∫s-(Δs/2)s+(3Δs/2)‍hs(t)[A(t)-A(s)]dt∥X≤((2p)α+δ(2p-1)α+δδ2δ+3δδ2δ)∥A∥α,δ(Δs)δ(s-a)α+δ.
Sincet-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),
(224)∥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≤∥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.
By
(225)∫s+(3Δs/2)s+Δs+((s-a)/p)‍(t-s-Δs)δ-2dt=∫Δs/2(s-a)/p‍uδ-2du=(Δs)δ-1(1-δ)2δ-1-(s-a)δ-1(1-δ)pδ-1≤(Δs)δ-1(1-δ)2δ-1
and (224),
(226)∥Δs∫s+(3Δs/2)s+Δs+((s-a)/p)‍hs+Δs(t)hs(t)[A(t)-A(s+Δs)]dt∥X≤∥A∥α,δ(Δs)δ(1-δ)2δ-1(s-a)α+δ,-∫s-((s-a)/p)s-(Δs/2)‍hs+Δs(t)[A(s+Δs)-A(s)]dt-∫s+(3Δs/2)s+Δs+((s-a)/p)‍hs(t)[A(s+Δs)-A(s)]dt=[A(s)-A(s+Δs)](∫-((s-a)/p)-Δs-3Δs/2‍duu-∫3Δs/2Δs+((s-a)/p)‍duu)=[A(s)-A(s+Δs)](∫3Δs/2Δs+((s-a)/p)‍duu-∫3Δs/2Δs+((s-a)/p)‍duu)=0.
Thus, by taking
(227)M2(α,δ,p)=2p+pα+δ(p-1)α+δ(1-δ)2δ-1+(2p)α+δ3δ(2p-1)α+δδ2δ+1δ2δ+(2p)α+δ(2p-1)α+δδ2δ+3δδ2δ+1(1-δ)2δ-1,
we obtain
(228)∥∫s-((s-a)/p)s+Δs+((s-a)/p)‍[hs+Δs(t)A(t)-hs(t)A(t)]dt∥X≤M2(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
from (203), (205), (209), (217), (223), (226). By (196), (199), and (228), we find
(229)∥Γ(A)(s+Δs)-Γ(A)(s)∥X≤(M2(α,δ,p)+M3(α,p))∥A∥α,δ(Δs)δ(s-a)α+δ
such that
(230)M3(α,p)=p1+α[(1+α)(p-1)1-α+1-α](1-α)(1+α).
By (190) and (229),
(231)∥Γ(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(s-a)α,∥Γ(A)(s+Δs)-Γ(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
with
(232)M(α,δ,p)=max{M1(α,δ,p),M2(α,δ,p)+M3(α,p)}
which terminates the proof.Hereafter, we assume unless stated otherwise thatM(α,δ,p) is defined by (232).Theorem 24.
The operatorT defined by
(233)T(A)(s)=p.v.∫ab‍A(t)t-sdt;A∈Hα,δ((a,b),X),s∈Ip
is the singular integral operator in the space B(Hα,δ((a,b),X),Hα,δ(Ip,X)), and ∥T∥ satisfies the inequality ∥T∥≤2pδM(α,δ,p).Proof.
If0≤Δs<(s-a)/p, we get by Lemma 23 that
(234)∥T(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(s-a)α,(235)∥T(A)(s+Δs)-T(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
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
(236)1s+Δs-a≤1s-a,1(Δs)δ≤pδ(s-a)δ
that
(237)∥ΔT(A)∥X≤∥T(A)(s+Δs)∥X+∥T(A)(s)∥X≤M(α,δ,p)∥A∥α,δ(s+Δs-a)α+M(α,δ,p)∥A∥α,δ(s-a)α≤2M(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α(Δs)δ≤2pδM(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ,
which holds for all s,s+Δs∈Ip, where ΔT(A)=T(A)(s+Δs)-T(A)(s). Thus, by (235) and (237),
(238)∥T(A)(s+Δs)-T(A)(s)∥X≤2pδM(α,δ,p)∥A∥α,δ(Δs)δ(s-a)α+δ
and by (234),
(239)∥T(A)(s)∥X≤2pδM(α,δ,p)∥A∥α,δ(s-a)α.
Since T(A)∈Hα,δ(Ip,X) and
(240)∥T(A)∥α,δ≤2pδM(α,δ,p)∥A∥α,δ
from (238) and (239), we get
(241)∥T∥=sup∥A∥α,δ=1∥T(A)∥α,δ≤2pδM(α,δ,p).
Also, T is linear. So, T∈B(Hα,δ((a,b),X),Hα,δ(Ip,X)). This also concludes the proof.Theorem 25.
The operatorT defined by
(242)T(A)(s)=p.v.∫ab‍K(s,t)A(t)dt,A∈Hα,δ((a,b),X),s∈(a,b)
is the singular integral operator in the space B(Hα,δ((a,b),X),Lq((a,b),X)) such that
(243)∥T∥≤2pδM(α,δ,p)[M(αq)]1/q,
where 1≤q<∞,α<1/q and K:(a,b)×(a,b)→ℝ is given by
(244)K(s,t)={1t-s,s∈Ip,t∈(a,b),s≠t0,s∉Ip,t∈(a,b).Proof.
Equation (242) may be rewritten as
(245)T(A)(s)={p.v.∫ab‍A(t)t-sdt,s∈Ip0,s∉Ip.
From (239) and (245),
(246)∥T(A)(s)∥X≤2pδM(α,δ,p)∥A∥α,δ(s-a)α
for all s∈(a,b). Thus, we have by (246) that
(247)∫ab‍∥T(A)(s)∥Xqds≤(2pδM(α,δ,p)∥A∥α,δ)q∫ab‍ds(s-a)αq=(2pδM(α,δ,p)∥A∥α,δ)qM(αq).
So, it is obvious by (247) that the inequality
(248)∥T(A)∥Lq((a,b),X)≤2pδM(α,δ,p)∥A∥α,δ[M(αq)]1/q
holds. From (248), we obtain ∥T∥≤2pδM(α,δ,p)[M(αq)]1/q and since T is linear, T∈B(Hα,δ((a,b),X),Lq((a,b),X)), and this completes the proof of theorem.



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*Source: 607204-2013-03-26.xml*