HYPERSTABILITY OF THE GENERAL LINEAR FUNCTIONAL EQUATION

Bull. Korean Math. So. 52 (2015, No. 6, pp. 1827 1838 http://dx.doi.org/10.4134/bkms.2015.52.6.1827 HYPERSTABILITY OF THE GENERAL LINEAR FUNCTIONAL EQUATION Magdalena Piszzek Abstrat. We give some results on hyperstability for the general linear equation. Namely, we show that a funtion satisfying the linear equation approximately (in some sense must be atually the solution of it. 1. Introdution LetX, Y be normedspaesoverfieldsf, K, respetively. Afuntion f: X Y is linear provided it satisfies the funtional equation (1 f(ax+by = Af(x+Bf(y, x,y X, where a,b F \ {0}, A,B K. We see that for a = b = A = B = 1 in (1 we get the Cauhy equation while the Jensen equation orresponds to a = b = A = B = 1 2. The general linear equation has been studied by many authors, in partiular the results of the stability an be found in [5], [6], [8], [9], [10], [13], [14]. We present some hyperstability results for the equation (1. Namely, we show that, for some natural partiular forms of ϕ, the funtional equation (1 is ϕ-hyperstable in the lass of funtions f: X Y, i.e., eah f: X Y satisfying the inequality f(ax+by Af(x Bf(y ϕ(x,y, x,y X, must be linear. In this way we expet to stimulate somewhat the further researh of the issue of hyperstability, whih seems to be a very promising subjet to study within the theory of Hyers-Ulam stability. The hyperstability results onerning the Cauhy equation an be found in [2], the general linear in [12] with ϕ(x,y = x p + y p, where p < 0. The Jensen equation was studied in [1] and there were reeived some hyperstability results for ϕ(x,y = x p y q, where 0, p,q R, p+q / {0,1}. Reeived Marh 29, 2014; Revised Marh 24, 2015. 2010 Mathematis Subjet Classifiation. 39B82, 39B62, 47H14, 47H10, 47J20. Key words and phrases. linear equation, hyperstability. 1827 2015 Korean Mathematial Soiety

1828 M. PISZCZEK The stability of the Cauhy equation involving a produt of powers of norms was introdued by J. M. Rassias in [15], [16] and it is sometimes alled Ulam- Gǎvruţa-Rassias stability. For more information about Ulam-Gǎvruţa-Rassias stability we refer to [7], [11], [17], [18], [19]. One of the method of the proof is based on a fixed point result that an be derived from [3] (Theorem 1. To present it we need the following three hypothesis: (H1 X is a nonempty set, Y is a Banah spae, f 1,...,f k : X X and L 1,...,L k : X R + are given. (H2 T : Y X Y X is an operator satisfying the inequality k T ξ(x T µ(x L i (x ξ(fi (x µ(f i (x, ξ,µ Y X, x X. i=1 (H3 Λ: R + X R + X is defined by Λδ(x := k L i (xδ(f i (x, δ R X +, x X. i=1 Nowweareinapositiontopresenttheabovementionedfixedpointtheorem. Theorem 1.1. Let hypotheses (H1 (H3 be valid and funtions ε: X R + and ϕ : X Y fulfil the following two onditions T ϕ(x ϕ(x ε(x, x X, ε (x := Λ n ε(x <, x X. Then there exists a unique fixed point ψ of T with Moreover, ϕ(x ψ(x ε (x, x X. ψ(x := lim n T n ϕ(x, x X. The next theorem shows that a linear funtion on X \{0} is linear on the whole X. Theorem 1.2. Let X,Y be normed spaes over F, K, respetively, a,b F\{0}, A,B K. If a funtion f: X Y satisfies (2 f(ax+by = Af(x+Bf(y, x,y X \{0}, then there exist an additive funtion g: X Y satisfying onditions (3 g(bx = Bg(x and g(ax = Ag(x, x X and a vetor β Y with (4 β = (A+Bβ

HYPERSTABILITY OF THE GENERAL LINEAR FUNCTIONAL EQUATION 1829 suh that (5 f(x = g(x+β, x X. Conversely, if a funtion f: X Y has the form (5 with some β Y, an additive g: X Y suh that (3 and (4 hold, then it satisfies the equation (1 (for all x,y X. Proof. Assume that f fulfils (2. Replaing x by bx and y by ax in (2 we get (6 f(0 = Af(bx+Bf( ax, x X \{0}. Next with x replaed by bx and y by ax in (2 we have (7 f(2abx = Af(bx+Bf(ax, x X \{0}. Letf = f e +f o, wheref e, f o denotethe evenandthe oddpartoff, respetively. It is obvious that f e,f o satisfy (2, (6 and (7. First we show that f o is additive. Aording to (6 and (7 for the odd part of f we have Af o (bx = Bf o (ax, x X and f o (2abx = Af o (bx+bf o (ax, x X. Thus ( x ( x (8 f o (x = 2Bf o = 2Af o, x X. 2b 2a By (8 and (2 ( x ( y ( f o (x+f o (y = 2Af o +2Bf o = 2f o a x 2a 2b 2a +b y (9 2b ( x+y = 2f o, x,y X \{0}. 2 Fix z X \{0} and write X z := {pz : p > 0}. Then X z is a onvex set, there exist an additive map g z : X z Y and a onstant β z Y suh that f o (x = g z (x+β z, x X z. We observe that ( 3pz pz g z (pz+β z = f o (pz = f o = f o(3pz f o (pz 2 2 = g z(3pz g z (pz = g z (pz, p > 0, 2 whih means that β z = 0. Hene ( 1 ( 1 f o 2 z = g z 2 z = 1 2 g z(z = 1 2 f o(z. Therefore with (9 we obtain ( x+y f o 2 = f o(x+f o (y, x,y X 2

1830 M. PISZCZEK and as f o (0 = 0, f o is additive. Using additivity of f o and (8 we obtain ( x f o (bx = 2Bf o = Bf o (x, x X 2 and ( x f o (ax = 2Af o = Af o (x, x X, 2 whih means that (3 holds with g = f o. Using (6 and (7 for the even part of f we obtain f e (0 = f e (2abx, x X \{0}, whih means that f e is a onstant funtion and (4 holds with β := f e (x. For the proof of the onverse, assume that a funtion f: X Y has the form (5 with some β Y, an additive g: X Y suh that (3 and (4 hold. Then for all x,y X whih finishes the proof. f(ax+by = g(ax+by+β = g(ax+g(by+β = Ag(x+Bg(y+(A+Bβ = Af(x+Bf(y, 2. Hyperstability results Theorem 2.1. Let X, Y be normed spaes over F, K, respetively, a,b F\{0}, A,B K\{0}, 0, p,q R, p+q < 0 and f: X Y satisfies (10 f(ax+by Af(x Bf(y x p y q, x,y X \{0}. Then f is linear. Proof. First we notie that without loss of generality we an assume that Y is a Banah spae, beause otherwise we an replae it by its ompletion. Sine p + q < 0, one of p, q must be negative. Assume that q < 0. We observe that there exists m 0 N suh that (11 1 a+bm p+q + B m p+q < 1 for m m 0. A A Fix m m 0 and replae y by mx in (10. Thus and (12 Write f(ax+bmx Af(x Bf(mx x p mx q, x X \{0} 1 A f((a+bmx B A f(mx f(x mq x p+q, x X \{0}. T ξ(x := 1 A ξ((a+bmx B A ξ(mx, ε(x := mq x p+q, x X \{0},

HYPERSTABILITY OF THE GENERAL LINEAR FUNCTIONAL EQUATION 1831 then (12 takes the form T f(x f(x ε(x, x X \{0}. Define Λη(x := 1 η((a+bmx+ B η(mx, x X \{0}. A A Then it is easily seen that Λ has the form desribed in (H3 with k = 2 and f 1 (x = (a+bmx, f 2 (x = mx, L 1 (x = 1, L 2(x = B A for x X \{0}. Moreover, for every ξ,µ Y X\{0}, x X \{0} T ξ(x T µ(x = 1 A ξ((a+bmx B A ξ(mx 1 A µ((a+bmx+ B A µ(mx 1 (ξ µ((a+bmx + B (ξ µ(mx A A = 2 L i (x (ξ µ(f i (x, i=1 so (H2 is valid. By (11 we have ε (x := = Λ n ε(x mq( 1 a+bm p+q + B m p+q n x p+q A A mq x p+q = 1 1 A a+bm p+q B x X \{0}. A mp+q, Hene,aordingtoTheorem1.1thereexistsauniquesolutionF: X \{0} Y of the equation suh that Moreover, (13 F(x = 1 A F((a+bmx B F(mx, x X \{0} A mq x p+q f(x F(x 1 1 A a+bm p+q B x X \{0}. A mp+q, We show that F(x := lim n (T n f(x, x X \{0}. T n f(ax+by AT n f(x BT n f(y ( 1 a+bm p+q + B m p+q n x p y q, x,y X \{0} A A

1832 M. PISZCZEK for every n N 0. If n = 0, then (13 is simply (10. So, take r N 0 and suppose that (13 holds for n = r. Then T r+1 f(ax+by AT r+1 f(x BT r+1 f(y = 1 A T r f((a+bm(ax+by B A T r f(m(ax+by A 1 A T r f((a+bmx+a B A T r f(mx B 1 A T r f((a+bmy+b B A T r f(my ( 1 a+bm p+q + B m p+q r 1 (a+bmx p (a+bmy q A A A ( 1 + a+bm p+q + B m p+q r B mx p my q A A A ( 1 = a+bm p+q + B m p+q r+1 x p y q, x,y X \{0}. A A Thus, by indution we have shown that (13 holds for every n N 0. Letting n in (13, we obtain that F(ax+by = AF(x+BF(y, x,y X \{0}. In this way, with Theorem 1.2, for every m m 0 there exists a funtion F satisfying the linear equation (1 suh that mq x p+q f(x F(x 1 1 A a+bm p+q B x X \{0}. A mp+q, It follows, with m, that f is linear. In similar way we an prove the following theorem. Theorem 2.2. Let X, Y be normed spaes over F, K, respetively, a,b F \ {0}, A,B K \ {0}, 0, p,q R, p + q > 0 and f: X Y satisfies (10. If q > 0 and a p+q, or p > 0 and b p+q B, then f is linear. Proof. We presentthe proofonlywhen q > 0beausethe seondaseissimilar. Let q > 0 and a p+q < 1. Replaing y by a bmx, where m N, in (10 we get (( f a a ( x Af(x Bf a m bm x x p a bm x q, x X \{0}, thus 1 (( A f a a x B ( m A f a bm x (14 f(x a q x p+q, x X \{0}. bm For x X \{0} we define T m ξ(x := 1 (( A ξ a a x B ( m A ξ a bm x,

HYPERSTABILITY OF THE GENERAL LINEAR FUNCTIONAL EQUATION 1833 ε m (x := a q x p+q, bm Λ m η(x := 1 (( η a a x + B ( η a A m A bm x, and as in Theorem 2.1 we observe that (14 takes the form T m f(x f(x ε m (x, x X \{0} and Λ m has the form desribed in (H3 with k = 2 and f 1 (x = (a a m x, f 2 (x = a bm x, L 1(x = 1, L 2(x = B A for x X \{0}. Moreover, for every ξ,µ Y X\{0}, x X \{0} 2 T m ξ(x T m µ(x L i (x (ξ µ(f i (x, so (H2 is valid. Next we an find m 0 N, suh that a p+q 1 1 p+q + B b p+q ( 1 p+q < 1 for m Nm0. m A a m Therefore ε m(x := Λ n mε m (x i=1 = a q ( a x p+q p+q 1 1 p+q + B b p+q ( 1 p+q n bm m A a m = a bm q x p+q 1 a p+q 1 1 m p+q B A b a p+q ( 1 m p+q, m N m 0, x X \{0}. Hene, aording to Theorem 1.1, for eah m N m0 there exists a unique solution F m : X \{0} Y of the equation F m (x = 1 (( A F m a a x B ( m A F m a bm x, x X \{0} suh that f(x F m (x a bm q x p+q x X \{0}. 1 a p+q 1 1 m p+q B A b a p+q ( 1 m p+q, Moreover, F m (ax+by = AF m (x+bf m (y, x,y X \{0}. In this way we obtain a sequene (F m m Nm0 of linear funtions suh that f(x F m (x a bm q x p+q x X \{0}. 1 a p+q 1 1 m p+q B A b a p+q ( 1 m p+q,

1834 M. PISZCZEK So, with m, f is linear on X \{0} and by Theorem 1.2 f is linear. Let q > 0 and a < 1. Replaing x by ( 1 p+q a 1 1 am x and y by bmx, where m N, in (10 we get ( ( 1 a 1 Whene f a x+b 1 am bm x ( 1 a 1 x p 1 am bm x q, x X \{0}. (( 1 f(x Af a 1 am 1 1 1 1 a p b q m (( 1 Af a 1 ( 1 x Bf am bm x ( 1 x Bf bm x p 1 q x p+q, x X \{0}. m For x X \{0} we define (( 1 T m ξ(x := Aξ a 1 ( 1 x +Bξ am bm x, ε m (x := 1 1 1 1 a p b q p 1 q x p+q, m m (( 1 Λ m η(x := η a 1 ( 1 x + B η am bm x, and as in Theorem 2.1 we observe that (14 takes form T m f(x f(x ε m (x, x X \{0} and Λ m has the form desribed in (H3 with k = 2 and f 1 (x = ( 1 a 1 am x, f 2 (x = 1 bm x, L 1(x =, L 2 (x = B for x X \{0}. Moreover, for every ξ,µ Y X\{0}, x X \{0} T m ξ(x T m µ(x 2 L i (x (ξ µ(f i (x, so (H2 is valid. Next we an find m 0 N, suh that 1 1 a p+q p+q + B 1 m b p+q p+q < 1 for m N m0. m Therefore ε m (x := Λ n m ε m(x = ε m (x i=1 ( 1 1 a p+q p+q + B 1 m b p+q p+q n m

HYPERSTABILITY OF THE GENERAL LINEAR FUNCTIONAL EQUATION 1835 = 1 a p 1 b q 1 1 m p 1 m q x p+q 1 a p+q 1 1 m p+q B b p+q 1 m p+q, m N m 0, x X \{0}. Hene, aording to Theorem 1.1, for eah m N m0 there exists a unique solution F m : X \{0} Y of the equation (( 1 F m (x = AF m a 1 ( 1 x +BF m am bm x, x X \{0} suh that 1 1 a (15 f(x F m (x p b 1 1 q m p 1 m q x p+q x X \{0}. 1 a 1 1 p+q m p+q B b 1 p+q m p+q, Moreover, F m (ax+by = AF m (x+bf m (y, x,y X. In this way we obtain a sequene (F m m Nm0 of linear funtions suh that (15 holds. It follows, with m, that f is linear. Theorem 2.3. Let X, Y be normed spaes over F, K, respetively, a,b F\{0}, A,B K\{0}, 0, p,q > 0, and f: X Y satisfies (16 f(ax+by Af(x Bf(y x p y q, x,y X. If a p+q or b p+q B, then f is linear. Proof. Of ourse this theorem follows from Theorem 2.2 but as p,q are positive we an set 0 in (16 and get an auxiliary equalities. In this way we obtain another proof whih we present in the first ase. Assume that a p+q <. Setting x = y = 0 in (16 we get (17 f(0(1 A B = 0. With y = 0 in (16 we have thus f(ax = Af(x+bf(0, x X ( x f(x = Af +Bf(0, x X. a Using the last equality, (16 and (17 we get ( ax+by ( x ( y Af AAf BAf x a a p y q, x,y X. a Replaing x by ax, y by ay and dividing the last inequality by we obtain f(ax+by Af(x Bf(y a p+q x p y q, x,y X. By indution it is easy to get ( a p+q n x f(ax+by Af(x Bf(y p y q, x,y X. Whene, with n, f(ax+by = Af(x+Bf(y for x,y X.

1836 M. PISZCZEK Inthease < a p+q, weusetheequationf(x = 1 A f(ax b a f(0together with (16 and (17. The following examples show that the assumption in the above theorems are essential. Example 2.4. Let f: R R be defined as f(x = x 2. Then f satisfies f(x+y f(x f(y 2 x y, x,y R, but f does not satisfy the Cauhy equation. Example 2.5. More generally a quadrati funtion f(x = x 2, x R satisfies f(ax+by Af(x Bf(y 2 ab x y, x,y R, where A = a 2, B = b 2, but f does not satisfy the linear equation (1. Example 2.6. A funtion f(x = x, x R satisfies ( x+y f f(x+f(y x 1 1 2 y 2, x,y R, 2 2 but f does not satisfy the Jensen equation. It is known that for p = q = 0 we have the stability result and a funtion f(x = x+, x R satisfies f(x+y f(x f(y, x,y R but it is not linear. To the end we show simple appliation of the above theorems. Corollary 2.7. Let X, Y be normed spaes over F, K, respetively, a,b F\{0}, A,B K\{0}, 0, p,q R, H: X 2 Y, H(w,z 0 for some z,w X and (18 H(x,y x p y q, x,y X \{0}, where 0, p,q R. If one of the following onditions (1 p+q < 0, (2 q > 0 and a p+q, (3 p > 0 and b p+q B holds, then the funtional equation (19 h(ax+by = Ah(x+Bh(y+H(x,y, x,y X has no solutions in the lass of funtions h: X Y. Proof. Suppose that h: X Y is a solution to (19. Then (10 holds, and onsequently, aording to the above theorems, h is linear, whih means that H(w,z = 0. This is a ontradition.

HYPERSTABILITY OF THE GENERAL LINEAR FUNCTIONAL EQUATION 1837 Example 2.8. The funtions f: R R defined as f(x = x 2 and H: R 2 R given by H(x,y = 2xy satisfy the equation f(x+y = f(x+f(y+h(x,y, x,y R and do not fulfill any ondition (1 (3 of Corollary 2.7. Remark 2.9. We notie that our results orrespond with the new results from hyperstability, for example in [4] was proved that linear equation is ϕ-hyperstabile with ϕ(x,y = x p y q, but there was onsidered only the ase when,p,q [0,+ (see Theorem 20. Referenes [1] A. Bahyryz and M. Piszzek, Hyperstability of the Jensen funtional equation, Ata Math. Hungar. 142 (2014, no. 2, 353 365. [2] J. Brzdȩk, Hyperstability of the Cauhy equation on restrited domains, Ata Math. Hungar. 41, (2013, no. 1-2, 58 67. [3] J. Brzdȩk, J. Chudziak, and Zs. Páles, A fixed point approah to stability of funtional equations, Nonlinear Anal. 74 (2011, no. 17, 6728 6732. [4] J. Brzdȩk and K. Ciepliński, Hyperstability and Superstability, Abstr. Appl. Anal. 2013 (2013, Art. ID 401756, 13 pp. [5] J. Brzdȩk and A. Pietrzyk, A note on stability of the general linear equation, Aequationes Math. 75 (2008, no. 3, 267 270. [6] P. Gǎvruţa, A generalization of the Hyers-Ulam-Rassias stability of approximately additive mapping, J. Math. Anal. Appl. 184 (1994, no. 3, 431 436. [7], An answer to a question of John M. Rassias onerning the stability of Cauhy equation, Advanes in Equations and Inequalities, Hadroni Math. Ser. 1999. [8] D. H. Hyers, G. Isa, and Th. M. Rassias, Stability of Funtional Equations in Several Variables, Birkhäuser, Boston-Basel-Berlin, 1998. [9] S. M. Jung, Hyers-Ulam-Rassias Stability of Funtional Equations in Mathematial Analysis, Hadroni Press, Palm Harbor, 2001. [10] M. Kuzma, An introdution to the Theory of Funtional Equation and Inequalities, PWN, Warszawa-Kraków-Katowie, 1985. [11] P. Nakmahahalasint, Hyers-Ulam-Rassias stability and Ulam-Gǎvruţa-Rassias stabilities of additive funtional equation in several variables, Int. J. Math. Si. 2007 (2007, Art. ID 13437, 6pp. [12] M. Piszzek, Remark on hyperstability of the general linear equation, Aequationes Math. 88 (2014, no. 1-2, 163 168. [13] D. Popa, Hyers-Ulam-Rassias stability of the general linear equation, Nonlinear Funt. Anal. Appl. 7 (2002, no. 4, 581 588. [14], On the stability of the general linear equation, Results Math. 53 (2009, no. 3-4, 383 389. [15] J. M. Rassias, On approximation of approximately linear mappings, J. Funt. Anal. 46 (1982, no. 1, 126 130. [16], Solution of a problem of Ulam, J. Approx. Theory 57 (1989, no. 3, 268 273. [17] K. Ravi and M. Arunkumar, On the Ulam-Gǎvruţa-Rassias stability of the orthogonally Euler-Lagrange type funtional equation, Int. J. Appl. Math. Stat. 7 (2007, 143 156. [18] K. Ravi and B. V. Senthil Kumar, Ulam-Gǎvruţa-Rassias stability of Rassias Reiproal funtional equation, Glob. J. App. Math. Si. 3 (2010, 57 79. [19] M. Ait Sibaha, B. Bouikhalene, and E. Elqorahi, Ulam-Gǎvruţa-Rassias stability of a linear funtional equation, Int. J. Appl. Math. Stat. 7 (2007, 157 166.

1838 M. PISZCZEK Institute of Mathematis Pedagogial University Podhora żyh 2 PL-30-084 Kraków, Poland E-mail address: magdap@up.krakow.pl