#### Abstract

Given a vector space , we investigate the solutions of the linear functional equation of third order , which is strongly associated with a well-known identity for the Fibonacci numbers. Moreover, we prove the Hyers-Ulam stability of that equation.

#### 1. Introduction

The problem of stability of functional equations was motivated by a question of Ulam  and a solution to it by Hyers . Since then, numerous papers have been published on that subject and we refer to  for more details, some discussions, and further references; for examples of very recent results, see, for example, .

In this paper, as usual, , , , and stand for the sets of complex numbers, real numbers, integers, and positive integers, respectively. For a nonempty subset of a vector space, let be a function. Moreover, , , and (only for bijective ) for and .

Jung has proved in  (see also ) some results on solutions and Hyers-Ulam stability of the functional equation in the case where and for .

If and , then solutions of the difference equation are called the Lucas sequences (see, e.g., ). In some special cases they are called with specific names, for example, the Fibonacci numbers (, , , and ), the Lucas numbers (, , , and ), the Pell numbers (, , , and ), the Pell-Lucas (or companion Lucas) numbers (, , , and ), and the Jacobsthal numbers (, , , and ).

For some information and further references concerning the functional equations in a single variable, we refer to . Let us mention yet that the problem of Hyers-Ulam stability of functional equations is connected to the notions of controlled chaos and shadowing (see ).

We remark that if is bijective, then (1) can be written in the following equivalent form: where .

In view of the last remark, the following Hyers-Ulam stability result concerning (1) can be derived from [14, Theorem 2] (see also ).

Theorem 1. Let be given with and let be a nonempty subset of a vector space. Assume that , are the complex roots of the quadratic equation with for . Moreover, assume that is either a real vector space if or a complex vector space if . Let be bijective. If a function satisfies the inequality for all and for some , then there exists a unique solution of (1) with for all .

In [16, Theorem 1.4], the method presented in  was modified so as to prove a theorem which is a complement of Theorem 1. Note that, for bijective , the following theorem improves the estimation (4) in some cases (e.g., , , or , ). However, in some other situations (e.g., , ), the estimation (4) is better than (5). The following theorem also complements Theorem 1, because can be quite arbitrary in the case of .

Theorem 2. Given with , assume that the distinct complex roots , of the quadratic equation satisfy one of the following two conditions: for ; for and is bijective.Moreover, assume that is either a real vector space if or a complex vector space if . If a function satisfies inequality (3), then there exists a solution of (1) such that for all . Moreover, if the condition is true, then the is the unique solution of (1) satisfying (5).

In this paper, we investigate the solutions of the functional equation where , , are real constants. Moreover, we also prove the Hyers-Ulam stability of that equation. Equation (6) is a kind of linear functional equations of third order because it is of the form for the case of , , , and .

#### 2. General Solution

In the following theorem, we apply [16, Theorem 1.1] for the investigation of general solutions of the functional equation (6).

Theorem 3. Let , , be real constants such that the cubic equation has the following properties:(i) and are two distinct nonzero real roots of the cubic equation (8);(ii)it holds true that either for or for .Let be either a real vector space if for or a complex vector space if for . Then, a function is a solution of the functional equation (6) if and only if there exist functions such that where denotes the largest integer not exceeding , and , are defined in (13) and (23).

Proof. Assume that is a solution of (6). If we define an auxiliary function by then it follows from (6) that satisfies for any . According to [16, Theorem 1.1] or [3, Theorem 2.1], there exists a function such that for all , where and , are the distinct roots of the quadratic equation that is,
Since is a root of the quadratic equation (14), we have We multiply both sides of (16) with and make use of (16) and (i) to get Similarly, we also obtain Using (13), (17), and (18), we have for all .
If we define an auxiliary function by then it follows from (6) that satisfies for any . According to [16, Theorem 1.1] or [3, Theorem 2.1], there exists a function such that for all , where and , are the distinct roots of the quadratic equation that is, As in the first part, we verify that for all .
We now multiply (12) with and (22) with , we subtract the former from the latter, and we then divide the resulting equation by to get (9).
We assume that a function is given by (9), where are arbitrarily given functions and , are given by (13) and (23), respectively. Then, by (9), (19), and (26), we have for all , which implies that is a solution of (6).

According to [17, p. 92], the Fibonacci numbers satisfy the identity for all integers . We can easily notice that the linear equation of third order is strongly related to identity (28).

Corollary 4. Let be a real vector space. A function is a solution of the functional equation (29) if and only if there exist functions such that where and are defined in (33).

Proof. If we set , , and in (8), then the cubic equation has three distinct nonzero roots including Moreover, it holds that and . By (13), (15), (23), and (25), we have where we make use of (15) and (25) to calculate
Finally, in view of Theorem 3, we conclude that the assertion of our corollary is true.

Corollary 5. If a function is a solution of functional equation (29), then there exist real constants , , , and such that for all , where and are defined in (33).

#### 3. Hyers-Ulam Stability

We apply the classical direct method to the proof of the following theorem. The classical direct method was first proposed by Hyers .

Theorem 6. Let , , be real constants with , let be a nonzero root of the cubic equation (8), and let , be the roots of the quadratic equation with and . Let be either a real Banach space if or a complex Banach space if . If a function satisfies the inequality for all and for some , then there exists a solution of (6) such that for all .

Proof. If we define an auxiliary function by then, as we did in (11), it follows from (36) that satisfies the inequality or for any .
If we replace with in the last inequality, then we have for all . Furthermore, we get for all and . By (42), we obviously have for and .
For any , (42) implies that the sequence is a Cauchy sequence (note that ). Therefore, we can define a function by since is complete. In view of the definition of and using the relations, and , we obtain for all . Since is a nonzero root of the cubic equation (8), it follows from (45) that for all . Hence, we conclude that is a solution of (6).
If tends to infinity, then (43) yields that for every .
On the other hand, it also follows from (36) that for all . Analogously to (42), replacing by in the last inequality and then dividing by both sides of the resulting inequality, then we have for all and . By using (49), we further obtain for and .
On account of (49), we see that the sequence is a Cauchy sequence for any fixed (note that ). Hence, we can define a function by Due to the definition of and the relations, and , we get for any . Similarly as in the first part, we can show that is a solution of (6).
If we let tend to infinity, then it follows from (50) that for .
It follows from (47) and (53) that for any .
Finally, if we define a function by for all , then is also a solution of (6). Moreover, the validity of (37) follows from the last inequality.

The following theorem is the main theorem of this paper.

Theorem 7. Given real constants , , with , let and be distinct nonzero roots of cubic equation (8) and let , be the roots of the quadratic equation with and for . Assume that either for all or for all . Let be either a real Banach space if or a complex Banach space if . If a function satisfies inequality (36) for all and for some , then there exists a solution of (6) such that for all .

Proof. According to Theorem 6, there exists a solution of (6) such that for any and . In view of the last inequalities, we have for all .
If we define a function by for each , then is also a solution of (6), and inequality (56) follows from the last inequality.

#### Conflict of Interests

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

#### Acknowledgments

This research paper was completed while Soon-Mo Jung was a visiting scholar at National Technical University of Athens during February 2014. He would like to express his cordial thanks to Professor Themistocles M. Rassias for his hospitality and kindness. This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (no. 2013R1A1A2005557). The authors would like to express their cordial thanks to the referees for useful remarks.