Research Article | Open Access
Some Bounded Linear Integral Operators and Linear Fredholm Integral Equations in the Spaces and
The spaces and 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 and by taking an arbitrary Banach space instead of , 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 and . Besides, the bounded linear integral operators on the spaces and , some of which are singular, are derived, and the solutions of the linear Fredholm integral equations of the form and are investigated in these spaces by analytical methods.
1. Preliminaries, Background, and Notation
The approximate solution of the singular integral equation was obtained in , where is a real-valued kernel, is a given function, and is the unknown function.
High-order numerical methods for the singular Fredholm integral equations of the form: were developed, where and are given functions, and 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, .
The polar kernel of integral equations was introduced in [3, 4]. This singular kernel is in the following form: where the first term of this kernel is weakly singular and and are bounded on the square and . With , we have the special case of the above kernel: see . 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: was examined in  by numerical methods, where and are given functions and is the unknown function to be determined.
The integral operator defined by was studied in .
An integral equation of the form: is called Fredholm integral equation of the second type. Here, is a given interval, is a function on which is unknown, and is a parameter. The kernel of the equation is a given function on the square and is a given function on .
Now, we may give some required definitions and theorems.
Definition 1 (see [10, page 41]). Let be a Banach space and be a finite measure space. Then is called measurable simple function if there exist and such that , where
Definition 3 (see [12, page 201]). Let be a Banach space with the norm and be a finite measure space. Then is called strongly measurable if there exists a sequence of -valued simple functions defined on such that
Theorem 4 (see [11, page 88]). All continuous functions are strongly measurable.
Theorem 5 (see [13, page 336]). Let be a Banach space with the norm and be a finite measure space. If is strongly measurable, then the scalar function is -measurable.
Definition 6 (see [11, page 88]). Let be a finite measure space and be a Banach space, and then the Bochner integral of a strongly measurable function is the strongly limit of the Bochner integral of an approximating sequence of simple functions satisfying (10). That is,
Theorem 7 (see [12, page 202]). Let be a finite measure space, be a Banach space with the norm , and be a strongly measurable function. If exists, then
Theorem 8 (see [12, page 203]). Let be a finite measure space and be a Banach space with the norm . The Bochner integral exists if and only if is strongly measurable and .
Theorem 9 (see [14, page 82]). Let be a real or complex Banach space with the norm . Let be the real numbers satisfying and be a nonnegative constant. The set consisting of all functions fulfilling the conditions:
for all is a linear space with the algebraic operations:
and is a Banach space with the norm
Again, let be a real or complex Banach space with the norm . Let be the real numbers with and also be a nonnegative constant. Let be an -valued function defined on such that for all with . By , we denote the set of all functions satisfying (18). is a linear space with the algebraic operations: and is a Banach space with the norm:
and are called a Hölder space. The functions spaces which are similar to and were investigated in [15, pages 2419–2426], [16, pages 25–51], and [17, pages 18–33]. The class of the functions satisfying the equalities and was introduced in , where functions are increasing, and is a natural number.
Theorem 10 (see [14, page 16]). Let be a bounded linear operator mapping a Banach space into itself with and denote the identity operator. Then has a bounded inverse operator on which is given by the Neumann series: which satisfies The iterated operators are defined by and . The series in the right of (24) is convergent in the norm on .
2. The Main Results
2.1. The Spaces and
In this section, we determine essentially the spaces and .
Theorem 11. Let be a real or complex Banach space with the norm , be real numbers with , and be a nonnegative constant. Then the set of all functions satisfying the inequalities for all with is a linear space with the usual algebraic operations addition and scalar multiplication defined by and is a Banach space with the norm where and are defined by
Proof. Let be a Banach space with the norm . It is known that the set is a linear space with operations in . Also, it is obvious that . On the other hand, since
is a linear space.
Furthermore, is a normed space with the norm . Indeed, consider the following. (N1) It is clear that for all . (N2) If and , then for all , since So . On the other hand, if , it is found that by (30), (31) and (32). Hence, the proposition “ if and only if ” is true.(N3) Let . Since by (31) and (32), (N4) If , by (31) and by (32). Hence, From the inequality which holds for all real numbers , we obtain .
As a result, is a normed space with the norm .
The space is a Banach space with respect to . To see this we consider an arbitrary Cauchy sequence in , and we show that converges to a function . Since is Cauchy, for every , there exists such that So from (41), where By (42), while or , we have for all . We see by (44) that is Cauchy in . Since is complete, there exists a unique such that The limit depends on the choice of . This defines a function: where Now, we want to show that and . By (48), the continuity of the norm gives together with (44) and (45) that for all with such that or . Since for all by (49) and (50), we derive . Furthermore, since is bounded, there exists the nonnegative constant such that which yields for all . Thus, it is obtained by (51) that for all with and . By (52) and (49), By (53) and (50), for all and . Therefore, we obtain from (54) and (55) that there exists the nonnegative constant such that Hence, . This step completes the proof.
Theorem 12. The inclusion and the inequality hold, where is defined by (16).
Theorem 14. The function is continuous with respect to the Euclidean metric on .
Proof. Let and be usual metric on . Then, we wish to show that the function is continuous. It is clear that Besides, since we have by (65) that and . Thus, since the equalities: hold, there exists the nonnegative constant by (28) that the inequalities: hold from (67) and (68), respectively. By (69), we have Hence, the function is continuous at the arbitrary point which means that it is continuous on .
Theorem 15. Let be a real or complex Banach space with the norm , be real numbers such that , and be a nonnegative constant. The set of all functions satisfying for all with is a linear space with the usual algebraic operations addition and scalar multiplication and is a Banach space with the norm: where .
Theorem 16. The function is continuous with respect to the Euclidean metric on .
Theorem 17. The inclusion and the inequality hold, where is the norm in the space .
Lemma 18. Let be a real or complex Banach algebra with the norm and define by where Then, if then and
Proof. We use the induction method. If , then
Thus, Lemma is true for .
Assume that if then and
Now, let and . Then, we must show that and