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Approximation of linear mappings in Banach modules over C -algebras

Journal of Inequalities and Applications20132013:185

https://doi.org/10.1186/1029-242X-2013-185

  • Received: 26 September 2012
  • Accepted: 5 April 2013
  • Published:

Abstract

Let X, Y be Banach modules over a C -algebra and let r 1 , , r n R be given. Using fixed-point methods, we prove the stability of the following functional equation in Banach modules over a unital C -algebra:

j = 1 n f ( 1 2 1 i n , i j r i x i 1 2 r j x j ) + i = 1 n r i f ( x i ) = n f ( 1 2 i = 1 n r i x i ) .

As an application, we investigate homomorphisms in unital C -algebras.

MSC:39B72, 46L05, 47H10, 46B03, 47B48.

Keywords

  • fixed point
  • Hyers-Ulam stability
  • super-stability
  • generalized Euler-Lagrange type additive mapping
  • homomorphism
  • C -algebra

1 Introduction and preliminaries

We say a functional equation ( ζ ) is stable if any function g satisfying the equation ( ζ ) approximately is near to the true solution of ( ζ ) . We say that a functional equation is superstable if every approximate solution is an exact solution of it (see [1]). The stability problem of functional equations was originated from a question of Ulam [2] concerning the stability of group homomorphisms. Hyers [3] gave a first affirmative partial answer to the question of Ulam in Banach spaces. Hyers’ theorem was generalized by Aoki [4] for additive mappings and by T.M. Rassias [5] for linear mappings by considering an unbounded Cauchy difference. A generalization of the T.M. Rassias theorem was obtained by Găvruta [6] by replacing the unbounded Cauchy difference by a general control function in the spirit of T.M. Rassias’ approach.

The functional equation
f ( x + y ) + f ( x y ) = 2 f ( x ) + 2 f ( y )
is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. A Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [7] for mappings f : X Y , where X is a normed space and Y is a Banach space. Cholewa [8] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik [9] proved the Hyers-Ulam stability of the quadratic functional equation. J.M. Rassias [10, 11] introduced and investigated the stability problem of Ulam for the Euler-Lagrange quadratic functional equation
f ( a 1 x 1 + a 2 x 2 ) + f ( a 2 x 1 a 1 x 2 ) = ( a 1 2 + a 2 2 ) [ f ( x 1 ) + f ( x 2 ) ] .
(1.1)

Grabiec [12] has generalized these results mentioned above.

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 [1343]).

Let X be a set. A function d : X × X [ 0 , ] is called a generalized metric on X if d satisfies the following conditions:
  1. (1)

    d ( x , y ) = 0 if and only if x = y ;

     
  2. (2)

    d ( x , y ) = d ( y , x ) for all x , y X ;

     
  3. (3)

    d ( x , z ) d ( x , y ) + d ( y , z ) for all x , y , z X .

     

We recall a fundamental result in fixed-point theory.

Theorem 1.1 [44, 45]

Let ( X , d ) be a complete generalized metric space and let J : X X be a strictly contractive mapping with Lipschitz constant L < 1 . Then, for each given element x X , either
d ( J n x , J n + 1 x ) =
for all nonnegative integers n or there exists a positive integer n 0 such that
  1. (1)

    d ( J n x , J n + 1 x ) < for all n n 0 ;

     
  2. (2)

    the sequence { J n x } converges to a fixed point y of J;

     
  3. (3)

    y is the unique fixed point of J in the set Y = { y X d ( J n 0 x , y ) < } ;

     
  4. (4)

    d ( y , y ) 1 1 L d ( y , J y ) for all y Y .

     

In 1996, Isac and T.M. Rassias [46] were the first to provide applications of stability theory of functional equations for the proof of new fixed-point theorems with applications. By using fixed-point methods, the stability problems of several functional equations have been extensively investigated by a number of authors (see [4758]).

Recently, Park and Park [59] introduced and investigated the following additive functional equation of Euler-Lagrange type:
(1.2)

whose solution is said to be a generalized additive mapping of Euler-Lagrange type.

In this paper, we introduce the following additive functional equation of Euler-Lagrange type which is somewhat different from (1.2):
j = 1 n f ( 1 2 1 i n , i j r i x i 1 2 r j x j ) + i = 1 n r i f ( x i ) = n f ( 1 2 i = 1 n r i x i ) ,
(1.3)

where r 1 , , r n R . Every solution of the functional equation (1.3) is said to be a generalized Euler-Lagrange type additive mapping.

Using fixed-point methods, we investigate the Hyers-Ulam stability of the functional equation (1.3) in Banach modules over a C -algebra. These results are applied to investigate C -algebra homomorphisms in unital C -algebras. Also, ones can get the super-stability results after all theorems by putting the product of powers of norms as the control functions (see for more details [60, 61]).

Throughout this paper, assume that A is a unital C -algebra with the norm A and the unit e, B is a unital C -algebra with the norm B , and X, Y are left Banach modules over a unital C -algebra A with the norms X and Y , respectively. Let U ( A ) be the group of unitary elements in A and let r 1 , , r n R .

2 Hyers-Ulam stability of the functional equation (1.3) in Banach modules over a C -algebra

For any given mapping f : X Y , u U ( A ) and μ C , we define D u , r 1 , , r n f and D μ , r 1 , , r n f : X n Y by
D u , r 1 , , r n f ( x 1 , , x n ) : = j = 1 n f ( 1 2 1 i n , i j r i u x i 1 2 r j u x j ) + i = 1 n r i u f ( x i ) n f ( 1 2 i = 1 n r i u x i )
and
D μ , r 1 , , r n f ( x 1 , , x n ) : = j = 1 n f ( 1 2 1 i n , i j μ r i x i 1 2 μ r j x j ) + i = 1 n μ r i f ( x i ) n f ( 1 2 i = 1 n μ r i x i )

for all x 1 , , x n X .

Lemma 2.1 Let X and Y be linear spaces and let r 1 , , r n be real numbers with k = 1 n r k 0 and r i 0 , r j 0 for some 1 i < j n . Assume that a mapping L : X Y satisfies the functional equation (1.3) for all x 1 , , x n X . Then the mapping L is additive. Moreover, L ( r k x ) = r k L ( x ) for all x X and 1 k n .

Proof One can find a complete proof at [62]. □

Lemma 2.2 Let X and Y be linear spaces and let r 1 , , r n be real numbers with r i 0 , r j 0 for some 1 i < j n . Assume that a mapping L : X Y with L ( 0 ) = 0 satisfies the functional equation (1.3) for all x 1 , , x n X . Then the mapping L is additive. Moreover, L ( r k x ) = r k L ( x ) for all x X and 1 k n .

Proof One can find a complete proof at [62]. □

We investigate the Hyers-Ulam stability of a generalized Euler-Lagrange type additive mapping in Banach modules over a unital C -algebra. Throughout this paper, let r 1 , , r n be real numbers such that r i 0 , r j 0 for fixed 1 i < j n .

Theorem 2.3 Let f : X Y be a mapping satisfying f ( 0 ) = 0 for which there is a function φ : X n [ 0 , ) such that
D e , r 1 , , r n f ( x 1 , , x n ) Y φ ( x 1 , , x n )
(2.1)
for all x 1 , , x n X . Let
φ i j ( x , y ) : = φ ( 0 , , 0 , x i th , 0 , , 0 , y j th , 0 , , 0 )
for all x , y X and 1 i < j n . If there exists 0 < C < 1 such that
φ ( 2 x 1 , , 2 x n ) 2 C φ ( x 1 , , x n )
for all x 1 , , x n X , then there exists a unique generalized Euler-Lagrange type additive mapping L : X Y such that
f ( x ) L ( x ) Y 1 4 4 C { φ i j ( 2 x r i , 2 x r j ) + 2 φ i j ( x r i , x r j ) + φ i j ( 2 x r i , 0 ) + 2 φ i j ( x r i , 0 ) + φ i j ( 0 , 2 x r j ) + 2 φ i j ( 0 , x r j ) }
(2.2)

for all x X . Moreover, L ( r k x ) = r k L ( x ) for all x X and 1 k n .

Proof For each 1 k n with k i , j , let x k = 0 in (2.1). Then we get the following inequality:
f ( r i x i + r j x j 2 ) + f ( r i x i r j x j 2 ) 2 f ( r i x i + r j x j 2 ) + r i f ( x i ) + r j f ( x j ) Y φ ( 0 , , 0 , x i i th , 0 , , 0 , x j j th , 0 , , 0 )
(2.3)
for all x i , x j X . Letting x i = 0 in (2.3), we get
f ( r j x j 2 ) f ( r j x j 2 ) + r j f ( x j ) Y φ i j ( 0 , x j )
(2.4)
for all x j X . Similarly, letting x j = 0 in (2.3), we get
f ( r i x i 2 ) f ( r i x i 2 ) + r i f ( x i ) Y φ i j ( x i , 0 )
(2.5)
for all x i X . It follows from (2.3), (2.4) and (2.5) that
f ( r i x i + r j x j 2 ) + f ( r i x i r j x j 2 ) 2 f ( r i x i + r j x j 2 ) + f ( r i x i 2 ) + f ( r j x j 2 ) f ( r i x i 2 ) f ( r j x j 2 ) Y φ i j ( x i , x j ) + φ i j ( x i , 0 ) + φ i j ( 0 , x j )
(2.6)
for all x i , x j X . Replacing x i and x j by 2 x r i and 2 y r j in (2.6), we get
f ( x + y ) + f ( x y ) 2 f ( x + y ) + f ( x ) + f ( y ) f ( x ) f ( y ) Y φ i j ( 2 x r i , 2 y r j ) + φ i j ( 2 x r i , 0 ) + φ i j ( 0 , 2 y r j )
(2.7)
for all x , y X . Putting y = x in (2.7), we get
2 f ( x ) 2 f ( x ) 2 f ( 2 x ) Y φ i j ( 2 x r i , 2 x r j ) + φ i j ( 2 x r i , 0 ) + φ i j ( 0 , 2 x r j )
(2.8)
for all x X . Replacing x and y by x 2 and x 2 in (2.7), respectively, we get
f ( x ) + f ( x ) Y φ i j ( x r i , x r j ) + φ i j ( x r i , 0 ) + φ i j ( 0 , x r j )
(2.9)
for all x X . It follows from (2.8) and (2.9) that
1 2 f ( 2 x ) f ( x ) Y 1 4 ψ ( x )
(2.10)
for all x X , where
ψ ( x ) : = φ i j ( 2 x r i , 2 x r j ) + 2 φ i j ( x r i , x r j ) + φ i j ( 2 x r i , 0 ) + 2 φ i j ( x r i , 0 ) + φ i j ( 0 , 2 x r j ) + 2 φ i j ( 0 , x r j ) .
Consider the set W : = { g : X Y } and introduce the generalized metric on W :
d ( g , h ) = inf { C R + : g ( x ) h ( x ) Y C ψ ( x ) , x X } .

It is easy to show that ( W , d ) is complete.

Now, we consider the linear mapping J : W W such that
J g ( x ) : = 1 2 g ( 2 x )
(2.11)

for all x X . By Theorem 3.1 of [44], d ( J g , J h ) C d ( g , h ) for all g , h W . Hence, d ( f , J f ) 1 4 .

By Theorem 1.1, there exists a mapping L : X Y such that
  1. (1)
    L is a fixed point of J, i.e.,
    L ( 2 x ) = 2 L ( x )
    (2.12)
     
for all x X . The mapping L is a unique fixed point of J in the set
Z = { g W : d ( f , g ) < } .
This implies that L is a unique mapping satisfying (2.12) such that there exists C ( 0 , ) satisfying
L ( x ) f ( x ) Y C ψ ( x )
for all x X .
  1. (2)
    d ( J n f , L ) 0 as n . This implies the equality
    lim n f ( 2 n x ) 2 n = L ( x )
     
for all x X .
  1. (3)

    d ( f , L ) 1 1 C d ( f , J f ) , which implies the inequality d ( f , L ) 1 4 4 C . This implies that the inequality (2.2) holds.

     
Since φ ( 2 x 1 , , 2 x n ) 2 C φ ( x 1 , , x n ) , it follows that
D e , r 1 , , r n L ( x 1 , , x n ) Y = lim k 1 2 k D e , r 1 , , r n f ( 2 k x 1 , , 2 k x n ) Y lim k 1 2 k φ ( 2 k x 1 , , 2 k x n ) lim k C k φ ( x 1 , , x n ) = 0

for all x 1 , , x n X . Therefore, the mapping L : X Y satisfies the equation (1.3) and L ( 0 ) = 0 . Hence, by Lemma 2.2, L is a generalized Euler-Lagrange type additive mapping and L ( r k x ) = r k L ( x ) for all x X and 1 k n . This completes the proof. □

Theorem 2.4 Let f : X Y be a mapping satisfying f ( 0 ) = 0 for which there is a function φ : X n [ 0 , ) satisfying
D u , r 1 , , r n f ( x 1 , , x n ) φ ( x 1 , , x n )
(2.13)
for all x 1 , , x n X and u U ( A ) . If there exists 0 < C < 1 such that
φ ( 2 x 1 , , 2 x n ) 2 C φ ( x 1 , , x n )

for all x 1 , , x n X , then there exists a unique A-linear generalized Euler-Lagrange type additive mapping L : X Y satisfying (2.2) for all x X . Moreover, L ( r k x ) = r k L ( x ) for all x X and 1 k n .

Proof By Theorem 2.3, there exists a unique generalized Euler-Lagrange type additive mapping L : X Y satisfying (2.2), and moreover L ( r k x ) = r k L ( x ) for all x X and 1 k n . By the assumption, for each u U ( A ) , we get
D u , r 1 , , r n L ( 0 , , 0 , x i th , 0 , , 0 ) Y = lim k 1 2 k D u , r 1 , , r n f ( 0 , , 0 , 2 k x i th , 0 , , 0 ) Y lim k 1 2 k φ ( 0 , , 0 , 2 k x i th , 0 , , 0 ) lim k C k φ ( 0 , , 0 , x i th , 0 , , 0 ) = 0
for all x X . So, we have
r i u L ( x ) = L ( r i u x )
for all u U ( A ) and x X . Since L ( r i x ) = r i L ( x ) for all x X and r i 0 ,
L ( u x ) = u L ( x )
for all u U ( A ) and x X . By the same reasoning as in the proofs of [63] and [64],
L ( a x + b y ) = L ( a x ) + L ( b y ) = a L ( x ) + b L ( y )

for all a , b A ( a , b 0 ) and x , y X . Since L ( 0 x ) = 0 = 0 L ( x ) for all x X , the unique generalized Euler-Lagrange type additive mapping L : X Y is an A-linear mapping. This completes the proof. □

Theorem 2.5 Let f : X Y be a mapping satisfying f ( 0 ) = 0 for which there is a function φ : X n [ 0 , ) such that
D e , r 1 , , r n f ( x 1 , , x n ) Y φ ( x 1 , , x n )
(2.14)
for all x 1 , , x n X . If there exists 0 < C < 1 such that
φ ( x 1 , , 2 n ) C 2 φ ( 2 x 1 , , 2 x n )
for all x 1 , , x n X , then there exists a unique generalized Euler-Lagrange type additive mapping L : X Y such that
f ( x ) L ( x ) Y C 4 4 C { φ i j ( 2 x r i , 2 x r j ) + 2 φ i j ( x r i , x r j ) + φ i j ( 2 x r i , 0 ) + 2 φ i j ( x r i , 0 ) + φ i j ( 0 , 2 x r j ) + 2 φ i j ( 0 , x r j ) }
(2.15)

for all x X , where φ i j is defined in the statement of Theorem  2.3. Moreover, L ( r k x ) = r k L ( x ) for all x X and 1 k n .

Proof It follows from (2.10) that
f ( x ) f ( x 2 ) Y 1 2 ψ ( x 2 ) C 4 ψ ( x )

for all x X , where ψ is defined in the proof of Theorem 2.3. The rest of the proof is similar to the proof of Theorem 2.3. □

Theorem 2.6 Let f : X Y be a mapping with f ( 0 ) = 0 for which there is a function φ : X n [ 0 , ) satisfying
D u , r 1 , , r n f ( x 1 , , x n ) φ ( x 1 , , x n )
(2.16)
for all x 1 , , x n X and u U ( A ) . If there exists 0 < C < 1 such that
φ ( x 1 , , 2 n ) C 2 φ ( 2 x 1 , , 2 x n )

for all x 1 , , x n X , then there exists a unique A-linear generalized Euler-Lagrange type additive mapping L : X Y satisfying (2.15) for all x X . Moreover, L ( r k x ) = r k L ( x ) for all x X and all 1 k n .

Proof The proof is similar to the proof of Theorem 2.4. □

Remark 2.7 In Theorems 2.5 and 2.6, one can assume that k = 1 n r k 0 instead of f ( 0 ) = 0 .

3 Homomorphisms in unital C -algebras

In this section, we investigate C -algebra homomorphisms in unital C -algebras. We use the following lemma in the proof of the next theorem.

Lemma 3.1 [64]

Let f : A B be an additive mapping such that f ( μ x ) = μ f ( x ) for all x A and μ S 1 n o 1 : = { e i θ ; 0 θ 2 π n o } . Then the mapping f : A B is -linear.

Note that a -linear mapping H : A B is called a homomorphism in C -algebras if H satisfies H ( x y ) = H ( x ) H ( y ) and H ( x ) = H ( x ) for all x , y A .

Theorem 3.2 Let f : A B be a mapping with f ( 0 ) = 0 for which there is a function φ : A n [ 0 , ) satisfying
(3.1)
(3.2)
(3.3)
for all x , x 1 , , x n A , u U ( A ) , k N and μ S 1 . If there exists 0 < C < 1 such that
φ ( 2 x 1 , , 2 x n ) 2 C φ ( x 1 , , x n )

for all x 1 , , x n A , then the mapping f : A B is a C -algebra homomorphism.

Proof Since | J | 3 , letting μ = 1 and x k = 0 for all 1 k n ( k i , j ) in (3.1), we get
f ( r i x i + r j x j 2 ) + f ( r i x i r j x j 2 ) + r i f ( x i ) + r j f ( x j ) = 2 f ( r i x i + r j x j 2 )
for all x i , x j A . By the same reasoning as in the proof of Lemma 2.1, the mapping f is additive and f ( r k x ) = r k f ( x ) for all x A and k = i , j . So, by letting x i = x and x k = 0 for all 1 k n , k i , in (3.1), we get f ( μ x ) = μ f ( x ) for all x A and μ S 1 . Therefore, by Lemma 3.1, the mapping f is -linear. Hence, it follows from (3.2) and (3.3) that
f ( u ) f ( u ) B = lim k 1 2 k f ( 2 k u ) f ( 2 k u ) B lim k 1 2 k φ ( 2 k u , , 2 k u n times ) lim k C k φ ( u , , u n times ) = 0 , f ( u x ) f ( u ) f ( x ) B = lim k 1 2 k f ( 2 k u x ) f ( 2 k u ) f ( x ) B lim k 1 2 k φ ( 2 k u x , , 2 k u x n times ) lim k C k φ ( u x , , u x n times ) = 0
for all x A and u U ( A ) . So, we have f ( u ) = f ( u ) and f ( u x ) = f ( u ) f ( x ) for all x A and u U ( A ) . Since f is -linear and each x A is a finite linear combination of unitary elements (see [65]), i.e., x = k = 1 m λ k u k , where λ k C and u k U ( A ) for all 1 k n , we have
f ( x ) = f ( k = 1 m λ ¯ k u k ) = k = 1 m λ ¯ k f ( u k ) = k = 1 m λ ¯ k f ( u k ) = ( k = 1 m λ k f ( u k ) ) = f ( k = 1 m λ k u k ) = f ( x ) , f ( x y ) = f ( k = 1 m λ k u k y ) = k = 1 m λ k f ( u k y ) = k = 1 m λ k f ( u k ) f ( y ) = f ( k = 1 m λ k u k ) f ( y ) = f ( x ) f ( y )

for all x , y A . Therefore, the mapping f : A B is a C -algebra homomorphism. This completes the proof. □

The following theorem is an alternative result of Theorem 3.2.

Theorem 3.3 Let f : A B be a mapping with f ( 0 ) = 0 for which there is a function φ : A n [ 0 , ) satisfying
(3.4)
(3.5)
for all x , x 1 , , x n A , u U ( A ) , k N and μ S 1 . If there exists 0 < C < 1 such that
φ ( x 1 , , 2 n ) C 2 φ ( 2 x 1 , , 2 x n )

for all x 1 , , x n A , then the mapping f : A B is a C -algebra homomorphism.

Remark 3.4 In Theorems 3.2 and 3.3, one can assume that k = 1 n r k 0 instead of f ( 0 ) = 0 .

Theorem 3.5 Let f : A B be a mapping with f ( 0 ) = 0 for which there is a function φ : A n [ 0 , ) satisfying (3.2), (3.3) and
D μ , r 1 , , r n f ( x 1 , , x n ) B φ ( x 1 , , x n )
(3.6)
for all x 1 , , x n A and μ S 1 . Assume that lim k 1 2 k f ( 2 k e ) is invertible. If there exists 0 < C < 1 such that
φ ( 2 x 1 , , 2 x n ) 2 C φ ( x 1 , , x n )

for all x 1 , , x n A , then the mapping f : A B is a C -algebra homomorphism.

Proof Consider the C -algebras A and B as left Banach modules over the unital C -algebra . By Theorem 2.4, there exists a unique -linear generalized Euler-Lagrange type additive mapping H : A B defined by
H ( x ) = lim k 1 2 k f ( 2 k x )
for all x A . By (3.2) and (3.3), we get
H ( u ) H ( u ) B = lim k 1 2 k f ( 2 k u ) f ( 2 k u ) B lim k 1 2 k φ ( 2 k u , , 2 k u n times ) = 0 , H ( u x ) H ( u ) f ( x ) B = lim k 1 2 k f ( 2 k u x ) f ( 2 k u ) f ( x ) B lim k 1 2 k φ ( 2 k u x , , 2 k u x n times ) = 0
for all u U ( A ) and x A . So, we have H ( u ) = H ( u ) and H ( u x ) = H ( u ) f ( x ) for all u U ( A ) and x A . Therefore, by the additivity of H, we have
H ( u x ) = lim k 1 2 k H ( 2 k u x ) = H ( u ) lim k 1 2 k f ( 2 k x ) = H ( u ) H ( x )
(3.7)
for all u U ( A ) and all x A . Since H is -linear and each x A is a finite linear combination of unitary elements, i.e., x = k = 1 m λ k u k , where λ k C and u k U ( A ) for all 1 k n , it follows from (3.7) that
H ( x y ) = H ( k = 1 m λ k u k y ) = k = 1 m λ k H ( u k y ) = k = 1 m λ k H ( u k ) H ( y ) = H ( k = 1 m λ k u k ) H ( y ) = H ( x ) H ( y ) , H ( x ) = H ( k = 1 m λ ¯ k u k ) = k = 1 m λ ¯ k H ( u k ) = k = 1 m λ ¯ k H ( u k ) = ( k = 1 m λ k H ( u k ) ) = H ( k = 1 m λ k u k ) = H ( x )
for all x , y A . Since H ( e ) = lim k 1 2 k f ( 2 k e ) is invertible and
H ( e ) H ( y ) = H ( e y ) = H ( e ) f ( y )

for all y A , it follows that H ( y ) = f ( y ) for all y A . Therefore, the mapping f : A B is a C -algebra homomorphism. This completes the proof. □

The following theorem is an alternative result of Theorem 3.5.

Theorem 3.6 Let f : A B be a mapping with f ( 0 ) = 0 for which there is a function φ : A n [ 0 , ) satisfying (3.4), (3.5) and
D μ , r 1 , , r n f ( x 1 , , x n ) B φ ( x 1 , , x n )
for all x 1 , , x n A and μ S 1 . Assume that lim k 2 k f ( e 2 k ) is invertible. If there exists 0 < C < 1 such that
φ ( x 1 , , 2 n ) C 2 φ ( 2 x 1 , , 2 x n )

for all x 1 , , x n A , then the mapping f : A B is a C -algebra homomorphism.

Remark 3.7 In Theorem 3.6, one can assume that k = 1 n r k 0 instead of f ( 0 ) = 0 .

Theorem 3.8 Let f : A B be a mapping with f ( 0 ) = 0 for which there is a function φ : A n [ 0 , ) satisfying (3.2), (3.3) and
D μ , r 1 , , r n f ( x 1 , , x n ) B φ ( x 1 , , x n )
(3.8)
for all x 1 , , x n A and μ = i , 1 . Assume that lim k 1 2 k f ( 2 k e ) is invertible and for each fixed x A the mapping t f ( t x ) is continuous in t R . If there exists 0 < C < 1 such that
φ ( 2 x 1 , , 2 x n ) 2 C φ ( x 1 , , x n )

for all x 1 , , x n A , then the mapping f : A B is a C -algebra homomorphism.

Proof Put μ = 1 in (3.8). By the same reasoning as in the proof of Theorem 2.3, there exists a unique generalized Euler-Lagrange type additive mapping H : A B defined by
H ( x ) = lim k f ( 2 k x ) 2 k
for all x A . By the same reasoning as in the proof of [58], the generalized Euler-Lagrange type additive mapping H : A B is -linear. By the same method as in the proof of Theorem 2.4, we have
D μ , r 1 , , r n H ( 0 , , 0 , x j th , 0 , , 0 ) Y = lim k 1 2 k D μ , r 1 , , r n f ( 0 , , 0 , 2 k x j th , 0 , , 0 ) Y lim k 1 2 k φ ( 0 , , 0 , 2 k x j th , 0 , , 0 ) = 0
for all x A and so
r j μ H ( x ) = H ( r j μ x )
for all x A . Since H ( r j x ) = r j H ( x ) for all x X and r j 0 ,
H ( μ x ) = μ H ( x )
for all x A and μ = i , 1 . For each λ C , we have λ = s + i t , where s , t R . Thus, it follows that
H ( λ x ) = H ( s x + i t x ) = s H ( x ) + t H ( i x ) = s H ( x ) + i t H ( x ) = ( s + i t ) H ( x ) = λ H ( x )
for all λ C and x A and so
H ( ζ x + η y ) = H ( ζ x ) + H ( η y ) = ζ H ( x ) + η H ( y )

for all ζ , η C and x , y A . Hence, the generalized Euler-Lagrange type additive mapping H : A B is -linear.

The rest of the proof is the same as in the proof of Theorem 3.5. This completes the proof. □

The following theorem is an alternative result of Theorem 3.8.

Theorem 3.9 Let f : A B be a mapping with f ( 0 ) = 0 for which there is a function φ : A n [ 0 , ) satisfying (3.4), (3.5) and
D μ , r 1 , , r n f ( x 1 , , x n ) B φ ( x 1 , , x n ) ,
for all x , x 1 , , x n A and μ = i , 1 . Assume that lim k 2 k f ( e 2 k ) is invertible and for each fixed x A the mapping t f ( t x ) is continuous in t R . If there exists 0 < C < 1 such that
φ ( x 1 , , 2 n ) C 2 φ ( 2 x 1 , , 2 x n )

for all x 1 , , x n A , then the mapping f : A B is a C -algebra homomorphism.

Proof We omit the proof because it is very similar to last theorem. □

Remark 3.10 In Theorem 3.9, one can assume that k = 1 n r k 0 instead of f ( 0 ) = 0 .

Declarations

Acknowledgements

The authors are grateful to the reviewers for their valuable comments and suggestions.

Authors’ Affiliations

(1)
Department of Mathematics, Research Institute for Natural Sciences, Hanyang University, Seoul, 133-791, Korea
(2)
Department of Mathematics Education and RINS, Gyeongsang National University, Chinju, 660-701, Korea
(3)
Department of Mathematics and Computer Science, Iran University of Science and Technology, Tehran, Iran

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