Int. Journal of Math. Analysis, Vol. 7, 013, no. 55, 745-75 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.1988/ijma.013.394 Hyers-Ulam-Rassias Stability of a Quadratic-Additive Type Functional Equation on a Restricted Domain Yang-Hi Lee Department of Mathematics Education Gongju National University of Education Gongju 314-711, Republic of Korea yanghi@hanmail.net Copyright c 013 Yang-Hi Lee. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract. In this paper, we prove Hyers-Ulam-Rassias stability of a functional equation f(x + y)+f(x y) f(x) f(y) f( y) =0 on a restricted domain by using the direct method in the sense of Hyers. Mathematics Subject Classification: Primary 39B5 Keywords: Hyers-Ulam-Rassias stability, quadratic-aditive type functional equation 1. Introduction In 1940, S. M. Ulam [9] presented a problem concerning the stability of group homomorphisms. The Ulam s problem for the case of approximately additive mappings was solved by Hyers []. Indeed, Hyers proved that each solution of the inequality f(x + y) f(x) f(y) ε for all x G 1 and y G under the assumption that G 1 and G are Banach spaces, can be approximated by an exact solution of the Cauchy additive functional equation (1.1) f(x + y) =f(x)+f(y).
746 Yang-Hi Lee In this case, the Cauchy additive functional equation is said to satisfy the Hyers-Ulam stability. Hyers result was generalized by T. Aoki[1] for additive mappings and by Th. M. Rassias[7] for linear mappings by allowing the Cauchy difference to be controlled by a sum of powers like f(x + y) f(x) f(y) ε( x p + y p ). In this case, the Cauchy additive functional equation is said to satisfy Hyers- Ulam-Rassias stability. In 1983, the stability theorem for the quadratic functional equation (1.) f(x + y)+f(x y) f(x) f(y) =0 was proved F. Skof [8]. We call a solution of (1.1) an additive mapping and a mapping satisfying (1.) is called a quadratic mapping. If a mapping is represented by the sum of an additive mapping and a quadratic mapping, we call the mapping a quadratic-additive mapping. For a functional equation Ef = 0 if all of the solutions of Ef = 0 are quadraticadditive mappings and all of quadratic-additive mappings are the solutions of Ef = 0, then we call the functional equation Ef = 0 a quadratic additive type functional equation. Now we consider the functional equation: (1.3) f(x + y)+f(x y) f(x) f(y) f( y) =0. The functional equations (1.3) is a quadratic-additive type functional equation (see Theorem.6 in [5]). In this paper, we will prove Hyers-Ulam-Rassias stability of quadratic-additive type functional equation (1.3)[3, 4, 6].. Main results Throughout this paper, let X be a normed space and Y a Banach space. For an arbitrary fixed p R, put s =sign( p) and t =sign(1 p). For a given mapping f : X Y, we use the following abbreviations J n f(x) := 4 sn (f(sn x)+f( sn x)) + tn ( f( tn x) f( tn x) ), Df(x, y) :=f(x + y)+f(x y) f(x) f(y) f( y) for all x, y X. Lemma.1. Let d>0be given. If f : X Y is a mapping such that Df(x, y) =0for all x, y X with x, y d, then f is a quadratic-additive mapping. Proof. Choose x d, then f(0) = 1 (Df(x, x)+df( x, x) Df(3x, x) Df(x, x)) = 0,
Hyers-Ulam-Rassias stability 747 i.e., f(0) = 0. If x d, y < d and z 3d + x, then we have z, y +z, y + z, x z, x + z d. Therefore we get Df(x, y) = Df(x, y +z)+df(x, z)+df(x z, y + z) + Df(x + z, y + z) Df(y + z, z) Df( y z, z) =0 for all x, y X with x d and y <d. By assumption we have Df(x, y) =0 for all x, y X with x d. If x <dand z 3d + y, then we have z, x +z, x + z d. Therefore we get Df(x, y) = Df(x +z, y)+df(z, y)+df(x + z, y z) + Df(x + z, y + z) Df(x + z, z)+df( z, y) =0 for all x, y X with x <d. By assumption we have Df(x, y) =0 for all x, y X. In the following theorem, we can prove the Hyers-Ulam-Rassias stability of the functional equation (1.3). Theorem.. Let d>0 and p<1 be given real numbers. If f : X Y is a mapping such that (.1) Df(x, y) x p + y p for all x, y X with x, y d, then there exists a unique quadratic-additive mapping F : X Y such that f(x) F (x) x p + (4 p +5+3 p )d p (.) if 0 p<1 p 6 for all x with x d and f(x) is itself a quadratic-additive mapping if p<0. Proof. Since f(0) = 1 Df(x, x)+df( x, x) Df(3x, x) Df(x, x) 1 (4 p +5+3 p ) x p for all x X with x d, we have f(0) = 0 if p < 0 and f(0) 1 (4 p +5+3 p )d p if p 0. Therefore we have lim J n f(0) = 0 for p<1. By the definitions of J n f(x) and Df(x, y) we get J n f(x) J n+1 f(x) = Df(n x, n x)+df( n x, n x) 4 n+1 (.3) Df(n x, n x) Df( n x, n x) + f(0) n+ 4 n+1
748 Yang-Hi Lee for all x X\{0} and all nonnegative integer n with n x d. It follows from (.1) and (.3) that J n f(x) J n+m f(x) (.4) = J j f(x) J j+1 f(x) ( ( j+1 +1)Df( j x, j x) 4 j+1 + ( j+1 1)Df( j x, j x) 4 j+1 ) + f(0) 4 { j+1 j j x p + f(0) 3 4 n if 0 p<1, j j x p if p<0 { np x p + (4 p +5+3 p )d p n 1 ( p ) if 0 p<1, 6 4 n np x p if p<0 n 1 ( p ) for all x X\{0} and all nonnegative integer n with n x d. So, it is easy to show that the sequence {J n f(x)} is a Cauchy sequence for all x X\{0}. Since Y is complete and lim J n f(0) = 0, the sequence {J n f(x)} converges for all x X. Hence, we can define a mapping F : X Y by F (x) := lim J n f(x) for all x X. Moreover, letting n = 0 and passing the limit m in (.4), we get the inequality (.) for all x with x d if 0 p<1 and (.5) f(x) F (x) x p p for all x with x d if p<0. From the definition of F, we get 4 n DF(x, y) = lim (Df(n x, n y)+df( n x, n y)) + n (Df(n x, n y) Df( n x, n y)) lim ( ( p)n + n(1 p) )( x p + y p )=0 for all x, y X\{0} and all nonnegative integer n with n x, n y d. By Lemma.1, F is a quadratic-additive mapping. Now, we are going to show that F is unique. Let F : X Y be another quadratic-additive mapping satisfying (.). By Lemma.1 and (.3), it is easy to show that F (0) = 0 and F (x) =J n F (x) for all n N and for all x X. Since F and F are quadratic-additive, replacing x by n x in (.), we
have Hyers-Ulam-Rassias stability 749 F (x) F (x) = J n F (x) J n F (x) 4 n ( (F f)( n x) + (f F )( n x) + (F f)( n x) + (f F )( n x) ) + n ( (F f)( n x) + (F f)( n x)) + (F f)( n x) + (F f)( n x) ) ( np p x p + (4 )( p +5+3 p )d p 3 4 + ) n n for all x X\{0} and all positive integers n with n x d. Taking the limit in the above inequality as n, we can conclude that F (x) =F (x) for all x X. This proves the uniqueness of F. In the case p<0, from (.5), we obtain the inequalities f(x) F (x) Df((k +1)x, kx) DF((k +1)x, kx) + F ((k +1)x) f((k +1)x) + f((k +1)x F ((k +1)x) + f(kx) F (kx) + f( kx) F ( kx) ((k +1) p + k p + (k ) +1)p +4(k +1) p +4k p x p p for all x 0 and all k>0 with kx d. So we conclude that f(x) =F (x) for all x 0 by taking the limit in the above inequality as k. Since f(0) = 0, f is itself a quadratic-additive mapping. Theorem.3. Let p be a given real number with 1 <p<. Iff : X Y is a mapping satisfying (.1) for all x, y X, then there exists a unique quadraticadditive mapping F : X Y such that (.6) for all x X. Proof. Since f(x) F (x) ( 4 + ) x p p p f(0) = 1 Df(0, 0) 0 p =0,
750 Yang-Hi Lee we have f(0) = 0. From the definitions of J n f(x) and Df(x, y), we get (.7) J n f(x) J n+1 f(x) = Df(n x, n x)+df( n x, n x) 4 n+1 ( x + (Df n 1, x ) ( )) x x Df, n+1 n+1 n+1 n+1 for all x X and all nonnegative integer n. It follows from (.1) and (.7) that (.8) J n f(x) J n+m f(x) = J j f(x) J j+1 f(x) ( j 1 j x p + j+1 j 1 x p) np x p n 1 (4 p ) + n+1 x p np ( p ) for all x X. So, it is easy to show that the sequence {J n f(x)} is a Cauchy sequence for all x X. Since Y is complete, we can define a mapping F : X Y by F (x) := lim J n f(x) for all x X. Moreover, letting n = 0 and passing the limit m in (.8), we get the inequality (.6). From the definition of F, we get 4 n DF(x, y) = lim (Df(n x, n y)+df( n x, n y)) ( Df( n x, n y) Df( n x, n y) ) + n lim ( (p )n + n(1 p) )( x p + y p )=0 for all x, y X. So, F is a quadratic-additive mapping. Now, to show that F is unique. Let F : X Y be another quadraticadditive mapping satisfying (.6). By Lemma.1 and (.7), it is easy to show that F (0) = 0 and F (x) =J n F (x) for all n N and all x X. From these
Hyers-Ulam-Rassias stability 751 and (.6), we have F (x) F (x) = J n F (x) J n F (x) 4 n ( (F f)( n x) + (f F )( n x) + (F f)( n x) + (f F )( n x) ) + n ( (F f)( n x) + (F f)( n x)) + (F f)( n x) + (F f)( n x) ) ( ) 4 4 p + 4 ( (p )n + (1 p)n ) x p p for all x X and all positive integer n. Taking the limit in the above inequality as n, we can conclude that F (x) =F (x) for all x X. This proves the uniqueness of F. Theorem.4. Let d>0 and p be given real numbers with p>. Iff : X Y is a mapping satisfying (.1) for all x, y X with x, y d, then there exists a unique quadratic-additive mapping F : X Y such that (.9) for all x d. Proof. Since f(x) F (x) x p p 4 f(0) = 1 Df(0, 0) 0 p =0, we have f(0) = 0. From the definitions of J n f(x) and Df(x, y), we get ( x J n f(x) J n+1 f(x) =( n 1 + n 1 )Df, x ) n+1 ( n+1 ) x +( n 1 n 1 x (.10) )Df, n+1 n+1 for all x X and all nonnegative integer n with n 1 x d. It follows from (.1) and (.10) that J n f(x) J n+m f(x) j+1 j 1 x p n+1 x p (.11) np ( p 4) for all x X and all nonnegative integer n with n 1 x d. So, it is easy to show that the sequence {J n f(x)} is a Cauchy sequence for all x X. Since Y is complete, we can define a mapping F : X Y by F (x) := lim J n f(x)
75 Yang-Hi Lee for all x X. Moreover, letting n = 0 and passing the limit m in (.11), we get the inequality (.9). From the definition of F, we get DF(x, y) = lim ( n 1 Df( n x, n y)+df( n x, n y) ) + ( n 1 Df( n x, n y) Df( n x, n y) ) lim ( ( p)n + (1 p)n )( x p + y p )=0 for all x, y X and all nonnegative integer n with n x, n y d. So, F is a quadratic-additive mapping. Now, to show that F is unique. Let F : X Y be another quadraticadditive mapping satisfying (.9). By Lemma.1 and (.10), it is easy to show that F (0) = 0 and F (x) =J n F (x) for all n N and for all x X. Since F and F are quadratic-additive, we have F (x) F (x) = J n F (x) J n F (x) n 1( (F f)( n x) + (f F )( n x) + (F f)( n x) + (f F )( n x) ) + n 1( (F f)( n x) + (F f)( n x)) + (F f)( n x) + (F f)( n x) ) ( ) 4 x p ( ( p)n + (1 p)n ) x p p 4 for all x X and all positive integer n with n x d. Taking the limit in the above inequality as n, we can conclude that F (x) =F (x) for all x X. This proves the uniqueness of F. References [1] T. Aoki, On the stability of the linear transformation in Banach spaces, J. Math. Soc. Japan (1950), 64 66. [] D. H. Hyers, On the stability of the linear functional equation, Proc. Natl. Acad. Sci. U.S.A. 7 (1941), 4. [3] Y.-H. Lee, On the Hyers-Ulam-Rassias stability of the generalized polynomial function of degree, J. Chuncheong Math. Soc., (009), 01 09. [4] Y.-H. Lee, On the stability of the monomial functional equation, Bull. Korean Math. Soc., 45 (008), 397 403. [5] Y.-H. Lee, On the quadratic additive type functional equations, Int. J. Math. Anal. (Ruse) 7 (013), 1935 1948. [6] Y.-H. Lee and S.-M. Jung, On the quadratic additive type functional equations, Abstr. Appl. Anal. 013 (013) Art. ID 47531, 5pages. [7] Th. M. Rassias, On the stability of the linear mapping in Banach spaces, Proc. Amer. Math. Soc. 7(1978), 97 300. [8] F. Skof, Local properties and approximations of operators, Rend. Sem. Mat. Fis. Milano 53 (1983), 113 19. [9] S.M. Ulam, A Collection of Mathematical Problems, Interscience, New York, 1960. Received: September 5, 013