eu ON THE STABILITY OF TRIBONACCI AND k-TRIBONACCI FUNCTIONAL EQUATIONS IN MODULAR SPACE

The purpose of this paper is to establish the Hyers-Ulam stability of the following Tribonacci and k-Tribonacci functional equations f(x) = f(x− 1) + f(x− 2) + f(x− 3), f(k, x) = kf(k, x− 1) + f(k, x− 2) + f(k, x− 3) in modular space. AMS Subject Classification: 39B82, 39B52, 39B72


Introduction
Stability is investigated when one is asking whether a small error of parameters in one problem causes a large deviation of its solution.Give an approximate homomorphism, is it possible to approximate it by a true homomorphism?In other words, we are looking for the situations when the homo-And in the second section, we denote by F k,n the nth k-Tribonacci number where with initial conditions F k,0 = 0, F k,1 = 1, F k,2 = 1.From this, we may derive a functional equation which if called the k-Tribonacci function equation if a function f : N × R → X satisfies the above equation for all x ∈ R, K ∈ N , characteristic equation of the k-Tribonacci sequence is x 3 − kx 2 − x − 1 = 0, and p, q, r denote the roots of characteristic equation where p is greater than one and q, r ∈ C and |q| = |r|.We know that p + q + r = k, pq + qr + pr = −1, pqr = 1.For each x ∈ R, [x] stands for the largest integer that does not exceed x.Finally, we prove the Hyers-Ulam stability of functional equations (3.1) and (3.2) respectively in modular space.

Preliminaries
In this section, we recall some definitions, basic notions and facts about Modular space.As, we know p + q + r = 1, pq + qr + pr = −1 and pqr = 1.Now it follows from that for all x = 0.By mathematical induction, we verify that for all x = 0 and all m belonging to the set {0, 1, 2, . ..}, we obtain, for all x = 0 and all m ∈ {0, 1, 2, . ..}.
And in the similar way we can define for equation .
(a) A functional ρ : X → [0, ∞] is called a modular if for arbitrary x, y ∈ X, (c) A modular ρ defines a corresponding modular space, i.e., the vector space X ρ given by 14]).Let ρ be a convex modular, the modular space X ρ can be equipped with a norm called the Luxemburg norm, defined by A function modular is said to be satisfy the ∆ 2 -condition if there exit k > 0 such that ρ(2x) ≤ kρ(x) for all x ∈ X ρ .

Example 2.3 ([23]
).Let (X, • ) be a norm space, then • is a convex modular on X.But converse is not true.
In general the modular ρ does not behave as a norm or as a distance because it is not sub-additive.But one can associate to a modular the F-norm (see [4]).

Definition 2.4 ([12]
).Let {x n } and x be in X ρ .Then (i) we say {x n } is a ρ-convegent to x and write x n ρx if and only if ρ(x n −x) → 0 as n → 0, (ii) the sequence {x n }, with (iii) a subset S of X ρ is called ρ-complete if and only if any ρ-Cauchy sequence is ρ-convergent to an element of S.
The modular ρ has the Fatou property if and only if any ρ(x) ≤ lim n→∞ inf ρ(x n ) whenever the sequence {x n } is ρ-convergent to x.For further details and proofs, we refer the reader to [14].

Stability of Tribonacci Functional Equation in Modular Space
In the following theorem, we prove the Hyers-Ulam stability of the Tribonacci functional equation .
Theorem 3.1.Let (X, ρ) be a Banach Modular space.If a function f : R → X satisfies the inequality for all x ∈ R, and for some ǫ > 0, then there exist a Tribonacci function Proof.It follows from that If we replace x by x − α in the last inequality, then we get for all x ∈ R and α ∈ N .Furthermore, we have . So by the completeness of X, we may define a function H 1 : R → X such that Applying the definition of H 1 , we introduce the Tribonacci function If m → ∞, then from , we obtain for all x ∈ R. Furthermore, it follows from that for all x ∈ R. Now, we replace x by x − α in above inequality, we have and now multiplying by r α on both sides.
for all x ∈ R, α ∈ Z. Now, we have for all x ∈ R and m ∈ N .We have is a cauchy sequence (|r| < 1) for all x ∈ R. Hence, we can define a function H 2 : R → X by for all x ∈ R. Using the above definition of H 2 , we have So, we can say that H 2 is also a Tribonacci function.If m tends to ∞, then from , we have for all x ∈ R. Finally, analogous to , we obtain for all x ∈ R. Now we replace x by x + α in above inequality, that we have for all x ∈ R and α ∈ Z. Applying , we obtain that for all x ∈ R, m ∈ N .We obviously have is a cauchy sequence by definition of completeness for a fixed x ∈ R. Hence, we may define a function H 3 : R → X by for all x ∈ R. In view of above definition of H 3 , we obtain Hence, we can say that H 3 is also a Tribonacci function.If we suppose, m tends to infinity in then we have for all x ∈ R. From , and , we observe that Now, we assume that Putting the value of |A| from we get the required result.Hence, for all x ∈ R. It is not difficult to show that H is a Tribonacci function satisfying .

Stability of k-Tribonacci Functional Equation in Modular Space
Throughout the following theorem, we prove the Hyers-Ulam stability of the k-Tribonacci functional equation .
Theorem 3.2.Let (X, ρ) be a Banach modular space.If a function f : R → X satisfies the inequality for all x ∈ R, k ∈ N and for some ǫ > 0, then there exist a k-Tribonacci Proof.Since, p + q + r = k, pq + qr + pr = −1 and pqr = 1.So from , we obtain for all x ∈ R, k ∈ N .Now it follows from that for all k ∈ N , x ≥ 0. If we replace x by x − α in inequality then we get for all x ∈ R, k ∈ N .Furthermore, we have ] is a cauchy sequence (|q| < 1).So by the completeness of X, we may define a function H 1 : R → X such that Applying the definition of H 1 , we introduce the k-Tribonacci function If m → ∞, then from we obtain for all x ∈ R, k ∈ N .Furthermore, it follows from that for all x ∈ R, k ∈ N .Now, we replace x by x − α in above inequality, we have and now multiplying by r α on both sides.
for all x ∈ R, α ∈ Z. Now, we have for all x ∈ R and m ∈ N .We have is a cauchy sequence (|r| < 1) for all x ∈ R. Hence, we can define a function H 2 : R → X by for all x ∈ R. Using the above definition of H 2 , we have So, we can say that H 2 is also a k-Tribonacci function.If m tends to ∞, then from , we have for all x ∈ R. Finally, analogous to , we obtain for all x ∈ R. Now we replace x by x + α in above inequality, that we have for all x ∈ R and α ∈ Z. Applying , we obtain that for all x ∈ R, m ∈ N .By using we see that is a cauchy sequence by definition of completeness for a fixed x ∈ R. Hence, we may define a function H 3 : R → X by for all x ∈ R. In view of above definition of H 3 , we obtain Hence, we can say that H 3 is also a k-Tribonacci function.If we suppose, m tends to infinity in then we have for all x ∈ R. From ( 22), ( 25) and ( 28), we observe that ρ f (k, x)− q 2 (r−p)H 1 (k, x) + r 2 (p−q)H 2 (k, x)−p 2 (q−r)H 3 (k, x) q 2 (r−p) + r 2 (p−q) + p 2 (q−r) = 1 |q 2 (r − p) + r 2 (p − q) + p 2 (q − r)| × ρ((q 2 (r − p) + r 2 (p − q) + p 2 (q − r)f (k, x) − q 2 (r − p)H 1 (k, x) − r 2 (p − q)H 2 (k, x) + p 2 (q − r)H 3 (k, x)) For convince, we assume that 1 |q 2 (r−p)+r 2 (p−q)+p 2 (q−r)| = 1 |A| (29) +q 2 (r−p)pr f (k, x−2)−q 2 (r−p)H 1 (k, x)) +(r 2 (p−q) f (k, x)−r 2 (p 2 −q 2 ) f (k, x−1)+r 2 (p−q)qp f (k, x−2) −r 2 (p−q)H 2 (k, x)) +(p 2 (q−r)f (k, x)−p 2 (q 2 −r 2 )f (k, x−1)+p 2 (q−r)qr f (k, x−2) Putting the value of |A| from we get the required result.Hence, H(k, x) = q 2 (r − p)H 1 (k, x) + r 2 (p − q)H 2 (k, x) − p 2 (q − r)H 3 (k, x) q 2 (r − p) + r 2 (p − q) + p 2 (q − r) for all x ∈ R. It is easy to show that H is a k-Tribonacci function satisfying .