#### Abstract

The Arzela-Ascoli theorem for demi-linear mappings is established. Some conclusions about preserving totally bounded sets are also obtained.

#### 1. Introduction

In 2009, Li et al. introduced the definition of demi-linear mappings which is a meaningful extension of linear operators [1]. They also established a new equicontinuity theorem and a uniform boundedness principle for demi-linear mappings. After that, many related theories (such as demi-linear duality, demi-distributions, and matrix transformations) are obtained for demi-linear mappings which can be found in [24].

Arzela-Ascoli theorem has a wide range of applications in many fields of mathematics. Many modern formulations of Arzela-Ascoli theorem have been obtained [510].

In this paper, we will give some results on Arzela-Ascoli theorem for the space of demi-linear mappings.

#### 2. Preliminary

Let , be topological vector spaces over the scalar field and the family of neighborhoods of . Let

Definition 1 ([1, Definition ]). A mapping is said to be demi-linear if and there exist a and a such that , , and yield for which , , and .

We denote by the family of demi-linear mappings related to and and by the subfamily of satisfying the following property: if , , and , then for some with .

Note that, in general, is a large extension of . For instance, if is a norm, then for every .

Definition 2. A family is said to be equicontinuous at if for every , there is a such that for all and is equicontinuous on or, simply, equicontinuous if is equicontinuous at each .

By the proof of [1, Theorem 3.1], the following conclusion can be easily obtained.

Theorem 3. Let . Then, is equicontinuous at 0 if and only If is equicontinuous on .

By the proof of [1, Theorem 3.2] and Definition 2, we also have the following result.

Theorem 4. Let and for every we have . Then, . If, in addition, is equicontinuous, then is continuous.

Definition 5. A family is said to be uniformly continuous if for any neighborhood of zero in , there exists a neighborhood of zero in such that implies .

Definition 6. Let be a nonempty set and a topological vector space, . A topology on is said to be -topology if there is an such that and is just the topology of uniform convergence on each ; that is, in if and only if for every , uniformly with respect to .

Let be the weakest -topology; that is, in if and only if , for all .

Let , and be as in Definition 6. For , define by . Then, and by Definition 6 we have the -topology on . Also, we denote the weakest -topology by , namely, the topology of pointwise convergence. That is,

#### 3. Main Results

Throughout this section, let and be topological vector spaces on . Denote by the family of demi-linear mappings related to and . For , there is a balanced closed such that . Then, . Without generality, we will assume that is balanced closed in .

Recall that a set in a topological vector space is called totally bounded if for each neighborhood of zero there is a finite set with .

Theorem 7. Let . If is continuous at zero, then preserves totally bounded sets.

Proof. Suppose that is a totally bounded subset of . Since is a topological vector space, for every balanced closed neighborhood , there exists a balanced closed neighborhood such that . For , there is a with such that by the continuity of at zero. For , there exists a finite subset of such that ; that is, for every there exists an with and such that . Then, where and . Let and be the balanced hull of . Obviously, the finite set is totally bounded and hence is totally bounded. From it follows that is totally bounded and hence there exists a finite subset of such that Then, from (3), (4), and (6), we have for all . Thus, is totally bounded.

Recall that a mapping with uniform continuity is also equicontinuous. The following theorem shows that the converse holds true for mappings in .

Theorem 8. Let . Then, is uniformly continuous if and only if is equicontinuous at zero.

Proof. We only need to prove the “if” part. For every , there exists a balanced closed with . For , since is equicontinuous at zero, there is a such that for all . For arbitrary with , we have where and . It follows that for all , which shows that is uniformly continuous.

Theorem 9 (Ascoli Theorem). Let be topological vector spaces, be Hausdorff, and be the family of all compact subsets of . Let be the topology on of uniform convergence on each . If is equicontinuous and is compact, then is equicontinuous and compact.

Proof. Since is equicontinuous, for every balanced closed , there exists a balanced closed such that for all . For every , there is a net such that . Since the singleton is compact, that is, for all , we have for all . Then, by the balanced closed property of , we have that is, . As runs over the set , we have that is equicontinuous at zero and hence is equicontinuous on by Theorem 3.
Let be a net in . Since the topology is weaker than , we have and hence is also a net in . Since is compact, the net has a convergent subnet of , which we also denote by . Let be the limit of in ; that is, Then, we have and is continuous by Theorem 4.
To prove is compact, we only need to prove . Assume this is not true; then, there exists a such that is not uniform for . Then, there are , increasing sequence , and such that For , there exists a balanced such that . Since is compact, the sequence has a convergent subnet with . For , there exists a such that by the equicontinuity of . Then, For , there exists a such that , for all . Then, Since is continuous, there is a such that Since , there exists an such that Then, (16) and (18) imply that It follows from (17) that This contracts (13) and hence , which yields the desired result.

In fact, the converse of Theorem 9 is also true. The following theorem gives the Arzela Theorem for demi-linear mappings.

Theorem 10 (Arzela Theorem). Let be topological vector spaces, be Hausdorff, and be the family of all compact subsets of . Let be the topology on of uniform convergence on each . If is a family of continuous mappings in and is compact, then(I) is equicontinuous;(II) is compact.

Proof. For conclusion (I), we just need to prove that is equicontinuous at zero. Suppose that is not equicontinuous at zero; that is, there exists a such that for every there are and such that Since is directed by the relation “,” we have that and are nets, respectively, in and . Since and is compact, the net has a convergent subnet, which we also denote by . Let be the limit of in ; that is, for each we have uniformly for . Obviously, and hence is compact; that is, . Therefore, we have uniformly for . For , there exists a such that Then for , there exists a such that It follows that By (22) and the continuity of , for , there is a such that From (23), (25), and (26), we have which is in contradiction with (21) and hence the desired result holds.
For conclusion (II), we will prove that . And we just need to show that . For each , there exists a net in such that , for all . We only need to prove that . If this is not true, then there is a such that is not uniform for . Then, there are , increasing sequence , and such that Since is compact, there exist an and a convergent subnet of such that . By the equicontinuity of , the continuity of , and (28), we have However, which leads a contradiction to (29) and hence the desired result holds.

Theorem 11. Let be topological vector spaces and be of second category. If is a family of continuous mappings in and for every , the set is totally bounded, then for every totally bounded subset of , we have that is totally bounded.

Proof. Since totally bounded set is bounded, it follows that is equicontinuous on by the equicontinuity theorem [1, Theorem 3.1]. For every balanced closed , there exists a balanced closed such that . For , there is a with such that For , there exist such that ; that is, for every , there exists an with and such that . Then, where , , and . Let . Since for every , the set is totally bounded, we see that is totally bounded. Let be the balanced hull of ; that is, . Then, is totally bounded. It follows from that is totally bounded. Hence, there exist such that Then, (31), (32), and (34) imply that for all and . Thus, is totally bounded.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work is supported by National Natural Science Foundation of China (Project no. 11101108), China Postdoctoral Science Foundation (Project no. 2011M500646), and Natural Scientific Research Innovation Foundation in Harbin Institute of Technology (Project no. 2011008).