Ulam type stability problems for alternative homomorphisms

Transcription

1 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 R E S E A R C H Open Access Ulam type stability problems for alternative homomorphisms Sin-Ei Takahasi 1*,MakotoTsukada 1, Takeshi Miura 2, Hiroyuki Takagi 3 and Kotaro Tanahashi 4 * Correspondence: sin_ei1@yahoo.co.jp 1 Department of Information Sciences, Toho University, Funabashi, , Japan Full list of author information is available at the end of the article Abstract We introduce an alternative homomorphism with respect to binary operations and investigate the Ulam type stability problem for such a mapping. The obtained results apply to Ulam type stability problems for several important functional equations. MSC: Primary 39B82; secondary 47H10 Keywords: Ulam type stability; homomorphism; binary operation; fixed point theorem 1 Introduction In 1940, SM Ulam proposed the following stability problem: Given an approximately additive mapping, can one find the strictly additive mapping near it? A year later, DH Hyers gave an affirmative answer to this problem for additive mappings between Banach spaces. Subsequently many mathematicians came to deal with this problem (cf. [1 5]). We introduce an alternative homomorphism from a set X with two binary operations and to another set E with two binary operations and defined by f (x y) f (x y)=f (x) f (y) ( x, y X), and we investigate the Ulam type stability problem for such a mapping when E is a complete metric space. In particular, if s t = s for all s, t E, then our results imply the stability results obtained in [6]. Also the method used in the paper have already applied for some other equations (cf. [7 15]). One consequence of Banach s fixed point theorem A fixed point theorem has played an important role in the stability problem (cf. [16]). The authors used an easy consequence of Banach s fixed point theorem in [6]. It will serve again in this paper. Here we review it. Let X be a set and (E, d) a complete metric space. Fix two mappings f : X E and ϕ : X R +,wherer + denotes the set of all nonnegative real numbers. Denote by f,ϕ the set of all mappings u : X E such that there exists a finite constant K u satisfying d ( u(x), f (x) ) K u ϕ(x) ( x X) Takahasi et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 2 of 13 For any u, v f,ϕ,wedefine ρ f,ϕ (u, v)=inf { K 0:d ( u(x), v(x) ) Kϕ(x)( x X) }. Then ( f,ϕ, ρ f,ϕ ) is a complete metric space which contains f. Now, fix three mappings σ : X X, τ : E E and ε : X X R +. For any mapping u : X E,wedefinethemappingT σ,τ u : X E by (T σ,τ u)(x)=τ ( u(σ x) ) (x X). Also, we consider three quantities: α σ,ε = inf { K 0:ε(σ x, σ y) Kε(x, y)(x, y X) }, β σ,ϕ = inf { K 0:ϕ(σ x) Kϕ(x)(x X) }, γ τ = inf { K 0:d(τs, τt) Kd(s, t)(s, t E) }. If α σ,ε <, β σ,ϕ < and γ τ <,thenwehave ε(σ x, σ y) α σ,ε ε(x, y) ( x, y X), ϕ(σ x) β σ,ϕ ϕ(x) d(τs, τt) γ τ d(s, t) ( x X), ( s, t E), respectively. We will use these inequalities throughout this paper. We now state our fixed point theorem. Lemma A ([6, Proposition 2.1]) Let X be a set and (E, d) a complete metric space. Suppose that four mappings f : X E, ϕ : X R +, σ : X Xandτ : E Esatisfy T σ,τ f f,ϕ, β σ,ϕ <, γ τ < and β σ,ϕ γ τ <1. Then T σ,τ ( f,ϕ ) f,ϕ and T σ,τ has a unique fixed point f in f,ϕ. Moreover, lim d(( T n n σ,τ f ) (x), f (x) ) =0 and d ( f (x), f (x) ) ρ f,ϕ(t σ,τ f, f ) ϕ(x) 1 β σ,ϕ γ τ for all x X. 2 A stability of alternative homomorphisms Let (X,, ) beasetx with two binary operations and. Let(E, d,, ) beacomplete metric space (E, d) with two binary operations and.givenf : X E, we consider the following commutative diagram: X X f f E E (f ) (f ) E E E. (1)

3 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 3 of 13 This means that f (x y) f (x y)=f (x) f (y) ( x, y X). (2) In particular, if s t = s for all s, t E,then(1)and(2)become and X X f f X E E f E f (x y)=f (x) f (y) ( x, y X). In other words, f is a homomorphism from X to E. Thus, if a mapping f : X E satisfies (2), then we say that f isan alternative homomorphism. In this section, we establish two general settings, on which we can give an affirmative answertotheulamtypestabilityproblemforthecommutativediagram (1). These settings have a property such as duality, that is, each of them works as a complement of the other. Let us describe the first setting. For ε : X X R + and δ : X R +, we consider the following three conditions: (i) The square operator x x x is an automorphism of X with respectto and. We denote by σ the inverse mapping of this automorphism. (ii) The binary operations and on E are continuous. The square operator τ : s s s is an endomorphism of E with respect to and. (iii) α α σ,ε <, β β σ,δ <, γ γ τ < and γ max{α, β} <1. Under the above conditions, we show the Ulam type stability for the commutative diagram (1), as follows. Theorem 1 Let (X,, ) and (E, d,, ) be as above. Suppose that four mappings σ : X X, τ : E E, ε : X X R + and δ : X R + satisfy (i), (ii), and (iii). If a mapping f : X E satisfies d ( f (x y) f (x y), f (x) f (y) ) ε(x, y) ( x, y X), (3) d ( f (x) f (σ x σ x), f (x) ) δ(x) ( x X), (4) then there exists a mapping f : X Esuchthat f (x y) f (x y)=f (x) f (y) ( x, y X), (5) f (x) f (σ x σ x)=f (x) ( x X), (6) d ( f (x), f (x) ) αε(x, x)+δ(x) 1 γ max{α, β} ( x X). (7) Moreover, if a mapping g : X E satisfies (5), (6), and K g 0:d ( f (x), g(x) ) { } K g αε(x, x)+δ(x) ( x X), (8) then g = f.

4 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 4 of 13 Proof For simplicity, we write T = T σ,τ.wenotethatα, β,andγ are finite by (iii). Suppose that f : X E satisfies (3)and(4). Put ϕ(x)=αε(x, x)+δ(x) for all x X. To apply Lemma A to f and ϕ, wefirstobservethattf f,ϕ.fixx X. Replacingx and y in (3) byσ x, we get d ( f (σ x σ x) f (σ x σ x), f (σ x) f (σ x) ) ε(σ x, σ x). Since σ x σ x = σ 1 (σ x)=x, f (σ x) f (σ x)=τ ( f (σ x) ) =(Tf )(x), and ε(σ x, σ x) αε(x, x), it follows that d ( f (x) f (σ x σ x), (Tf )(x) ) αε(x, x). Using this and (4), we have d ( (Tf )(x), f (x) ) d ( (Tf )(x), f (x) f (σ x σ x) ) + d ( f (x) f (σ x σ x), f (x) ) αε(x, x)+δ(x) = ϕ(x). Hence Tf f,ϕ and ρ f,ϕ (Tf, f ) 1. We next estimate the quantity β σ,ϕ.forx X,wehave ϕ(σ x)=αε(σ x, σ x)+δ(σ x) α 2 ε(x, x)+βδ(x) max{α, β} ( αε(x, x)+δ(x) ) = max{α, β}ϕ(x). Hence β σ,ϕ max{α, β} and β σ,ϕ γ τ γ max{α, β} < 1 by (iii). Thus we can apply Lemma A. As a consequence, T has a unique fixed point f f,ϕ. Moreover, and lim d(( T n f ) (x), f (x) ) =0 (9) n d ( f (x), f (x) ) ρ f,ϕ(tf, f ) 1 β σ,ϕ γ τ ϕ(x) (10) for all x X.Sinceρ f,ϕ (Tf, f ) 1andβ σ,ϕ γ τ γ max{α, β} <1,(10)implies(7).

5 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 5 of 13 Here we show (5). If x, y X and n N,thenwehave d ( f (x y) f (x y), f (x) f (y) ) d ( f (x y) f (x y), ( T n f ) (x y) ( T n f ) (x y) ) + d (( T n f ) (x y) ( T n f ) (x y), ( T n f ) (x) ( T n f ) (y) ) + d (( T n f ) (x) ( T n f ) (y), f (x) f (y) ). (11) We will see that the right hand side of (11)tendsto0asn. The first and third terms on the right hand side tend to 0 as n,becauseof(9) and the continuity of and in (ii). Moreover, the second term, say A n (x, y), is estimated as follows: By (i), (ii), and (3), we have A n (x, y)=d ( τ n( f ( σ n (x y) )) τ n( f ( σ n (x y) )), τ n( f ( σ n x )) τ n( f ( σ n y ))) = d ( τ n( f ( σ n x σ n y )) τ n( f ( σ n x σ n y )), τ n( f ( σ n x ) f ( σ n y ))) = d ( τ n( f ( σ n x σ n y ) f ( σ n x σ n y )), τ n( f ( σ n x ) f ( σ n y ))) γ n d ( f ( σ n x σ n y ) f ( σ n x σ n y ), f ( σ n x ) f ( σ n y )) γ n ε ( σ n x, σ n y ) γ n α n ε(x, y), where τ n and σ n denote the n-fold compositions of endomorphisms τ and σ,respectively. Since γα< 1 by (iii), it follows that A n (x, y) 0asn. Thus the right hand side of (11) tends to 0, and we obtain (5). Next, we show (6). For x X,wereplacex and y in (5)byσxto get f (σ x σ x) f (σ x σ x)=f (σ x) f (σ x). Since σ x σ x = x and f (σ x) f (σ x)=τ ( f (σ x) ) =(Tf )(x)=f (x), we obtain (6). Finally, we show the last statement. Since g satisfies (5)and(6), we have (Tg)(x)=τ ( g(σ x) ) = g(σ x) g(σ x) = g(σ x σ x) g(σ x σ x) = g(x) g(σ x σ x) = g(x) for all x X.Thissaysthatg is a fixed point of T.Also,by(8), we have g f,ϕ.thusthe uniqueness of a fixed point of T in f,ϕ implies that g = f.

6 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 6 of 13 The next corollary is obtained in [6]. Corollary 1 ([6, Corollary 3.2]) LetXbeasetwithabinaryoperation such that the square operation x x x is an automorphism of X with respect to and E a complete metric space with a continuous binary operation such that the square operation τ : s s s is an endomorphism of E with respect to. Let ε : X X R + and suppose that α α σ,ε <, γ γ τ < and γα<1,where σ denotes the inverse mapping of the square operation x x x. If a mapping f : X E satisfies d ( f (x y), f (x) f (y) ) ε(x, y) ( x, y X), then there exists a unique mapping f : X Esuchthat f (x y)=f (x) f (y) and d ( f (x), f (x) ) α ε(x, x) 1 αγ for all x, y X. Proof Consider the case that = and s t = s for s, t E,inTheorem1.Inthiscase,τ is clearly an endomorphism of E with respect to. Therefore the corollary follows immediately from Theorem 1 with δ =0. Now we turn to another setting. Let (X,, ) and(e, d,, ) be as in the first part of this section. For ε : X X R + and δ : X R +, we consider the following three conditions: (iv) The square operator σ : x x x is an endomorphism of X with respect to and. (v) The binary operations and on E are continuous. The square operator s s s is an automorphism of E with respectto and. We denote by τ the inverse mapping of this automorphism. (vi) α α σ,ε <, β β σ,δ <, γ γ τ <,and γ max{ α, β} <1. Under the above conditions, we show the Ulam type stability for the commutative diagram (1), as follows. Theorem 2 Let (X,, ) and (E, d,, ) be as above. Suppose that four mappings σ : X X, τ : E E, ε : X X R + and δ : X R + satisfy (iv), (v), and (vi). If a mapping f : X E satisfies (3) and d ( f (x x) f (x x), f (x x) ) δ(x) ( x X), (12) then there exists a mapping f : X Esatisfying(5) f (x x) f (x x)=f (x x) ( x X), (13) d ( f (x), f (x) ) γ {ε(x, x)+δ(x)} 1 γ max{ α, β} ( x X). (14) Moreover, if a mapping g : X E satisfies (13), (14), and K g 0:d ( f (x), g(x) ) K g γ { ε(x, x)+δ(x) } ( x X), (15) then g = f.

7 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 7 of 13 Proof For simplicity, we write T = T σ, τ,thatis,( Tf)(x) = τ(f ( σ x)) for x X. Wenote that α, β and γ are finite by (vi). Suppose that f : X E satisfies (3) and(12). Put ϕ(x)= γ {ε(x, x)+δ(x)} for all x X. To apply Lemma A to f and ϕ, wefirstobservethat Tf f, ϕ.fixx X.Since τ(f (x) f (x)) = f (x), it follows from (3)and(12)that d ( ( Tf)(x), f (x) ) = d ( τ ( f ( σ x) ), f (x) ) = d ( τ ( f (x x) ), τ ( f (x) f (x) )) γ d ( f (x x), f (x) f (x) ) γ { d ( f (x x), f (x x) f (x x) ) + d ( f (x x) f (x x), f (x) f (x) )} γ { δ(x)+ε(x, x) } = ϕ(x). Hence Tf f, ϕ and ρ f, ϕ ( Tf, f ) 1. We next estimate the quantity β σ, ϕ.forx X,wehave ϕ( σ x)= γ { ε( σ x, σ x)+δ( σ x) } γ { αε(x, x)+ βδ(x) } γ max{ α, β} { ε(x, x)+δ(x) } = max{ α, β} ϕ(x). Hence β σ, ϕ max{ α, β} and β σ, ϕ γ τ γ max{ α, β} <1by(vi). Thus we can apply Lemma A. As a consequence, T has a unique fixed point f f, ϕ. Moreover, and lim d(( T n f ) (x), f (x) ) =0 (16) n d ( f (x), f (x) ) ρ f, ϕ( Tf, f ) 1 β σ, ϕ γ τ ϕ(x) (17) for all x X.Sinceρ f, ϕ ( Tf, f ) 1andβ σ, ϕ γ τ γ max{ α, β} <1,(17)implies(14). Here we show (5). If x, y X and n N,thenwehave d ( f (x y) f (x y), f (x) f (y) ) d ( f (x y) f (x y), ( T n f ) (x y) ( T n f ) (x y) ) + d (( T n f ) (x y) ( T n f ) (x y), ( T n f ) (x) ( T n f ) (y) ) + d (( T n f ) (x) ( T n f ) (y), f (x) f (y) ). Letting n, the first and third terms on the right hand side tend to 0, because of (16) and the continuity of and in (v). Moreover, the second term, say à n (x, y), is estimated

8 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 8 of 13 as follows: By (iv), (v), and (3), à n (x, y)=d ( τ n( f ( σ n (x y) )) τ n( f ( σ n (x y) )), τ n( f ( σ n x )) τ n( f ( σ n y ))) = d ( τ n( f ( σ n x σ n y )) τ n( f ( σ n x σ n y )), τ n( f ( σ n x ) f ( σ n y ))) = d ( τ n( f ( σ n x σ n y ) f ( σ n x σ n y )), τ n( f ( σ n x ) f ( σ n y ))) γ n d ( f ( σ n x σ n y ) f ( σ n x σ n y ), f ( σ n x ) f ( σ n y )) γ n ε ( σ n x, σ n y ) γ n α n ε(x, y), where τ n and σ n denote the n-fold compositions of endomorphisms τ and σ,respectively. Since γ α < 1 by (vi), it follows that à n (x, y) 0asn.Thusweobtain(5). Next, we show (13). Replacing y in (5)byx,wehave f (x x) f (x x)=f (x) f (x). (18) Also since τ ( f (x x) ) = τ ( f ( σ x) ) =( Tf )(x)=f (x)= τ ( f (x) f (x) ), it follows that f (x x)=f (x) f (x). Combining with (18), we obtain (13). Finally, we show the last statement. Since g satisfies (14) and(13), we have g( σ x)=g(x x)=g(x x) g(x x)=g(x) g(x)= τ 1( g(x) ), that is, ( Tg)(x)=g(x) for all x X. Thissaysthatg is a fixed point of T. Also,by(15), we have g f, ϕ. Hence the uniqueness of a fixed point of T in f, ϕ implies that g = f. The next corollary is obtained in [6]. Corollary 2 ([6, Corollary 3.5]) Let X be a set with a binary operation such that the square operation σ : x x x is an endomorphism of X with respect to and E a complete metric space with a continuous binary operation such that the square operation s s s is an automorphism of E with respect to. Let ε : X X R + and suppose that α α σ,ε <, γ γ τ < and γ α <1,where τ denotes the inverse mapping of the square operation s s s. If a mapping f : X E satisfies d ( f (x y), f (x) f (y) ) ε(x, y) ( x, y X), then there exists a unique mapping f : X Esuchthat f (x y)=f (x) f (y) and d ( f (x), f (x) ) for all x, y X. γ ε(x, x) 1 α γ

9 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 9 of 13 Proof Consider the case that = and s t = s for s, t E,inTheorem2.Then τ is clearly an endomorphism of E with respect to. Therefore the corollary follows immediately from Theorem 2 with δ =0. 3 Application I The Ulam type stability problem for Euler-Lagrange type additive mappings has been investigated in [17]. Here we take up the following Euler-Lagrange type mapping f : X E satisfying f (ax + by)+f (bx + ay)+(a + b) ( f ( x)+f ( y) ) =0 ( x, y X), (19) where X is a complex normed space, E a complex Banach space and a, b C with a +b 0. The following is an Ulam type stability result for this mapping. Corollary 3 (cf. [17, Theorem 2.1]) Let ε : X X R + and suppose that (vii) K 0: a + b K <1and ε(x, y) Kε( (a + b)x, (a + b)y) ( x, y X). If a mapping f : X E satisfies f (ax + by)+f (bx + ay)+(a + b) ( f ( x)+f ( y) ) ε(x, y) ( x, y X), (20) then there exists a unique mapping f : X Esatisfying(19) and f (x) f (x) K ε( x, x) ( x X). (21) 2(1 a + b K) Proof Put u = x, v = y for each x, y X. Under these transformations, (20)changesinto the following estimate: 1 { } a + b { } f ( au bv)+f ( bu av) + f (u)+f(v) ε 1 (u, v) ( u, v X), (22) 2 2 where ε 1 (u, v)= 1 ε( u, v)( u, v X). 2 Now we define u v = au bv, u v = bu av for each u, v X. Inthiscase,wecan easily see that the square operator u u u is an endomorphism of X with respect to and.alsosincea+b 0, this endomorphism is bijective and so automorphic. We denote by σ the inverse mapping of this automorphism. Moreover, we define s t = 1 2 (a+b)(s+t), s t = 1 (s + t) foreachs, t E. Then we can also see that the binary operations and 2 on E are continuous and the square operator τ : s s s is an automorphism of E with respect to and.note that(22) changes into the following: f (u v) f (u v) f (u) f (v) ε1 (u, v) ( u, v X). (23) Since x x = x x for all x X, it follows that σ x σ x = σ x σ x = σ 1 σ x = x for all x X. Also, since s s = s for all s E, it follows that f (x) f (σ x σ x)=f (x) f (x)=f(x) for all x X and then (4) holdswithδ =0.Moreover,β σ,δ =0mustholdwithδ =0.Itisalso obvious that γ τ = a + b from the definition of τ.wealsonotethatα σ,ε1 K from the

10 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 10 of 13 second condition of (vii) and hence γ τ α σ,ε1 a + b K < 1 from the first condition of (vii). Therefore, by Theorem 1, there exists a unique mapping f : X E such that f (u v) f (u v)=f (u) f (v) ( u, v X), namely, (19) holds and f (u) f (u) α σ,ε1 ε 1 (u, u) 1 γ τ max{α σ,ε1, β σ,δ } K ε( u, u) ( u X), 2(1 a + b K) and so (21)holds. The following is also an Ulam type stability result for the mapping satisfying (19). Corollary 4 (cf. [17, Theorem 2.2]) Let ε : X X R + and suppose that (viii) K 0:K < a + b and ε( (a + b)x, (a + b)y) Kε(x, y) ( x, y X). If a mapping f : X E satisfies (20), then there exists a unique mapping f : X Esatisfying (19) and f (x) f (x) 1 ε( x, x) ( x X). (24) 2( a + b K) Proof As observed in the proof of Corollary 3, (20) changesinto(22). Now we define u v = au bv, u v = bu av for each u, v X.Inthiscase,wecaneasilyseethatthe square operator σ : u u u is an endomorphism of X with respect to and.moreover, we define s t = 1 2 (a + b)(s + t), s t = 1 (s + t)foreachs, t E. Then we can also see that 2 the binary operations and on E are continuous and the square operator s s s is an endomorphism of E with respect to and. Alsosincea + b 0, this endomorphism is bijective and so automorphic. We denote by τ the inverse mapping of this automorphism. Note that (22)changesinto(23). Since x x = x x ( x X)ands s = s ( s E), it follows that f (x x) f (x x)=f (x x) for all x X and then (12)holdswithδ =0. Moreover, β σ,δ =0mustholdwithδ =0.Itisalsoobviousthatγ τ = a + b 1 from the definition of τ.wealsonotethatα σ,ε1 K from the second condition of (viii) and hence γ τ α σ,ε1 a + b 1 K < 1 from the first condition of (viii). Therefore, by Theorem 2, there exists a unique mapping f : X E such that f (u v) f (u v)=f (u) f (v) ( u, v X), namely, (19) holds and f (u) f (u) γ τ ε 1 (u, u) 1 γ τ max{α σ,ε1, β σ,δ } = a + b 1 2(1 a + b 1 K) ε( u, u) 1 2( a + b K) ε( u, u) ( u X), and so (24)holds.

11 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 11 of 13 Corollary 5 (cf. [17, Corollary 2.3]) Suppose that a + b 1,δ, p, q 0 and p + q 1.If a mapping f : X E satisfies f (ax + by)+f (bx + ay)+(a + b) { f ( x)+f ( y) } δ x p y q for all x, y X, then there exists a unique mapping f : X Esatisfying(19) and f (x) f (x) δ 2( a + b p+q a + b x p+q ( x X). Proof Put ε(x, y)=δ x p y q for each x, y X. (a) The case where either { a + b >1, p + q >1, or { a + b <1, p + q <1. Put K = a + b (p+q).thenk satisfies (vii). Note also that K 2(1 a + b K) ε( x, x)= δ 2( a + b p+q a + b ) x p+q for all x X. Then the desired result follows from Corollary 3. (b) The case where either { a + b >1, p + q <1, or { a + b <1, p + q >1. Put K = a + b p+q.thenk satisfies (viii). Note also that 1 2( a + b K) ε( x, x)= δ 2( a + b a + b p+q ) x p+q for all x X. Then the desired result follows from Corollary 4. 4 Application II Let (X, +) be an Abelian group. In [18], the following result has been shown by A. Simon and P. Volkmann.

12 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 12 of 13 Lemma B ([18,Théorème1)] A mapping f : X R satisfies max { f (x + y), f (x y) } = f (x)+f (y) ( x, y X), (25) if and only if f (x)= π(x) ( x X) for some additive function π : X R. In this section, we deal with the Ulam type stability problem for Equation (25). Put x y = x + y and x y = x y for each x, y X. Moreover,puts t = s + t and s t = max{s, t} for each s, t R.Then(25)changesinto(2). Also we can easily see that the square operation σ : x x x is endomorphic with respect to and and that the square operator s s s is automorphic with respect to and. Denoteby τ the inverse mapping of this automorphism. In this case, it is obvious that τ(s)= 1 2 s for each s R and hence γ τ = 1/2. Now let ε be a nonnegative constant and suppose that f : X R satisfies max { f (x + y), f (x y) } { f (x)+f (y) } ε ( x, y X). (26) Putting x = y =0in(26), we obtain f (0) ε. (27) Also, putting x = y in (26), we obtain ε + f (0) ε + max { f (x + x), f (0) } 2f (x) ( x X). (28) Combining (27)and(28), we obtain ε f (x) ( x X). (29) Put δ =2ε.By(27)and(28), we obtain 0 max { f (x + x), f (0) } f (x + x) ε + ε = δ ( x X), and hence (12) holds. Moreover, note that α σ,ε = β σ,δ =1sinceε and δ are constant. Then Lemma B and Theorem 2 easily imply the following. Corollary 6 Let X be an Abelian group and ε a nonnegative constant. If f : X R satisfies (26), then there exists an additive mapping π : X R such that f (x) π(x) 3ε ( x X). For the related results, see [19, 20]. Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally to this paper. They read and approved the final manuscript.

13 Takahasi et al. Journal of Inequalities and Applications 2014, 2014:228 Page 13 of 13 Author details 1 Department of Information Sciences, Toho University, Funabashi, , Japan. 2 Department of Mathematics, Niigata University, Niigata, , Japan. 3 Department of Mathematical Sciences, Faculty of Science, Shinshu University, Matsumoto, , Japan. 4 Department of Mathematics, Tohoku Pharmaceutical University, Sendai, , Japan. Acknowledgements The authors are deeply grateful to the referees for careful reading of the paper and for the helpful suggestions and comments. All authors are partially supported by Grant-in-Aid for Scientific Research, Japan Society for the Promotion of Science. Received: 29 January 2014 Accepted: 27 May 2014 Published: 04 Jun 2014 References 1. Brillouët-Belluot, N, Brzdȩk, J, Ciepliński, K: On some recent developments in Ulam s type stability. Abstr. Appl. Anal. 2012, Article ID (2012) 2. Brzdȩk, J: Hyperstability of the Cauchy equation on restricted domains. Acta Math. Hung. 141, (2013) 3. Brzdȩk,J, Ciepliński, K: Hyperstability and superstability. Abstr. Appl. Anal. 2013, Article ID (2013) 4. Gajda, Z: On stability of additive mappings. Int. J. Math. Math. Sci. 14, (1991) 5. Jung, S-M: Hyers-Ulam-Rassias Stability of Functional Equations in Nonlinear Analysis. Springer Optimization and Its Applications, vol. 48. Springer, New York (2011) 6. Takahasi, S-E, Miura, T, Takagi, H: On a Hyers-Ulam-Aoki-Rassias type stability and a fixed point theorem. J. Nonlinear Convex Anal. 11, (2010) 7. Bahyrycz, A, Piszczek, M: Hyperstability of the Jensen functional equation. Acta Math. Hung. 142, (2014) 8. Brzdȩk, J: Stability of the equation of the p-wright affine functions. Aequ. Math. 85, (2013) 9. Brzdȩk, J: A hyperstability result for the Cauchy equation. Bull. Aust. Math. Soc. 89, (2014) 10. Gilányi, A, Kaiser, Z, Páles, Z: Estimates to the stability of functional equations. Aequ. Math. 73, (2007) 11. Kim, GH: On the stability of functional equations with square-symmetric operation. Math. Inequal. Appl. 4, (2001) 12. Kim, GH: Addendum to On the stability of functional equations on square-symmetric groupoid. Nonlinear Anal. 62, (2005) 13. Páles, Z: Hyers-Ulam stability of the Cauchy functional equation on square-symmetric groupoids. Publ. Math. (Debr.) 58, (2001) 14. Páles, Z, Volkmann, P, Luce, RD: Hyers-Ulam stability of functional equations with square symmetric operations. Proc. Natl. Acad. Sci. USA 95, (1998) 15. Piszczek, M: Remark on hyperstability of the general linear equation. Aequ. Math. (2013). doi: /s x 16. Ciepliński, K: Applications of fixed point theorems to the Hyers-Ulam stability of functional equations-a survey. Ann. Funct. Anal. 3, (2012) 17. Kim, H-M, Jun, K-W, Rassias, JM: Extended stability problem for alternative Cauchy-Jensen mappings. J. Inequal. Pure Appl. Math. 8,120 (2007) 18. Simon, A, Volkmann, P: Caractérisation du module d une fonction á l aide d une équation fonctionnelle. Aequ. Math. 47, (1994) 19. Gilányi, A, Nagatou, K, Volkmann, P: Stability of a functional equation coming from the characterization of the absolute value of additive functions. Ann. Funct. Anal. 1,1-6 (2010) 20. Jarczyk, W, Volkmann, P: On functional equations in connection with the absolute value of additive functions. Series Mathematicae Catoviciensis et Debreceniensis 32, 1-11 (2010) / X Cite this article as: Takahasi et al.: Ulam type stability problems for alternative homomorphisms. Journal of Inequalities and Applications 2014, 2014:228