Noname manuscript No. (will be inserted by the editor) Cubic Derivations on Banach Algebras Abasalt Bodaghi 3 Accepted inActaMathematica Vietnamica 1 0 2 Abstract Let A be a Banach algebra and X be a Banach A-bimodule. A n mapping D : A −→ X is a cubic derivation if D is a cubic homogeneous a mapping, that is D is cubic and D(λa) = λ3D(a) for any complex number J λ and all a ∈ A, and D(ab) = D(a)·b3 +a3 ·D(b) for all a,b ∈ A. In this 4 1 paper,weprovethestabilityofacubicderivationwithdirectmethod.Wealso employafixedpointmethodtoestablishofthestabilityandthesuperstability ] for cubic derivations. A F Keywords Banach algebra · Cubic derivation · Stability · Superstability h. Mathematics Subject Classification (2010) 39B52 · 47B47 · 39B72 · t 46H25 a m [ 1 Introduction 1 v In 1940, Ulam [19] posed the following question concerning the stability of 8 group homomorphisms: Under what condition does there is an additive map- 8 ping near an approximately additive mapping between a group and a metric 8 group? One year later, Hyers [9] answeredthe problem of Ulam under the as- 2 sumptionthatthegroupsareBanachspaces.Thisproblemforlinearmapping . 1 onBanachspaceswassolvedbyJ.M.Rassiasin[15].Ageneralizedversionof 0 the theorem of Hyers for approximately linear mappings was givenby Th. M. 3 Rassias [16]. Subsequently, the stability problems of various functional equa- 1 : tion have been extensively investigated by a number of authors (for example, v [2],[12]and[14]).Inparticular,oneofthefunctionalequationswhichhasbeen i X studied frequently is the cubic functional equation: r f(2x+y)+f(2x−y)=2f(x+y)+2f(x−y)+12f(x) (1) a AbasaltBodaghi DepartmentofMathematics, GarmsarBranch,IslamicAzadUniversity,Garmsar,Iran Tel.:+98-232-4225010 E-mail:[email protected] 2 AbasaltBodaghi The cubic function f(x) = ax3 is a solution of this functional equation. The stability of the functional equation (1) has been considered on different spaces by a number of writers (for instance, [11] and [17]). In 2003,Ca˘dariu and Radu applied a fixed point method to the investiga- tion of the Jensen functional equation. They presented a short and a simple proof for the Cauchy functional equation and the quadratic functional equa- tion in [4] and [3], respectively. After that, this method has been applied by many authors to establish of miscellaneous functional equations (see [1], [7] and [13]). In [8], Eshaghi Gordji et al. introduced the concept of a cubic derivation which is a different notion of the currentpaper. In fact, they did not consider the homogeneous property of such derivations. In that paper, the authors studied the stability of cubic derivations on commutative Banach algebras. The stability and the superstability of cubic double centralizers and cubic multipliers on Banach algebras has been earlier proved in [10]. In this paper, we prove the stability of cubic derivations on Banach alge- bras. An example of such derivations is indicated as well. Using a fixed point theorem, we also show that a cubic derivation can be superstable. 2 Stability of cubic derivations Let A be a Banach algebra. A Banach space X which is also a left A-module is said to be a left Banach A-module if there is k >0 such that ka·xk≤kkakkxk. Similarly,arightBanachA-moduleandaBanachA-bimodule aredefined. Throughoutthis paper,weassume thatAisa Banachalgebra,X is aBanach n−times A-bimodule and denote A×A×...×A by An. For a natural number n , we z }| { 0 define T 1 := {eiθ ; 0 ≤ θ ≤ 2nπ} and denote T 1 by T when n0 = 1. We n0 0 n0 alsodenote thesetofallpositiveintegersnumbers,realnumbersandcomplex numbers by N, R and C, respectively. First let us show by a example that the cubic derivations exist on Banachalgebras.Indeed, the following example is taken from [8] with the non-trivial module actions while in the mentioned paper the left module action is zero. Example Let A be a Banach algebra. Consider 0 AAA   0 0 AA T := . 0 0 0 A   0 0 0 0 CubicDerivationsonBanachAlgebras 3 Then T is a Banachalgebrawith the sum andproduct being givenby the usual 4×4 matrix operations and with the following norm: 0 a a a 1 2 3   6 0 0 a a k 4 5 k= ka k (a ∈A). 0 0 0 a6 Xj=1 j j 0 0 0 0  So 0 A∗ A∗ A∗ ∗ 0 0 A∗ A∗ T =0 0 0 A∗,   0 0 0 0  is the dual of T equipped with the following norm: 0f f f 1 2 3   k 0 0 f4 f5 k=Max{kf k:0≤j ≤6} (f ∈A∗). 0 0 0 f  j j  6 0 0 0 0  0a a a 0x x x 1 2 3 1 2 3     0 0 a a 0 0 x x Suppose that A = 4 5 ,X = 4 5 ∈ T and F = 0 0 0 a6 0 0 0 x6 0 0 0 0  0 0 0 0  0f f f 1 2 3   0 0 f f 4 5 ∈T∗ inwhichf ∈A∗ anda ,x ∈A (0≤j ≤6).Considerthe 0 0 0 f  j j j  6 0 0 0 0  ∗ module actions of T on T as follows: 6 6 hF ·A,Xi= f(a x ), hA·F,Xi= f(x a ). X j j X j j j=1 j=1 0g g g 1 2 3   0 0 g g Then T∗ is a Banach T-bimodule. Let G = 4 5 ∈ T∗. Define 0 0 0 0 g   6 0 0 0 0  D :T −→T∗ via D(A)=G ·A3−A3·G (A∈T). 0 0 0a a a 0 b b b 1 2 3 1 2 3     0 0 a a 0 0 b b Given A= 4 5 ,B = 4 5 ∈T, we have 0 0 0 a6 0 0 0 b6 0 0 0 0  0 0 0 0  4 AbasaltBodaghi hD(2A+B),Xi= hG ·(2A+B)3−(2A+B)3·G ,Xi 0 0 = g ((2a +b )(2a +b )(2a +b )x ) 3 1 1 4 4 6 6 3 − g (x (2a +b )(2a +b )(2a +b )). (2) 3 3 1 1 4 4 6 6 Similarly, hD(2A−B),Xi= g ((2a −b )(2a −b )(2a −b )x ) 3 1 1 4 4 6 6 3 − g (x (2a −b )(2a −b )(2a −b )). (3) 3 3 1 1 4 4 6 6 On the other hand, h2D(A+B),Xi= h2G ·(A+B)3−2(A+B)3·G ,Xi 0 0 = 2g ((a +b )(a +b )(a +b )x ) 3 1 1 4 4 6 6 3 − 2g (x (a +b )(a +b )(a +b )), (4) 3 3 1 1 4 4 6 6 and h2D(A−B),Xi= h2G ·(A−B)3−2(A−B)3·G ,Xi 0 0 = 2g ((a −b )(a −b )(a −b )x ) 3 1 1 4 4 6 6 3 − 2g (x (a −b )(a −b )(a −b )). (5) 3 3 1 1 4 4 6 6 Also, h12D(A),Xi =h12G ·A3−12A3·G ,Xi 0 0 =12g (a a a x )−12g (x a a a ). (6) 3 1 4 6 3 3 3 1 4 6 If follows from (2)-(6) that D(2A+B)+D(2A−B)=2D(A+B)+2D(A−B)+12D(A) for all A,B ∈T. This shows that D is a cubic mapping. It is easy to see that D(λA) = λ3D(A) for all A ∈ T and λ ∈ C. Thus D is a cubic homogeneous mapping. Since T4 = {0}, we have D(AB) = D(A)·B3+A3·D(B) = 0 for all A,B ∈T. Hence, D is a cubic derivation. Now, we are going to prove the stability of cubic derivations on Banach algebras. Theorem 1 Suppose that f :A−→X is a mapping with f(0)=0 for which there exists a function ϕ:A4 −→[0,∞) such that ∞ 1 ϕ(a,b,c,d):= ϕ(2ka,2kb,2kc,2kd)<∞ (7) X 8k e k=0 kf(2λa+λb)+f(2λa−λb)−2λ3f(a+b)−2λ3f(a−b)−12λ3f(a)k CubicDerivationsonBanachAlgebras 5 ≤ϕ(a,b,0,0) (8) kf(cd)−f(c)·d3−c3·f(d)k≤ϕ(0,0,c,d) (9) for all λ ∈ T 1 and all a,b,c,d ∈ A. Also, if for each fixed a ∈ A the map- n0 pings t 7→ f(ta) from R to X is continuous, then there exists a unique cubic derivation D :A−→X satisfying 1 kf(a)−D(a)k≤ ϕ(a,0,0,0) (10) 16 e for all a∈A. Proof Putting b=0 and λ=1 in (8) , we have 1 1 k f(2a)−f(a)k≤ ϕ(a,0,0,0) (11) 8 16 for all a ∈ A (indeed 1 ∈ T1 for all n ∈ N). We replace a by 2a in (11) and n continue this method to get f(2na) 1 n−1ϕ(2ka,0,0,0) (cid:13) −f(a)(cid:13)≤ (12) (cid:13) 8n (cid:13) 16 X 8k (cid:13) (cid:13) k=0 (cid:13) (cid:13) On the other hand, we can use induction to obtain f(2na) f(2ma) 1 n−1 ϕ(2ka,0,0,0) (cid:13) − (cid:13)≤ (13) (cid:13) 8n 8m (cid:13) 16 X 8k (cid:13) (cid:13) k=m (cid:13) (cid:13) for all a ∈ A, and n > m ≥ 0. It follows from (7) and (13) that sequence f(2na) is Cauchy. Since A is complete, this sequence convergence to the n 8n o map D, that is f(2na) lim =D(a) (14) n→∞ 8n Takingthe limit asntendto infinity in(12)andapplying(14),wecansee that the inequality (10) holds. Now, replacing a,b by 2na,2nb, respectively in (8), we get f(2n(2λa+b)) f(2n(2λa−b)) f(2n(a+b)) k − −2λ3 8n 8n 8n f(2n(a−b)) f(2na) ϕ(2na,2nb,0,0) −2λ3 −12λ3 k≤ . 8n 8n 8n Letting the limit as n−→∞, we obtain D(2λa+λb)+D(2λa−λb)=2λ3D(a+b)+2λ3D(a−b)+12λ3f(a)(15) 6 AbasaltBodaghi foralla,b∈Aandallλ∈T 1 .Iffollowsfrom(15)thatD isacubicmapping n0 when λ =1. Letting b =0 in (15), we get D(λa) = λ3D(a) for all a ∈A and iθ λ∈T 1 .Now,letλ=eiθ ∈T. We setλ0 =en0,thusλ0 belongsto T 1 and n0 n0 D(λa)=D(λn0a)=λ3n0D(a)=λ3D(a) for all a∈A. Under the assumption 0 0 that f(ta) is continuous in t∈R for each fixed a∈A, by the same reasoning as in the proof of [5], D(λa)=λ3D(a) for all λ∈R and a∈A. So, λ λ3 λ3 D(λa)=D( |λ|a)= D(|λ|a)= |λ|3D(a)=λ3D(a), |λ| |λ|3 |λ|3 foralla∈Aandλ∈C (λ6=0).Therefore,D iscubichomogeneousmapping. If we replace c,d by 2nc,2nd respectively in (9), we have f(22ncd) f(2nc) f(2nd) ϕ(0,0,2nc,2nd) ϕ(0,0,2nc,2nd) k − ·d3−c3· k≤ ≤ . 82n 8n 8n 82n 8n forallc,d∈A.Takingthelimitasn−→∞,wegetD(cd)=D(c)·d3+c3·D(d), for all c,d∈A. This shows that D is a cubic derivation. Now, let D′ : A −→ X be another cubic derivation satisfying (10). Then we have 1 kD(a)−D′(a)k = kD(2na)−D′(2na)k 8n 1 ≤ (kD(2na)−f(2na)k+kf(2na)−D′(2na)k) 8n 1 ≤ ϕ(2na,0,0,0) 8n+1 ∞ e 1 1 = ϕ(2n+ka,0,0,0) 8X8n+k k=0 ∞ 1 1 = ϕ(2ka,0,0,0) 8 X 8k k=n for all a∈A. By letting n−→∞ in the preceding inequality, we immediately find the uniqueness of D. This completes the proof. Corollary 1 Let δ,r be positive real numbers with r <3, and let f :A−→X be a mapping with f(0)=0 such that kf(2λa+λb)+f(2λa−λb)−2λ3f(a+b)−2λ3f(a−b)−12λ3f(a)k ≤δ(kakr+kbkr) (16) kf(cd)−f(c)·d3−c3·f(d)k≤δ(kckr+kdkr) (17) CubicDerivationsonBanachAlgebras 7 for all λ∈T 1 and all a,b,c,d∈A. Then there exists a unique cubic deriva- n0 tion D :A−→X satisfying δ kf(a)−D(a)k≤ kakr (18) 2(8−2r) for all a∈A. Proof It follows from Theorem 1 by taking ϕ(a,b,c,d)=δ(kakr+kbkr+kckr+kdkr). Theorem 2 Suppose that f :A−→X is a mapping with f(0)=0 for which there exists a function ϕ:A4 −→[0,∞) satisfying (8), (9) and ∞ ϕ(a,b,c,d):= 8kϕ(2−ka,2−kb,2−kc,2−kd)<∞, X e k=1 for all a,b,c,d∈A. Also, if for each fixed a∈A the mappings t7→f(ta) from R to A is continuous, then there exists a unique cubic derivation D :A−→X satisfying 1 kf(a)−D(a)k≤ ϕ(a,0,0,0) (19) 16 e for all a∈A. Proof Putting b=0 and λ=1 in (8), we have 1 kf(2a)−8f(a)k≤ ϕ(a,0,0,0) (20) 2 for all a∈A. We replace a by a in (20) to obtain 2 a 1 a kf(a)−8f( )k≤ ϕ( ,0,0,0) (21) 2 2 2 Using triangular inequality and proceeding this way, we have n a 1 a f(a)−8nf( ) ≤ 8kϕ( ,0,0,0) (22) (cid:13)(cid:13) 2n (cid:13)(cid:13) 16X 2k (cid:13) (cid:13) k=1 If we show that the sequence 8nf( a ) is Cauchy, then it will be con- vergent by the completeness of A.(cid:8)For th2ins,(cid:9)replace a by a in (22) and then 2m multiply both side by 8m, we get n a a 1 a 8mf( )−8m+nf( ) ≤ 8k+mϕ( ,0,0,0) (cid:13)(cid:13) 2m 2m+n (cid:13)(cid:13) 16X 2k+m (cid:13) (cid:13) k=1 m+n 1 a = 8kϕ( ,0,0,0) 16 X 2k k=m+1 8 AbasaltBodaghi for all a ∈ A, and n > m ≥ 0. Thus the mentioned sequence is convergent to the map D, i.e., a D(a)= lim 8nf( ). n→∞ 2n Now, similar to the proof of Theorem 1, we can continue the rest of the proof. Corollary 2 Let δ,r be positive real numbers with r >3, and let f :A−→X be a mapping with f(0) = 0 satisfying (16), (17). Then there exists a unique cubic derivation D :A−→X satisfying δ kf(a)−D(a)k≤ kakr (23) 2(2r−8) for all a∈A. Proof The result follows from Theorem 2 by putting ϕ(a,b,c,d)=δ(kakr+kbkr+kckr+kdkr). 3 A fixed point approach In this section, we prove the stability and the superstability for cubic deriva- tions on Banach algebras by using a fixed point theorem. First, we bring the following fixed point theorem which is proved in [6]. This theorem plays a fundamental role to achieve our purpose in this section (an extension of the result was given in [18]). Theorem 3 (The fixedpoint alternative) Let (∆,d) be a complete generalized metric space and J :∆−→∆ be a mapping with a Lipschitz constant L<1. Then, for each element α∈∆, either d(Jnα,Jn+1α)=∞ for all n≥0, or there exists a natural number n such that: 0 (i) d(Jnα,Jn+1α)<∞ for all n≥n ; 0 (ii) the sequence {Jnα} is convergent to a fixed point β∗ of J; (iii) β∗ is the unique fixed point of J in the set ∆ = {β ∈ ∆ : d(Tn0α,β) < 1 ∞}; (iv) d(β,β∗)≤ 1 d(β,Jβ) for all β ∈∆ . 1−L 1 Theorem 4 Let f :A−→X be a continuous mapping with f(0)=0 and let φ:A2 −→[0,∞) be a continuous function such that kf(2λa+λb)+f(2λa−λb)−2λ3f(a+b)−2λ3f(a−b)−12λ3f(a)k ≤φ(a,b) (24) kf(ab)−f(a)·b3−a3·f(b)k≤φ(a,b) (25) CubicDerivationsonBanachAlgebras 9 for all λ∈T 1 and all a,b∈A. If there exists a constant k ∈(0,1), such that n0 φ(2a,2b)≤8kφ(a,b) (26) for all a,b ∈ A, then there exists a unique cubic derivation D : A −→ X satisfying 1 kf(a)−D(a)k≤ φ(a,0) (27) 16(1−k) for all a∈A. Proof Toachieveourgoal,wemakethe conditionsofTheorem3.Weconsider the set ∆={g :A−→X| g(0)=0} and define the mapping d on ∆×∆ as follows: d(g,h):=inf{c∈(0,∞):kg(a)−h(a)k≤cφ(a,0), (∀a∈A)}, ifthereexistsuchconstantc,andd(g,h)=∞,otherwise.Similartothe proof of [2, Theorem 2.2], we can show that d is a generalized metric on ∆ and the metric space (∆,d) is complete. Now, we define the mapping J :∆−→∆ by 1 Jh(a)= h(2a), (a∈A). (28) 8 If g,h∈∆ such that d(g,h)<c, by definition of d and J, we have 1 1 1 (cid:13) g(2a)− h(2a)(cid:13)≤ cφ(2a,0) (cid:13)8 8 (cid:13) 8 (cid:13) (cid:13) (cid:13) (cid:13) for all a∈A. Applying (26), we get 1 1 (cid:13) g(2a)− h(2a)(cid:13)≤ckφ(a,0) (cid:13)8 8 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) for all a ∈ A. The above inequality shows that d(Jg,Jh) ≤ kd(g,h) for all g,h ∈ ∆. Hence, J is a strictly contractive mapping on ∆ with a Lipschitz constant k. Here, we prove that d(Jf,f) < ∞. Putting b = 0 and λ = 1 in (24), we obtain k2f(2a)−16f(a)k≤φ(a,0) for all a∈A. Hence 1 1 k f(2a)−f(a)k≤ φ(a,0) (29) 8 16 for all a ∈ A. We conclude from (29) that d(Jf,f) ≤ 1 . It follows from 16 Theorem 3 that d(Jng,Jn+1g)<∞ for all n≥0, and thus in this Theorem we have n = 0. Therefore the parts (iii) and (iv) of Theorem 3 hold on the 0 10 AbasaltBodaghi whole ∆. Hence there exists a unique mapping D :A−→X such that D is a fixed point of J and that Jnf →D as n→∞. Thus f(2na) lim =D(a) (30) n→∞ 8n for all a∈A, and so 1 1 d(f,D)≤ d(Jf,f)≤ . 1−k 16(1−k) The above equalities show that (27) is true for all a ∈ A. Now, it follows from (26) that φ(2na,2nb) lim =0. (31) n→∞ 8n Replacing a and b by 2na and 2nb respectively in (24), we get f(2n(2λa+b)) f(2n(2λa−b)) f(2n(a+b)) k − −2λ3 8n 8n 8n f(2n(a−b)) f(2na) φ(2na,2nb) −2λ3 −12λ3 k≤ . 8n 8n 8n Taking the limit as n−→∞, we obtain D(2λa+λb)+D(2λa−λb)=2λ3D(a+b)+2λ3D(a−b)+12λ3D(a)(32) for all a,b ∈ A and all λ ∈ T 1 . By (32), D is a cubic mapping when λ = 1. n0 Letting b = 0 in (32), we get D(λa) = λ3D(a) for all a ∈ A and λ ∈ T 1 . n0 Similar to the proof of Theorem 1, we have D(λa) = λ3D(a) for all a ∈ A and λ ∈ T. Since D is a cubic mapping, D(ra) = r3D(a) for any rational number r. It follows from the continuity of f and φ that for each λ ∈ R, D(λa)=λ3D(a).TheproofofTheorem1indicatesthatD(λa)=λ3D(a), for alla∈Aandλ∈C (λ6=0).Therefore,D is acubic homogeneousmap.If we replace a,b by 2na,2nb respectively in (25), we have f(22nab) f(2na) f(2nb) φ(2na,2nb) φ(2na,2nb) k − ·b3−a3· k≤ ≤ . 82n 8n 8n 82n 8n for all a,b ∈ A. Taking the limit as n tend to infinity, we get D(ab) = D(a)· b3+a3·D(b), for all a,b∈A. Therefore D is a unique cubic derivation. Inthe followingresult,we getagainCorollary1 whichis a directconsequence of the above Theorem.