Journal of Mathematical Extension Vol. 9, No., (205), 5-63 ISSN: 735-8299 URL: http://www.ijmex.com Quintic Functional Equations in Non-Archimedean Normed Spaces A. Bodaghi Garmsar Branch, Islamic Azad University Abstract. The Hyers-Ulam stability of a quintic functional equation in the normed spaces and non-archimedean normed spaces by direct method are proved. AMS Subject Classification: 39B52; 39B72; 39B82; 46B03 Keywords and Phrases: Banach space, Hyers-Ulam stability, non- Archimedean normed space, quintic functional equation. Introduction In 940, S. M. Ulam [20] asked the first question on the stability problem. In 94, D. H. Hyers [5] solved the problem of Ulam. This result was generalized by Aoki [] for additive mappings and by Rassias [9] for linear mappings by considering an unbounded Cauchy difference. The paper of Rassias [9] has provided a lot of influence in the development of what we now call Hyers-Ulam- Rassias stability of functional equations. Since then, several stability problems for various functional equations have been investigated by numerous mathematicians; cf e.g. [2], [3], [4], [5], [8], [2], [5], [6], [7] and [23]. In [22], Xu et al. obtained the general solution and investigated the Ulam stability problem for the following quintic functional equation f(x + 3y) 5f(x + 2y) + 0f(x + y) 0f(x) + 5f(x y) f(x 2y) = 20f(y) in quasi-β-normed spaces via fixed point method. This method which is different from the direct method, initiated by Hyers in [5], had been applied by Received: August 204; Accepted: October 204 5
52 A. BODAGHI Cădariu and Radu for the first time in [0]. In other words, they employed this fixed point method to the investigation of the Cauchy functional equation [] and for the quadratic functional equation [0] (see also [6] and [7]). Recently, in [8], Park et al. introduced the following new form of quintic functional equation f(3x + y) 5f(2x + y) + f(2x y) + 0f(x + y) 5f(x y) = 0f(y) + f(3x) 3f(2x) 27f(x). () It is easily verified the function f(x) = αx 5 satisfies the functional equation (). In other words, every solution of the quintic functional equation is called a quintic mapping. In [8], the authors applied the fixed point method to establish the Hyers-Ulam stability of the orthogonally quintic functional equation () in Banach spaces and in non-archimedean Banach spaces. In this paper, we prove the Hyers-Ulam stability of the quintic functional equation () in the normed spaces and non-archimedean normed spaces via direct way. 2. Stability of () in Real Normed Spaces Throughout this paper, we use the abbreviation for the given mapping f : X as follows: D q f(x, y) := f(3x + y) 5f(2x + y) + f(2x y) + 0f(x + y) 5f(x y) 0f(y) f(3x) + 3f(2x) + 27f(x) for all x, y X. Throughout this section, we assume X is a real normed space with norm. X and is a real Banach space with norm.. We are going to prove the stability of the quintic functional equation () in real normed spaces. Theorem 2.. Let α be a real number and let f : X be a mapping for which there exists a function φ : X X [ α, ) such φ(x, y) := n=0 n φ(2n x, 2 n y) <, (2) D q f(x, y) α + φ(x, y), (3) for all x, y X. Then there exists a unique quintic mapping Q : X such 0 f(x) Q(x) 3 f(0) α 3 + φ(x, 0), (4)
QUINTIC FUNCTIONAL EQUATIONS IN... 53 for all x X. Proof. Setting y = 0 in (3), we have for all x X. Hence f(x) f(2x) 0f(0) α + φ(x, 0), f(x) 0 f(2x) f(0) α + φ(x, 0), (5) for all x X. Replacing x by 2x in (5) and continuing this method, we obtain f(x) f(2n x) n 0 n k=0 k f(0) Also, we can use induction and triangular inequality to get f(2 m x) m f(2n x) n 0 + n α k + n φ(2 k x, 0) k. (6) k=0 k=0 n k=m n k=m k f(0) for all x X, and n > m 0. Thus the sequence α k + n φ(2 k x, 0) k, (7) k=m { } f(2 n x) n and (7). Since is a Banach space, there exists a map Q so is Cauchy by (2) f(2 n x) lim n n = Q(x). (8) Letting the limit as n tends to infinity in (6) and applying (8), we can observe the inequality (4) is true. Now, by replacing x, y by 2 n x, 2 n y, respectively in (2), we deduce D q Q(x, y) = lim n n D qf(x, y) lim n n α + lim φ(2 n x, 2 n y) n n = 0, for all x, y X. It implies Q : X is a quintic mapping. For the uniqueness of Q, let Q : X be another quintic mapping satisfying
54 A. BODAGHI (4). Then we have Q(x) Q (x) = n Q(2n x) Q (2 n x) n Q(2n x) f(2 n x) + 0 3 f(0) + n f(2n x) Q (2 n x) 0 3 f(0) [ α n 3 + φ(x, ] 0) = α 3 n + k=0 = α 3 n + k=n n+k φ(2n+k x, 0) k φ(2k x, 0), for all x X. Taking n in the preceding inequality, we have Q = Q. This completes the proof. Corollary 2.2. Let α, β, γ, r and s be non-negative real numbers such s > 0 and r, s < 5. Suppose f : X is a mapping fulfilling D q f(x, y) α + β x r X + γ y s X, (9) for all x, y X. Then there exists a unique quintic mapping Q : X such f(x) Q(x) 0α 65 + β x r X 2 r, (0) for all x X, and all x X \{0} if 5 < r < 0. Also, if for each fixed x X the mappings t f(tx) from R to is continuous, then Q(tx) = t 5 Q(x) for all x X and all t R. Proof. Letting φ(x, y) = β x r X + γ y s X in Theorem 3., we have f(x) Q(x) 0 3 f(0) α 3 + β x r X 2 r. It follows from (9) f(0) α 2. By these statements we can get the result. Obviously, if 5 < r < 0, the inequality (0) holds for all x X \{0}. Now, suppose F is any continuous linear functional on X and x is a fixed element
QUINTIC FUNCTIONAL EQUATIONS IN... 55 in X. Define the mapping h : R R via h(t) = F[Q(tx)] for each t R. It is easily verified h is a quintic function. Under the hypothesis f(tx) is continuous in t R for each fixed x, the function h is the pointwise limit of the sequence of continuous functions {h n } in which h n (t) = F(2n tx), where n n N and t R. Thus h is a contiuous function and has the form h(t) = t 5 h() for all t R. Therefore F[Q(tx)] = h(t) = t 5 h() = t 5 F[Q(x)] = F[t 5 Q(x)]. Since F is an arbitrary continuous linear functional on X, we have Q(tx) = t 5 Q(x) for all t R and x X. We have the following result which is analogous to Theorem 3. for the quintic functional equation (). The proof is similar but we include its proof. Theorem 2.3. Suppose f : X is a mapping for which there exists a function φ : X X [0, ) such φ(x, y) := k= k φ( x 2 k, y ) <, () 2k D q f(x, y) φ(x, y), (2) for all x, y X. Then there exists a unique quintic mapping Q : X such for all x X. f(x) Q(x) φ(x, 0), (3) Proof. It follows from () φ(0, 0) = 0. Hence from (2) we get f(0) = 0. Putting y = 0 in (2), we have f(x) f(2x) φ(x, 0), for all x X. Interchanging x into x 2 in the above inequality, we get f( x 2 ) f(x) φ( x 2, 0). Using triangular inequality and proceeding this method, we obtain n f( x 2 n ) f(x) n k= k φ( x, 0), (4) 2k
56 A. BODAGHI for all x X. If we show the sequence { n f( x 2 ) } is Cauchy, then it will n be convergent by the completeness of. For this, we replace x by x 2 in (4) m and then multiply both side by m, we get m+n x f( 2 m+n ) m f( x 2 m ) = n k+m x φ(, 0) 2k+m k= m+n k φ( x 2 k, 0), k=m+ for all x X, and n > m > 0. Thus the above sequence converges to the mapping Q. In fact, Q(x) := lim n n f( x 2 n ). Now, similar to the proof of Theorem 3., we can complete the rest of the proof. Corollary 2.4. Let β, γ, r and s be non-negative real numbers such r, s > 5. Suppose f : X is a mapping fulfilling D q f(x, y) β x r X + γ y s X, (5) for all x, y X. Then there exists a unique quintic mapping Q : X such f(x) Q(x) β 2 r x r X, (6) for all x X, and all x X \{0} if r < 5. Also, if for each fixed x X the mappings t f(tx) from R to is continuous, then Q(tx) = t 5 Q(x) for all x X and all t R. Proof. First, we note if we put x = y = 0 in (5), we have f(0) = 0. Defining φ(x, y) = β x r X + γ y s X in Theorem 3.3, we can obtain (6). The rest of the proof is obvious by the proof of Corollary 3.4. The quintic functional equation () can be superstable under some conditions. It is shown in the next result. Recall a functional equation is called superstable if every approximately solution is an exact solution of it. Corollary 2.5. Let r, s and α be a non-negative real numbers such r +s 5. If f : X is a mapping such D q f(x, y) α y s (or α x r y s ), (7)
QUINTIC FUNCTIONAL EQUATIONS IN... 57 for all x, y X, then the mapping f is quintic. Also, if for each fixed x X the mapping t f(tx) from R to is continuous, then f(tx) = t 5 f(x) for all x X and all t R. Proof. The inequality (7) shows f(0) = 0. Putting y = 0 in (7), we get f(2x) = f(x) (x X ), and so f(x) = f(2n x) for all x X and n N. n Letting φ(x, y) = α y s (or φ(x, y) = α x r y s ) in Theorems 2. and 2.3, we can see Q = f is a quintic mapping. 3. Stability of () in Non-Archimedean Normed Spaces We first recall some definitions and basic facts in the non-archimedean normed spaces setting. By a non-archimedean field we mean a field K equipped with a function (valuation) from K into [0, ) such r = 0 if and only if r = 0, rs = r s, and r + s max{ r, s } for all r, s K. Clearly = = and n for all n N. Let X be a vector space over a scalar field K with a non-archimedean nontrivial valuation. A function : X R is a non-archimedean norm (valuation) if it satisfies the following conditions: (i) x = 0 if and only if x = 0; (ii) rx = r x, (x X, r K); (iii) the strong triangle inequality (ultrametric); namely, x + y max{ x, y }, (x, y X). Then (X, ) is called a non-archimedean normed space. Due to the fact x n x m max{ x j+ x j ; m j n }, (n m), a sequence {x n } is Cauchy if and only if {x n+ x n } converges to zero in a non-archimedean normed space X. By a complete non-archimedean normed space we mean one in which every Cauchy sequence is convergent. In [4], Hensel discovered the p-adic numbers as a number theoretical analogue of power series in complex analysis. The most interesting example of non-archimedean normed spaces is p-adic numbers. A key property of p-adic numbers is they do not satisfy the Archimedean axiom: for all x, y > 0, there exists an integer n such x < ny.
58 A. BODAGHI Let p be a prime number. For any non-zero rational number x = p r m n in which m and n are coprime to the prime number p. Consider the p-adic absolute value x p = p r on Q. It is easy to check is a non-archimedean norm on Q. The completion of Q with respect to which is denoted by Q p is said to be the p-adic number field. One should remember if p > 2, then 2 n = in for all integers n. In [2], the stability of some functional equations in non-archimedean normed spaces are investigated (see also [9] and [2]). From now on, we assume X is a real vector space and is a complete non-archimedean normed space unless otherwise stated explicitly. In the upcoming theorem, we prove the stability of the functional equation () in non- Archimedean normed spaces. Theorem 3.. Let φ : X X [0, ) be a function such lim k k ϕ(2k x, 2 k y) = 0, (8) for all x, y X. Suppose f : X is a mapping satisfying the inequality D q f(x, y) ϕ(x, y), (9) for all x, y X. Then there exists a unique quintic mapping Q : X such f(x) Q(x) ϕ(x), (20) { } for all x X where ϕ(x) = sup ϕ(2 j x,0) : j N {0}. j Proof. Putting y = 0 in (9), we have f(2x) f(x) ϕ(x, 0), (2) for all x X. Replacing x by 2 n x in (2) and then dividing both sides by n+, we get n f(2n x) n+ f(2n+ x) n+ ϕ(2n x, 0), (22) for all x X and all non-negative integers n. Thus the sequence { } f(2 n x) n Cauchy by (8) and (22). Completeness of non-archimedean space allows us to assume there exists a mapping Q so lim n f(2 n x) n = Q(x). (23) is
QUINTIC FUNCTIONAL EQUATIONS IN... 59 For each x X and non-negative integers n, we have f(x) f(2n n x) n = f(2 j x) j f(2j+ x) j+ j=0 { f(2 j } x) max j f(2j+ x) j+ : 0 j < n { ϕ(2 j } max x, 0) j : 0 j < n. (24) Taking n tends to approach infinity in (24) and applying (23), we can see the inequality (20) holds. It follows from (8), (9) and (23) for all x, y X, D q Q(x, y) = lim n n D qf(2 n x, 2 n y) lim n n ϕ(2n x, 2 n y) = 0, Hence, the mapping Q satisfies () and thus it is quintic. Now, let Q : X be another quintic mapping satisfying (20). Then we have Q(x) Q(x) = lim k k Q(2k x) Q(2 k x) lim k k max{ Q(2k x) f(2 k x), = lim lim k lim sup k { ϕ(2 j max x, 0) n for all x X. This shows the uniqueness of Q. f(2 k x) Q(2 k x) } } j : k j < n + k } = 0. { ϕ(2 j x, 0) j : k j < Corollary 3.2. Let α > 0, X be a non-archimedean normed space and let f : X be a mapping satisfying the inequality D q f(x, y) α (Γ( x ) + Γ( y )), (25) for all x, y X. If Γ : [0, ) [0, ) is a function satisfying Γ( r s) Γ( r )Γ(s) for all r, s [0, ) for which Γ( 2 ) <, then there exists a unique quintic mapping Q : X such f(x) Q(x) αγ( x ),
60 A. BODAGHI for all x X. Proof. Defining ϕ : X X [0, ) via φ(x, y) = α (Γ( x ) + Γ( y )), we get ( ) n Γ( 2 ) lim n n φ(2n x, 2 n y) lim ϕ(x, y) = 0. n for all x, y X. We also have { ϕ(2 j } x, 0) ϕ(x) = sup j : 0 j < = ϕ(x, 0) = α (Γ( x )). for all x X. Now, Theorem 3. implies the desired result. We have the following result which is analogous to Theorem 3. for the functional equations (). We bring the proof for the sake of completeness. Theorem 3.3. Let ϕ : X X [0, ) be a function such lim k k φ( x 2 k, y ) = 0, (26) 2k for all x, y X. Suppose f : X is a mapping satisfying the inequality D q f(x, y) ϕ(x, y), (27) for all x, y X. Then there exists a unique quintic mapping Q : X such f(x) Q(x) ϕ(x), (28) for all x X where ϕ(x) = sup { j ϕ( x 2 j+, 0) : j N {0} }. Proof. Similar to the proof of Theorem 3., we have f(x) f(2x) ϕ(x, 0), (29) x for all x X. If we replace x by 2 in the inequality (29) and then multiply n+ both sides of the result to n, we get n+ x f( 2 n+ ) n f( x 2 n ) x n ϕ(, 0), (30) 2n+ for all x X and all non-negative integers n. Thus, we conclude from (26) and (30) the sequence { n f( x 2 ) } is Cauchy. Since is a non-archimedean n Banach space, this sequence converges in to the mapping Q. Indeed, Q(x) = lim n n f( x ), (x X). (3) 2n
QUINTIC FUNCTIONAL EQUATIONS IN... 6 Using induction and (29), one can show n f( x 2 n ) f(x) max { j x } ϕ( 2 j+, 0) : 0 j < n, () for all x X and non-negative integers n. Since the right hand side of the inequality () tends to 0 as n to approach infinity, by applying (3), we deduce the inequality (28). Now, similar to the proof of Theorem 3., we can complete the rest of the proof. Corollary 3.4. Let α, r and s be positive real numbers such r, s 5 and 2 <. Suppose X is a non-archimedean normed space and f : X is a mapping fulfilling D q f(x, y) α( x r + y s ), for all x, y X. Then there exists a unique quintic mapping Q : X such α x r r, s > 5 f(x) Q(x) α x r 2 r, s < 5 r for all x X. Proof. The result follows from Theorems 3. and 3.3 by letting ϕ(x, y) = α( x r + y s ). In the next result, we prove the superstability of the functional equations () under some conditions. Corollary 3.5. Let α, r and s be positive real numbers such r + s 5 and 2 <. Suppose X is a non-archimedean normed space and f : X is a mapping fulfilling D q f(x, y) α x r y s, for all x, y X. Then f is a quintice mapping. Proof. Taking ϕ(x, y) = α x r y s in Theorems 3. and 3.3, we can obtain the required result. Acknowledgements The author expresses his sincere thanks to the referees for the careful and detailed reading of the manuscript and very helpful comments.
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