#### Abstract

Wardowski (2012) introduced a new type of contractive mapping and proved a fixed point result in complete metric spaces as a generalization of Banach contraction principle. In this paper, we introduce a notion of generalized *F*-contraction mappings which is used to prove a fixed point result for generalized nonexpansive mappings on star-shaped subsets of normed linear spaces. Some theorems on invariant approximations in normed linear spaces are also deduced. Our results extend, unify, and generalize comparable results in the literature.

#### 1. Introduction and Preliminaries

One of the most basic and important results in metric fixed point theory is the Banach contraction principle due to Banach [1]. It states that if is a complete metric space and satisfies for all , with , then has a unique fixed point. This theorem that has been extended in many directions (see, e.g., [2–6]) has many applications in mathematics and other related disciplines as well (see, e.g., [7–9]).

Meinardus [10] and Brosowski [11] employed fixed point theory to obtain invariant approximation results in normed linear spaces. A number of authors generalized their results (see [12–18] and the references therein). On the other hand, Dotson [19] extended Banach's contraction principle for nonexpansive mappings on star-shaped subsets of Banach spaces and proved Brosowski-Meinardus type theorems on invariant approximations. L. A. Khan and A. R. Khan [20] generalized Dotson's results on star shaped subsets of p-normed spaces.

Recently, Wardowski [21] introduced a new contractive mapping called an -contraction and proved some fixed point theorems in complete metric spaces. In this paper we introduce a notion of generalized -contraction mappings which is used to prove a fixed point result for generalized nonexpansive mappings on star shaped subsets of normed linear spaces. Some theorems on invariant approximations in normed linear spaces are deduced. Our results extend, unify, and generalize comparable results in [10, 11, 19, 20]. Some illustrative examples are also presented.

Next we give some definitions which will be used in the sequel. The letters , will denote the set of positive real numbers and the set of real numbers, respectively.

Let be the collection of all mappings which satisfy the following conditions:(C1) is strictly increasing, that is, for all such that ;(C2)for every sequence of positive numbers if and only if (C3)there exists a such that

*Definition 1 (see [21]). *Let be a metric space and . A mapping is said to be an -contraction on if there exists a such that
for all .

*Remark 2 (see [21]). *Every -contraction mapping is continuous.

Motivated by the work of Wardowski [21] and by Theorem 4 of [22], we give the following definition.

*Definition 3. *Let be a metric space and . A mapping is said to be a generalized -contraction if there exists a such that
for all .

*Definition 4. *Let be a metric space and . A mapping is said to be -nonexpansive if
for all .

*Remark 5. *It follows from condition (C1) that if and is an -nonexpansive self-mapping of a metric space , then is nonexpansive (recall that is nonexpansive provided that for all ). Conversely, it is clear, by (C1), that if is a nonexpansive self-mapping of a metric space , then is -nonexpansive for all .

By considering different choices of mappings in (2), (3), and (4), we obtain a variety of contractions. For details we refer to [21] and the following examples.

*Example 6. *Let be a metric space, , and let be given by , where . It is clear that . Now, if is a generalized -contraction, then it is a generalized -contraction because for any with , we have
Similarly, if is an -contraction, then it is a -contraction. Furthermore, if is -nonexpansive then
whenever , which shows that is -nonexpansive. Finally, note that taking in (6), we deduce that is nonexpansive.

*Example 7. *Let be a metric space, let be given by , and let be a generalized -contraction. Since , then (3) becomes
whenever , which implies
and thus . Hence our definition is more general than those given in [18, 20].

If we take , it is clear that , and then (2) becomes whenever , which implies that that is,

*Definition 8. *Let be a closed subset of metric space . Then is called compact if for every bounded subset of , is compact in .

*Definition 9. *If is a mapping with , then is called an -invariant subset of .

*Definition 10. *Let be a subset of metric space . As usual, for any , we define
is called the set of best approximations of from . If for each , is nonempty, then is called proximal. Observe that if is closed, then is also closed.

*Definition 11. *Let be a linear space over . A subset of is called star-shaped if there exists at least one point such that for all and . In this case is called a star centre of .

Let be a metric space, a closed subset of , and a self-mapping. For each , the set is called the orbit of (compare [23]). The mapping is called orbitally continuous at if implies and is orbitally continuous on a set if is orbitally continuous for all .

#### 2. Main Results

In the following a normed linear space will be simply denoted by if no confusion arises. Furthermore, by a complete subset of a normed linear space we will mean a subset of such that the restriction to of the metric induced on by its norm is complete. Of course, every complete subset of a normed linear space is closed, and every closed subset of a Banach space is complete.

Our main result (Theorem 13 below) will be proved with the help of the following re-formulation of Theorem 4 of [22].

Theorem 12 (see [22]). *Let be a complete metric space, , and an orbitally continuous generalized -contraction. Then has a fixed point.*

Theorem 13. *Let be a normed linear space, a complete and star-shaped subset of, and . If is an -nonexpansive mapping and is compact, then has a fixed point.*

*Proof. *We first note that, by Remark 5, is nonexpansive on , so it is continuous on .

Now let be a star centre of . For each , define by
for all , where and . From the fact that is continuous on it immediately follows that each is continuous on .

For any fixed and any , we have

Since is strictly increasing, with for each , and is -nonexpansive, we have
Hence
This implies that there exists , such that
Therefore,
Hence, is a generalized -contraction for each . By Theorem 12, for each there exists such that . Since is compact, there exist a subsequence of the sequence , and an such that
In fact because and is closed.

Since and for all , we deduce that
Therefore,
We conclude that .

*Remark 14. *Note that, by Remark 5, we can restate the preceding theorem as follows. Let be a normed linear space (Banach space, resp.) and a complete (closed, resp.) and star-shaped subset of. If is a nonexpansive mapping and is compact, then has a fixed point.

The following is an example where we can apply Theorem 13 to every .

*Example 15. *Let be the linear space of all summable sequences of real numbers. Then, the pair is a (classical) Banach space, where is the norm on such that, for each ,

In the following any element of will be also denoted as .

Let be the closed unit ball of ; that is,
It is well known that is a (noncompact) closed subset of . Moreover is star-shaped with a star center of .

Now let be constant, and define as
for all . Clearly is nonexpansive on , and hence it is -nonexpansive for any , by Remark 5.

Furthermore, is homeomorphic to the compact subset of ,
so that is compact. Hence and thus is compact. We have shown that all conditions of Theorem 13 (compare Remark 14) are satisfied. Thus has a fixed point. In fact, the fixed points of are the elements of such that whenever .

Now we give an example of a discontinuous self-mapping on a compact metric space which is a generalized -contractionbut not an -contraction. So the class of generalized -contraction mappings is bigger than the class of -contraction.

*Example 16. *Let , and let be endowed with usual metric. Define a mapping as if , and if .

Let be defined by for . Note that
is satisfied for all whenever . Hence, is a generalized -contraction. Clearly is not continuous at and at . Hence, is not an -contraction (see Remark 2.1 in [21] ) or let and ; then
while
and thus
Hence, there does not exist any , such that, for and ,
is satisfied. Now let ; then , and . Hence, the orbit of is the set which is compact and
that is is orbitally continuous at . Hence, by Theorem 12, has a fixed point (in fact is the only fixed point of ).

In the last part of the paper we discuss nonemptiness and existence of fixed points for the set of best approximations of closed subsets of metric spaces and of normed spaces, respectively.

Theorem 17. *Let be a metric space. Let be a continuous mapping and let be -nonexpansive with a fixed point . If is a closed -invariant subset of such that is compact on , then the set of best approximations is nonempty.*

*Proof. *Let . Then, there is a sequence in such that . Since is a bounded subset of and is compact on , the set is a compact subset of , and so there exist a subsequence of and an such that . Now,
This implies
Since is strictly increasing, we get . Hence is nonempty.

As an application of Theorems 13 and 17, we deduce the following.

Theorem 18. *Let be a normed linear space. Let be a continuous mapping and let be -nonexpansive with a fixed point . If is a complete -invariant subset of such that is compact on, and is a star-shaped set, then has a fixed point in .*

*Proof. *By Theorem 17, is nonempty. We show that is -invariant. To this end, let and set ; then
This implies
Since is strictly increasing, we get . So . This proves that is -invariant, so is -nonexpansive. Now observe that if is complete, then is also complete. Hence, is star-shaped and complete, and is compact, so, by Theorem 13, there exists such that .

*Remark 19. *As in the case of Theorem 13 (see Remark 14), the preceding theorem can be restated as follows. Let be a normed linear space (Banach space, resp.) and let be nonexpansive with a fixed point . If is a complete (closed, resp.) -invariant subset of such that is compact on and is a star-shaped set, then has a fixed point in .

We conclude the paper illustrating Theorem 18 with the following example.

*Example 20. *Let be the Banach space of Example 15. Define as
for all , with . Let be continuous. Since is nonexpansive, it follows from Remark 5 that it is -nonexpansive. Of course, has fixed points. In fact

As in Example 15, let be the closed unit ball of . We know that is a closed and thus complete, -invariant subset of such that is compact on .

Then, if we choose such that , we deduce that if , and if . Therefore, is trivially star-shaped. Thus, all conditions of Theorem 18 (compare Remark 19) have been verified.

#### Conflict of Interests

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

#### Acknowledgments

The authors are very grateful to the referees for their valuable comments and suggestions, and, in particular, to one of them for calling our attention on the crucial fact stated in the first part of Remark 5 and for the elegant reformulation of Theorem 13 stated in Remark 14. Salvador Romaguera acknowledges the support of the Universitat Politècnica de València, Grant PAID-06-12-SP20120471.