/compile/output.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 8, No. 2, 2015, 283-293 ISSN 1307-5543 – www.ejpam.com Generalized Hyers-Ulam-Rassias Stability of a System of Bi-Reciprocal Functional Equations K. Ravi1,!, B.V. Senthil Kumar2 1 PG & Research Department of Mathematics, Sacred Heart College, Tirupattur - 635 601, Tamil Nadu, India 2 Department of Mathematics, C. Abdul Hakeem College of Engineering & Technology, Melvisharam - 632 509, Tamil Nadu, India Abstract. In this paper, we find the generalized Hyers-Ulam-Rassias stability of the system of bi-reciprocal functional equations r(x + u, y) = r(x , y)r(u, y) r(x , y) + r(u, y) , r(x , y + v) = r(x , y)r(x , v) r(x , y) + r(x , v) in the setting of Fréchet spaces. 2010 Mathematics Subject Classifications: 39B82, 39B72 Key Words and Phrases: Reciprocal function, Bi-reciprocal functional equation, Generalized Hyers- Ulam-Rassias stability. 1. Introduction In functional analysis and related areas of mathematics, Fréchet spaces, named after Mau- rice Fréchet, are special topological vector spaces. They are generalizations of Banach spaces (normed vector spaces which are complete with respect to the metric induced by the norm). Many vector spaces of holomorphic, differentiable or continuous functions which arise in connection with various problems in analysis and its applications are defined by (at most) countably many conditions, whence they carry a natural Fréchet topology (if they are, in ad- dition, complete). In particular, each Banach space is a Fréchet space and so has a countable basis of absolutely convex zero neighborhoods. A topological vector space X is a Fréchet space if and only if it satisfies the following three properties: !Corresponding author. Email addresses: shckravi@yahoo.co.in (K. Ravi), bvssree@yahoo.co.in (B. Kumar) http://www.ejpam.com 283 c" 2015 EJPAM All rights reserved. K. Ravi, B. Kumar / Eur. J. Pure Appl. Math, 8 (2015), 283-293 284 (i) it is complete as a uniform space (ii) it is locally convex (iii) its topology can be induced by a translation invariant metric, i.e. a metric d : X #X $ ! such that d(x , y) = d(x + a, y + a) for all a, x ,% X . This means that a subset U of X is open if and only if for every u in U , there exists an !> 0 such that {v : d(u, v)< !} is a subset of U . Note that there is no natural notion of distance between two points of a Fréchet space: many different translation-invariant metrics may induce the same topology. The vector space C&([0,1]) of all infinitely often differentiable functions f : [0,1] $ ! becomes a Fréchet space with the seminorms || f ||k = sup{| f (k)(x)| : x % [0,1]} for every integer k ' 0. Here, f (k) denotes the kth derivative of f , and f (0) = f . The spaces C&(!) for ! ( !n open, )(K), C&(K) for K ( !n, )(!n), * (!) for ! ( "n open are other examples of Fréchet spaces. More generally, if M is a compact C& manifold and B is a Banach space, then the set of all infinitely often differentiable functions f : M $ B can be turned into a Fréchet space; the seminorms are given by the suprema of the norms of all partial derivatives. The space " of real valued sequences becomes a Fréchet space if we define the kth semi- norm of a sequence to be the absolute value of the kth element of the sequence. Convergence in this Fréchet space is equivalent to element-wise convergence. Not all vector spaces with complete translation-variant metrics are Fréchet spaces. An example is Lp with p < 1. Of course, such spaces fail to be locally convex. The topology of a Fréchet space E can be given by a sequence of seminorms + +1 , + +2 , . . . in the following way: a basis of neighborhoods of zero are the sets Uk,! = {x % E : +x+k , !}. Such a system is called a fundamental system of seminorms. It is by no means uniquely determined by the topology. In fact, two systems + +1 , + +2 , . . . and + +-1 , + + - 2 , . . . give the same topology if and only if there exist constants Ck and n(k) % # such that + +k , Ck + + - n(k) and + +-k , Ck + +n(k) for all k. In this case the systems of seminorms are called equivalent. A Fréchet space equipped with a fixed fundamental system of seminorms is called a graded Fréchet space. This concept is important in connection with many problems in analysis, where the index of a norm indicates e.g. the order of derivatives involved. It is a classical fact that the Fréchet spaces are characterized by the existence of a countable, sufficient and increasing family of semi-norms {pi}i%# (that is pi(x) = 0 implies x = 0 and pi(x), pi+1(x) for all x % X and i % #), which define the pseudo-norm "(x) = & ! i=0 1 2i pi(x) 1+ pi(x) and the metric d(x , y) ="(x.y) invariant with respect to translations, such that d generates a complete topology equivalent to that of locally convex space. Also, notice that since pi(x) 1+pi(x) , 1 and "& i=0 1 2i = 1, it follows that "(x), 1 for all x % X (see [7]). K. Ravi, B. Kumar / Eur. J. Pure Appl. Math, 8 (2015), 283-293 285 Moreover, d has the properties given by the following. Theorem 1 ([7]). Let (X , {pi}i%#, d) be a Fréchet space, then (i) d(cx , c y) , d(x , y) for x , y % X , |c|< 1; (ii) d(x + u, y + v), d(x , y) + d(u, v) for x , y,u, v % X ; (iii) d(kx , k y), d(r x , r y) if k, r % !, 0< k , r; (iv) d(kx , k y), kd(x , y) for x , y % X , k % #, k ' 2; (v) d(cx , c y) , (|c|+ 1)d(x , y) for all x , y % X and c % !. Fréchet spaces are studied because even though their topological structure is more com- plicated due to the lack of a norm, many important results in functional analysis, like the open mapping theorem and the Banach-Steinhaus theorem, still hold. For further concepts on Fréchet spaces, one can refer to ([6, 14, 17]). The stability theory of functional equations basically deals with the following question: “Given an approximately linear mapping f , when does a linear mapping T estimating f exist?” This problem was raised by S.M. Ulam [22] in the year 1940 and D.H. Hyers [11] in the year 1941, gave a first affirmative partial answer to the question of Ulam in the case of Banach spaces. Hyers’ theorem was generalized by T. Aoki [2] for additive mappings in the year 1950 and by Th.M. Rassias [18] for linear mappings by considering an unbounded Cauchy difference in the year 1978. The type of stability investigated by Th.M. Rassias is known as "Hyers-Ulam- Rassias stability" of functional equation. A generalized form of the theorem given by Th.M. Rassias was advocated by P. Gavruta [8] who replaced the unbounded Cauchy difference in Rassias’ theorem by a general control function. This type of stability is called "Generalized Hyers-Ulam-Rassias Stability". The stability problems of several functional equations have been extensively investigated by a number of authors and there are many interesting results concerning this problem (see [1, 4, 5, 10, 12, 13, 15, 16, 21]). Let A and B be vector spaces. A mapping J : A$ B is called Jensen mapping if J satisfies the functional equation 2J # x + y 2 $ = J(x) + J(y). Definition 1 ([3]). Let A and B be vector spaces. A mapping f : A# A$ B is called a bi-Jensen mapping if f satisfies the system of functional equations 2 f # x + y 2 , z $ = f (x , z) + f (y, z) 2 f # x , y + z 2 $ = f (x , y) + f (x , z). % & ' & ( (1) When A= B = !, the function f : !#!$ ! given by f (x , y) = ax y + bx + c y + d is a solution of (1). K. Ravi, B. Kumar / Eur. J. Pure Appl. Math, 8 (2015), 283-293 286 Definition 2 ([9]). Let A and B be vector spaces. A mapping f : A#A$ B is called a bi-quadratic mapping if f satisfies the system of functional equations f (x1 + x2, y) + f (x1 . x2, y) = 2 f (x1, y) + 2 f (x2, y) f (x , y1 + y2) + f (x , y1 . y2) = 2 f (x , y1) + 2 f (x , y2). ) (2) When A= B = !, the function f : !#!$ ! given by f (x , y) = x2 y2 is a solution of (2). In the year 2010, K. Ravi and B.V. Senthil Kumar [19] investigated the generalized Hyers- Ulam-Rassias stability for the reciprocal functional equation r(x + y) = r(x)r(y) r(x) + r(y) (3) where r : !! $ ! is a mapping with R! as the space of non-zero real numbers and with the assumptions x+ y /= 0, r(x)+ r(y) /= 0 and r(x) /= 0, for all x , y % !!. The reciprocal function r(x) = 1 x is a solution of the functional equation (3). K.Ravi, J.M. Rassias and B.V. Senthil Kumar [20] obtained the general solution and in- vestigated the generalized Hyers-Ulam-Rassias stability of a 2-variable reciprocal functional equation F(x + u, y + v) = F(x , y)F(u, v) F(x , y) + F(u, v) (4) where F : !! #!! $ ! is a mapping with R! as the space of non-zero real numbers and with the conditions x+ y /= 0, u+ v /= 0, x+u /= 0, y+ v /= 0, F(x , y) /= 0 and F(x , y)+ F(u, v) /= 0 for all x ,u, y, v % !!. The 2-variable reciprocal function F(x , y) = 1 x+y is a solution of the functional equation (4). Motivated by the system of functional equations (1) and (2), we say that a mapping r : !+ #!+$ !+ is bi-reciprocal if r satisfies the system of functional equations r(x + u, y) = r(x , y)r(u, y) r(x , y) + r(u, y) r(x , y + v) = r(x , y)r(x , v) r(x , y) + r(x , v) . % & & ' & & ( (5) It is easy to see that r(x , y) = 1 x y is a solution of the system of functional equations (5). In this paper, we investigate the generalized Hyers-Ulam-Rassias stability problem for the system of functional equations (5). Throughout this paper, we assume that E is a real normed space and F is a real Banach space. We also assume that X is the space of non-zero real numbers and Y is a real Fréchet space with metric d with the conditions x + u /= 0, y + v /= 0, r(x , y) /= 0, r(x , y) + r(u, y) /= 0 and r(x , y) + r(x , v) /= 0 for all x ,u, y, v % X . For notational convenience, let us denote for a given mapping r : X $ Y , the difference operators #r : X # X # X $ Y and "r : X # X # X $ Y by #r(x ,u, y) = d * r(x + u, y), r(x , y)r(u, y) r(x , y) + r(u, y) + , K. Ravi, B. Kumar / Eur. J. Pure Appl. Math, 8 (2015), 283-293 287 "r(x , y, v) = d * r(x , y + v), r(x , y)r(x , v) r(x , y) + r(x , v) + for all x ,u, y, v % X . 2. Generalized Hyers-Ulam-Rassias Stability of the System of Functional Equations (5) Theorem 2. Let G, H : X # X # X $ [0,&) be mappings satisfying & ! i=0 4iG(2i x , 2i x , 2i y)<&, & ! i=0 4iH(2i+1 x , 2i y, 2i y)<& % & & & ' & & & ( (6) for all x , y % X . Let r : X # X $ Y be a mapping such that #r(x ,u, y), G(x ,u, y) (7) "r(x , y, v) , H(x , y, v) (8) for all x ,u, y, v % X . Then there exists a unique bi-reciprocal mapping R : X # X $ Y satisfying (5) and d(R(x , y), r(x , y)), 2 & ! i=0 4iG(2i x , 2i x , 2i y) + 4 & ! i=0 4iH(2i+1 x , 2i y, 2i y) (9) for all x , y % X . The mapping R(x , y) is defined by R(x , y) = lim n$& 4nr(2n x , 2n y), for all x , y % X . Proof. Setting u= x in (7) and then multiplying by 2 on both sides, we get d(2r(2x , y), r(x , y)), 2G(x , x , y) (10) for all x , y % X . Putting v = y in (8) and then multiplying by 2 on both sides, we obtain d(2r(x , 2y), r(x , y)), 2H(x , y, y) (11) for all x , y % X . Replacing x by 2x in (11) and then multiplying by 2 on both sides, yields d(4r(2x , 2y), 2r(2x , y)), 4H(2x , y, y) (12) for all x , y % X . Combining (10) and (12) and using triangle inequality, we get d(4r(2x , 2y), r(x , y)), 2G(x , x , y) + 4H(2x , y, y) (13) K. Ravi, B. Kumar / Eur. J. Pure Appl. Math, 8 (2015), 283-293 288 for all x , y % X . Now, substituting (x , y) by (2x , 2y) in (13) and then multiplying by 4 on both sides, we have d(16r(4x , 4y), 4r(2x , 2y)), 8G(2x , 2x , 2y) + 16H(4x , 2y, 2y) (14) for all x , y % X . Combining (13) and (14), we see that d , 42r(22 x , 22 y), r(x , y) - , 2 1 ! i=0 4iG(2i x , 2i x , 2i y) + 4 1 ! i=0 4iH(2i+1 x , 2i y, 2i y) for all x , y % X . Using induction arguments, we conclude that d(4nr(2n x , 2n y), r(x , y)) ,2 n.1 ! i=0 4iG(2i x , 2i x , 2i y) + 4 n.1 ! i=0 4iH(2i+1 x , 2i y, 2i y) ,2 & ! i=0 4iG(2i x , 2i x , 2i y) + 4 & ! i=0 4iH(2i+1 x , 2i y, 2i y) (15) for all x , y % X . In order to prove the convergence of the sequence {4nr(2n x , 2n y)}, replace (x , y) by (2m x , 2m y) in (15) and multiply by 4m to get d(4n+mr(2n+m, 2n+m y), 4mr(2m x , 2m y)) =4md(4nr(2n+m x , 2n+m y), r(2m x , 2m y)) ,2 & ! i=0 4m+iG(2m+i x , 2m+i x , 2m+i y) + 4 & ! i=0 4m+iH(2m+i+1 x , 2m+i y, 2m+i y). Using (6), the right-hand side of the above inequality tends to zero as m$&. This shows that {4nr(2n x , 2n y)} is a Cauchy sequence in Y . Since Y is a Fréchet space, it follows that the sequence {4nr(2n x , 2n y)} converges. Define R : X # X $ Y by R(x , y) = lim n$& 4nr(2n x , 2n y) for all x , y % X . It follows from (7) that #R(x ,u, y) = lim n$& 4n #r(2 n x , 2nu, 2n y) , lim n$& 4nG(2n x , 2nu, 2n y) = 0 for all x ,u, y, v % X . Also it follows from (8) that "R(x , y, v) = lim n$& 4n"r(2 n x , 2n y, 2nv) , lim n$& 4nH(2n x , 2n y, 2nv) = 0 K. Ravi, B. Kumar / Eur. J. Pure Appl. Math, 8 (2015), 283-293 289 for all x , y, v % X , which shows that R is bi-reciprocal. To prove R is unique bi-reciprocal mapping, let us consider another bi-reciprocal mapping R0 : X # X $ Y which satisfies (5) and (9). Since 4nR(2n x , 2n y) = R(x , y) and 4nR0(2n x , 2n y) = R0(x , y) for all x , y % X , we conclude that d(R(x , y),R0(x , y)) =4nd(R(2n x , 2n y),R0(2n x , 2n y)) ,4n . d(R(2n x , 2n y), r(2n x , 2n y)) + d(r(2n x , 2n y),R0(2n x , 2n y)) / ,4 & ! i=0 4n+iG(2n+i x , 2n+i x , 2n+i x) + 8 & ! i=0 4n+iH(2n+i+1 x , 2n+i y, 2n+i y) for all x , y % X . Using (6), letting n $& in the right-hand side of the above inequality, it follows that R(x , y) = R0(x , y) for all x , y % X , which completes the proof of the theorem. Theorem 3. Suppose the mapping G, H : E # E # E $ [0,&) satisfy (6) for all x , y % E. If r : E # E$ F is a mapping such that 0 0 0 0 r(x + u, y). r(x , y)r(u, y) r(x , y) + r(u, y) 0 0 0 0 , G(x ,u, y), (16) 0 0 0 0 r(x , y + v). r(x , y)r(x , v) r(x , y) + r(x , v) 0 0 0 0 , H(x , y, v) (17) for all x ,u, y, v % E, then there exists a unique bi-reciprocal mapping R : E # E $ F satisfying (5) and 0 0R(x , y). r(x , y) 0 0, 2 & ! i=0 4iG(2i x , 2i x , 2i y) + 4 & ! i=0 4iH(2i+1 x , 2i y, 2i y) for all x , y % E. Proof. Putting d(a, b) = +a. b+, for all a, b % E in Theorem 2, the proof follows immedi- ately. We investigate the Hyers-Ulam-Rassias stability of the system of functional equations (5) in the corollary presented below. Corollary 1. Let c1 > 0 be fixed and p < .2. If a mapping r : E#E$ F satisfies the inequalities 0 0 0 0 r(x + u, y). r(x , y)r(u, y) r(x , y) + r(u, y) 0 0 0 0 , c1 1 +x+p + +u+p + 0 0y 0 0 p 2 , 0 0 0 0 r(x , y + v). r(x , y)r(x , v) r(x , y) + r(x , v) 0 0 0 0 , c1 1 +x+p + 0 0y 0 0 p + +v+p 2 % & & ' & & ( (18) for all x ,u, y, v % E, then there exists a unique bi-reciprocal mapping R : E # E $ F satisfying (5) and 0 0R(x , y). r(x , y) 0 0, # 2c1 1. 2p+2 $ . 2 (2p + 1)+x+p + 0 0y 0 0 p / (19) for all x , y % E. K. Ravi, B. Kumar / Eur. J. Pure Appl. Math, 8 (2015), 283-293 290 Proof. Considering G(x , y, z) = H(x , y, z) = c1 1 +x+p + 0 0y 0 0 p + +z+p 2 , for all x , y, z % E in Theorem 3, we get 0 0R(x , y). r(x , y) 0 0 ,2c1 & ! i=0 4i 10 02i x 0 0 p + 0 02i x 0 0 p + 0 02i y 0 0 p 2 + 4c1 & ! i=0 4i 10 02i+1 x 0 0 p + 0 02i y 0 0 p + 0 02i y 0 0 p 2 ,2c1 & ! i=0 4i2pi 1 2+x+p + 0 0y 0 0 p 2 + 4c1 & ! i=0 4i2pi 1 2p +x+p + 2 0 0y 0 0 p 2 ,2c1 & ! i=0 2(p+2)i 1 2+x+p + 0 0y 0 0 p 2 + 4c1 & ! i=0 2(p+2)i 1 2p +x+p + 2 0 0y 0 0 p 2 ,2c1 # 1 1. 2p+2 $ 1 2+x+p + 0 0y 0 0 p 2 + # 4c1 1. 2p+2 $ 1 2p +x+p + 2 0 0y 0 0 p 2 , # 2c1 1. 2p+2 $ . 2 (2p + 1)+x+p + 5 0 0y 0 0 p / , for all x , y % E. Theorem 4. Let G, H : X # X # X $ [0,&) be mappings satisfying & ! i=0 1 4i G # x 2i+1 , x 2i+1 , y 2i+1 $ <&, & ! i=0 1 4i H # x 2i , y 2i+1 , y 2i+1 $ <& % & & & ' & & & ( (20) for all x , y % X . Let r : X #X $ Y be a mapping such that (7) and (8) hold for all x ,u, y, z % X . Then there exists a unique bi-reciprocal mapping R : X # X $ Y satisfying (5) and d(r(x , y),R(x , y)), 1 2 & ! i=0 1 4i G # x 2i+1 , x 2i+1 , y 2i+1 $ + & ! i=0 1 4i H # x 2i , y 2i+1 , y 2i+1 $ (21) for all x , y % X . The mapping R(x , y) is defined by R(x , y) = lim n$& 1 4n r # x 2n , y 2n $ , for allx , y % X . Proof. Replacing (x ,u, y) by , x 2 , x 2 , y 2 - in (7) and then dividing by 2, we obtain d # 1 2 r # x , y 2 $ , 1 4 r # x 2 , y 2 $$ , 1 2 G # x 2 , x 2 , y 2 $ (22) for all x , y % X . Now, replacing (x , y, v) by , x 2 , y 2 , y 2 - in (8), we get d # r # x 2 , y $ , 1 2 r # x 2 , y 2 $$ , H # x 2 , y 2 , y 2 $ (23) K. Ravi, B. Kumar / Eur. J. Pure Appl. Math, 8 (2015), 283-293 291 for all x , y % X . Putting x = 2x in (23), we lead to d # r(x , y), 1 2 r # x , y 2 $$ , H # x , y 2 , y 2 $ (24) for all x , y % X . Combining (22) and (24), applying triangle inequality, yields d # r(x , y), 1 4 r # x 2 , y 2 $$ , 1 2 G # x 2 , x 2 , y 2 $ + H # x , y 2 , y 2 $ for all x , y % X . Proceeding further and using induction arguments on a positive integer n, we have d # r(x , y), 1 4n r # x 2n , y 2n $$ , 1 2 n.1 ! i=0 1 4i G # x 2i+1 , x 2i+1 , y 2i+1 $ + n.1 ! i=0 1 4i H # x 2i , y 2i+1 , y 2i+1 $ , 1 2 & ! i=0 1 4i G # x 2i+1 , x 2i+1 , y 2i+1 $ + & ! i=0 1 4i H # x 2i , y 2i+1 , y 2i+1 $ for all x , y % X . The rest of the proof is obtained by similar arguments as in Theorem 2. Theorem 5. Suppose the mappings G, H : E # E # E $ [0,&) satisfy (23) and (24) for all x , y % E. If r : E # E $ F is a mapping such that (16) and (17) hold for all x ,u, y, v % E, then there exists a unique bi-reciprocal mapping R : E # E$ F satisfying (5) and 0 0r(x , y). R(x , y) 0 0, 1 2 & ! i=0 1 4i G # x 2i+1 , x 2i+1 , y 2i+1 $ + & ! i=0 1 4i H # x 2i , y 2i+1 , y 2i+1 $ for all x , y % E. Proof. By taking d(a, b) = +a. b+, for all a, b % E in Theorem 4, we arrive at the desired result. Corollary 2. Let !> 0 be fixed. If r : E # E$ F satisfies 0 0 0 0 r(x + u, y). r(x , y)r(u, y) r(x , y) + r(u, y) 0 0 0 0 , ! 2 , 0 0 0 0 r(x , y + v). r(x , y)r(x , v) r(x , y) + r(x , v) 0 0 0 0 , ! 2 for all x ,u, y, v % E, then there exists a unique bi-reciprocal mapping R : E # E$ F such that 0 0r(x , y). R(x , y) 0 0, !, for all x , y % E. REFERENCES 292 Proof. Letting G(x , y, z) = H(x , y, z) = ! 2 , for all x , y, z % E in Theorem 5, we lead to 0 0r(x , y). R(x , y) 0 0, 1 2 & ! i=0 1 4i ! 2 + & ! i=0 1 4i ! 2 , 3! 4 & ! i=0 1 4i = 3! 4 # 4 3 $ = !, for all x , y % E. Corollary 3. Let c1 > 0 be fixed and p > .2. If a mapping r : E#E$ F satisfies the inequalities (18), for all x ,u, y, v % E, then there exists a unique bi-reciprocal mapping R : E # E $ F satisfying (5) and 0 0r(x , y). R(x , y) 0 0, # 2c1 2p+2 . 1 $ . 2 (2p + 1)+x+p + 5 0 0y 0 0 p / , for all x , y % E. Proof. 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