Open Access

Additive functional inequalities in 2-Banach spaces

Journal of Inequalities and Applications20132013:447

https://doi.org/10.1186/1029-242X-2013-447

Received: 26 February 2013

Accepted: 30 August 2013

Published: 4 November 2013

Abstract

We prove the Hyers-Ulam stability of the Cauchy functional inequality and theCauchy-Jensen functional inequality in 2-Banach spaces.

Moreover, we prove the superstability of the Cauchy functional inequality and theCauchy-Jensen functional inequality in 2-Banach spaces under someconditions.

MSC: 39B82, 39B52, 39B62, 46B99, 46A19.

Keywords

Hyers-Ulam stabilitylinear 2-normed spaceadditive mappingadditive functional inequalitysuperstability

1 Introduction and preliminaries

In 1940, Ulam [1] suggested the stability problem of functional equations concerning thestability of group homomorphisms as follows: Let ( G , ) be a group and let ( H , , d ) be a metric group with the metric d ( , ) . Given ε > 0 , does there exist a δ = δ ( ε ) > 0 such that if a mapping f : G H satisfies the inequality
d ( f ( x y ) , f ( x ) f ( y ) ) < δ
for all x , y G , then a homomorphism F : G H exists with
d ( f ( x ) , F ( x ) ) < ε

for all x G ?

In 1941, Hyers [2] gave a first (partial) affirmative answer to the question of Ulam forBanach spaces. Thereafter, we call that type the Hyers-Ulam stability.

Hyers’ theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. Ageneralization of the Rassias theorem was obtained by Gǎvruta [5] by replacing the unbounded Cauchy difference by a general controlfunction.

Gähler [6, 7] introduced the concept of linear 2-normed spaces.

Definition 1.1 Let be a real linear space with dim X > 1 , and let , : X × X R 0 be a function satisfying the following properties:
  1. (a)

    x , y = 0 if and only if x and y are linearly dependent,

     
  2. (b)

    x , y = y , x ,

     
  3. (c)

    α x , y = | α | x , y ,

     
  4. (d)

    x , y + z x , y + x , z

     

for all x , y , z X and α R . Then the function , is called 2-norm on and the pair ( X , , ) is called a linear 2-normed space.Sometimes condition (d) is called the triangle inequality.

See [8] for examples and properties of linear 2-normed spaces.

White [9, 10] introduced the concept of 2-Banach spaces. In order to definecompleteness, the concepts of Cauchy sequences and convergence are required.

Definition 1.2 A sequence { x n } in a linear 2-normed space is called a Cauchy sequence if
lim m , n x n x m , y = 0

for all y X .

Definition 1.3 A sequence { x n } in a linear 2-normed space is called a convergent sequenceif there is an x X such that
lim n x n x , y = 0

for all y X . If { x n } converges to x, write x n x as n and call x the limit of { x n } . In this case, we also write lim n x n = x .

The triangle inequality implies the following lemma.

Lemma 1.4[11]

For a convergent sequence { x n } in a linear 2-normed space ,
lim n x n , y = lim n x n , y

for all y X .

Definition 1.5 A linear 2-normed space, in which every Cauchy sequence is aconvergent sequence, is called a 2-Banach space.

Eskandani and Gǎvruta [12] proved the Hyers-Ulam stability of a functional equation in 2-Banachspaces.

In [13], Gilányi showed that if f satisfies the functionalinequality
2 f ( x ) + 2 f ( y ) f ( x y 1 ) f ( x y ) ,
(1.1)
then f satisfies the Jordan-von Neumann functional equation
2 f ( x ) + 2 f ( y ) = f ( x y ) + f ( x y 1 ) .

See also [14]. Gilányi [15] and Fechner [16] proved the Hyers-Ulam stability of functional inequality (1.1).

Park et al.[17] proved the Hyers-Ulam stability of the following functional inequalities:
f ( x ) + f ( y ) + f ( z ) f ( x + y + z ) ,
(1.2)
f ( x ) + f ( y ) + 2 f ( z ) 2 f ( x + y 2 + z ) .
(1.3)

In this paper, we prove the Hyers-Ulam stability of Cauchy functional inequality(1.2) and Cauchy-Jensen functional inequality (1.3) in 2-Banach spaces.

Moreover, we prove the superstability of Cauchy functional inequality (1.2) andCauchy-Jensen functional inequality (1.3) in 2-Banach spaces under someconditions.

Throughout this paper, let be a normed linear space, and let be a 2-Banach space.

2 Hyers-Ulam stability of Cauchy functional inequality (1.2) in 2-Banachspaces

In this section, we prove the Hyers-Ulam stability of Cauchy functional inequality(1.2) in 2-Banach spaces.

Proposition 2.1 Let f : X Y be a mapping satisfying
f ( x ) + f ( y ) + f ( z ) , w f ( x + y + z ) , w
(2.1)

for all x , y , z X and all w Y . Then the mapping f : X Y is additive.

Proof Letting x = y = z = 0 in (2.1), we get 3 f ( 0 ) , w f ( 0 ) , w and so f ( 0 ) , w = 0 for all w Y . Hence f ( 0 ) = 0 .

Letting y = x and z = 0 in (2.1), we get f ( x ) + f ( x ) , w f ( 0 ) , w = 0 and so f ( x ) + f ( x ) , w = 0 for all x X and all w Y . Hence f ( x ) + f ( x ) = 0 for all x X .

Letting z = x y in (2.1), we get
f ( x ) + f ( y ) + f ( x y ) , w f ( 0 ) , w = 0
and so
f ( x ) + f ( y ) + f ( x y ) , w = 0
for all x , y X and all w Y . Hence
0 = f ( x ) + f ( y ) + f ( x y ) = f ( x ) + f ( y ) f ( x + y )

for all x , y X . So, f : X Y is additive. □

Theorem 2.2 Let θ [ 0 , ) , p , q , r ( 0 , ) with p + q + r < 1 , and let f : X Y be a mapping satisfying
f ( x ) + f ( y ) + f ( z ) , w f ( x + y + z ) , w + θ x p y q z r w
(2.2)
for all x , y , z X and all w Y . Then there is a unique additive mapping A : X Y such that
f ( x ) A ( x ) , w 2 r θ 2 2 p + q + r x p + q + r w
(2.3)

for all x X and all w Y .

Proof Letting x = y = z = 0 in (2.2), we get 3 f ( 0 ) , w f ( 0 ) , w and so f ( 0 ) , w = 0 for all w Y . Hence f ( 0 ) = 0 .

Letting y = x and z = 0 in (2.2), we get f ( x ) + f ( x ) , w f ( 0 ) , w = 0 and so f ( x ) + f ( x ) , w = 0 for all x X and all w Y . Hence f ( x ) + f ( x ) = 0 for all x X .

Putting y = x and z = 2 x in (2.2), we get
f ( 2 x ) 2 f ( x ) , w f ( 0 ) , w + 2 r θ x p + q + r w = 2 r θ x p + q + r w
(2.4)
for all x X and all w Y . So, we get
f ( x ) 1 2 f ( 2 x ) , w 2 r θ 2 x p + q + r w
(2.5)
for all x X and all w Y . Replacing x by 2 j x in (2.5) and dividing by 2 j , we obtain
1 2 j f ( 2 j x ) 1 2 j + 1 f ( 2 j + 1 x ) , w 2 ( p + q + r 1 ) j + r 1 θ x p + q + r w
for all x X , all w Y and all integers j 0 . For all integers l, m with 0 l < m , we get
1 2 l f ( 2 l x ) 1 2 m f ( 2 m x ) , w j = l m 1 2 ( p + q + r 1 ) j + r 1 θ x p + q + r w
(2.6)
for all x X and all w Y . So, we get
lim l 1 2 l f ( 2 l x ) 1 2 m f ( 2 m x ) , w = 0
for all x X and all w Y . Thus the sequence { 1 2 j f ( 2 j x ) } is a Cauchy sequence in for each x X . Since is a 2-Banach space, the sequence { 1 2 j f ( 2 j x ) } converges for each x X . So, one can define the mapping A : X Y by
A ( x ) : = lim j 1 2 j f ( 2 j x )
for all x X . That is,
lim j 1 2 j f ( 2 j x ) A ( x ) , w = 0

for all x X and all w Y .

By (2.2), we get
lim j 1 2 j ( f ( 2 j x ) + f ( 2 j y ) + f ( 2 j z ) ) , w lim j ( 1 2 j f ( 2 j x + 2 j y + 2 j z ) , w + 2 ( p + q + r ) j 2 j θ x p y q z r w ) lim j 1 2 j f ( 2 j x + 2 j y + 2 j z ) , w
for all x , y , z X and all w Y . So,
A ( x ) + A ( y ) + A ( z ) , w A ( x + y + z ) , w

for all x , y , z X and all w Y . By Proposition 2.1, A : X Y is additive.

By Lemma 1.4 and (2.6), we have
f ( x ) A ( x ) , w = lim m f ( x ) 1 2 m f ( 2 m x ) , w 2 r θ 2 2 p + q + r x p + q + r w

for all x X and all w Y .

Now, let B : X Y be another additive mapping satisfying (2.3). Then wehave
A ( x ) B ( x ) , w = 1 2 j A ( 2 j x ) B ( 2 j x ) , w 1 2 j [ A ( 2 j x ) f ( 2 j x ) , w + f ( 2 j x ) B ( 2 j x ) , w ] 2 2 r θ 2 2 p + q + r x p + q + r w 2 ( p + q + r ) j 2 j ,

which tends to zero as j for all x X and all w Y . By Definition 1.1, we can conclude that A ( x ) = B ( x ) for all x X . This proves the uniqueness ofA. □

Theorem 2.3 Let θ [ 0 , ) , p , q , r ( 0 , ) with p + q + r > 1 , and let f : X Y be a mapping satisfying (2.2). Then there is a unique additivemapping A : X Y such that
f ( x ) A ( x ) , w 2 r θ 2 p + q + r 2 x p + q + r w

for all x X and all w Y .

Proof It follows from (2.4) that
f ( x ) 2 f ( x 2 ) , w θ 2 p + q x p + q + r w
(2.7)
for all x X and all w Y . Replacing x by x 2 j in (2.7) and multiplying by 2 j , we obtain
2 j f ( x 2 j ) 2 j + 1 f ( x 2 j + 1 ) , w 2 j θ 2 p + q 2 ( p + q + r ) j x p + q + r w
for all x X and all w Y and all integers j 0 . For all integers l, m with 0 l < m , we get
2 l f ( x 2 l ) 2 m f ( x 2 m ) , w j = l m 1 2 j θ 2 p + q 2 ( p + q + r ) j x p + q + r w
for all x X and all w Y . So, we get
lim l 2 l f ( x 2 l ) 2 m f ( x 2 m ) , w = 0
for all x X and all w Y . Thus the sequence { 2 j f ( x 2 j ) } is a Cauchy sequence in . Since is a 2-Banach space, the sequence { 2 j f ( x 2 j ) } converges. So, one can define the mapping A : X Y by
A ( x ) : = lim j 2 j f ( x 2 j )
for all x X . That is,
lim j 2 j f ( x 2 j ) A ( x ) , w = 0

for all x X and all w Y .

The further part of the proof is similar to the proof of Theorem2.2. □

Now we prove the superstability of the Cauchy functional inequality in 2-Banachspaces.

Theorem 2.4 Let θ [ 0 , ) , p , q , r , t ( 0 , ) with t 1 , and let f : X Y be a mapping satisfying
f ( x ) + f ( y ) + f ( z ) , w f ( x + y + z ) , w + θ x p y q z r w t
(2.8)

for all x , y , z X and all w Y . Then f : X Y is an additive mapping.

Proof Replacing w by sw in (2.8) for s R { 0 } , we get
f ( x ) + f ( y ) + f ( z ) , s w f ( x + y + z ) , s w + θ x p y q z r s w t
and so
f ( x ) + f ( y ) + f ( z ) , w f ( x + y + z ) , w + θ x p y q z r w t | s | t | s |
(2.9)

for all x , y , z X , all w Y and all s R { 0 } .

If t > 1 , then the right-hand side of (2.9) tends to f ( x + y + z ) , w as s 0 .

If t < 1 , then the right-hand side of (2.9) tends to f ( x + y + z ) , w as s + .

Thus
f ( x ) + f ( y ) + f ( z ) , w f ( x + y + z ) , w

for all x , y , z X and all w Y . By Proposition 2.1, f : X Y is additive. □

3 Hyers-Ulam stability of Cauchy-Jensen functional inequality (1.3) in 2-Banachspaces

In this section, we prove the Hyers-Ulam stability of Cauchy-Jensen functionalinequality (1.3) in 2-Banach spaces.

Proposition 3.1 Let f : X Y be a mapping satisfying
f ( x ) + f ( y ) + 2 f ( z ) , w 2 f ( x + y 2 + z ) , w
(3.1)

for all x , y , z X and all w Y . Then the mapping f : X Y is additive.

Proof Letting x = y = z = 0 in (3.1), we get 4 f ( 0 ) , w 2 f ( 0 ) , w and so f ( 0 ) , w = 0 for all w Y . Hence f ( 0 ) = 0 .

Letting y = x and z = 0 in (3.1), we get f ( x ) + f ( x ) , w 2 f ( 0 ) , w = 0 and so f ( x ) + f ( x ) , w = 0 for all x X and all w Y . Hence f ( x ) + f ( x ) = 0 for all x X .

Letting z = x + y 2 in (3.1), we get
f ( x ) + f ( y ) + 2 f ( x + y 2 ) , w 2 f ( 0 ) , w = 0
and so
f ( x ) + f ( y ) + 2 f ( x + y 2 ) , w = 0
for all x , y X and all w Y . Hence
0 = f ( x ) + f ( y ) + 2 f ( x + y 2 ) = f ( x ) + f ( y ) 2 f ( x + y 2 )

for all x , y X . Since f ( 0 ) = 0 , f : X Y is additive. □

Theorem 3.2 Let θ [ 0 , ) , p , q , r ( 0 , ) with p + q + r < 1 , and let f : X Y be a mapping satisfying
f ( x ) + f ( y ) + 2 f ( z ) , w 2 f ( x + y 2 + z ) , w + θ x p y q z r w
(3.2)
for all x , y , z X and all w Y . Then there is a unique additive mapping A : X Y such that
f ( x ) A ( x ) , w θ 2 2 p + q + r x p + q + r w

for all x X and all w Y .

Proof Letting x = y = z = 0 in (3.2), we get 4 f ( 0 ) , w 2 f ( 0 ) , w and so f ( 0 ) , w = 0 for all w Y . Hence f ( 0 ) = 0 .

Letting y = x and z = 0 in (3.2), we get f ( x ) + f ( x ) , w 2 f ( 0 ) , w = 0 and so f ( x ) + f ( x ) , w = 0 for all x X and all w Y . Hence f ( x ) + f ( x ) = 0 for all x X .

Letting y = x and z = x in (3.2), we get
2 f ( x ) f ( 2 x ) , w 2 f ( 0 ) , w + θ x p + q + r w = θ x p + q + r w
(3.3)
for all x X and all w Y . Replacing x by 2 j x in (3.3) and dividing by 3 j , we obtain
1 2 j f ( 2 j x ) 1 2 j + 1 f ( 2 j + 1 x ) , w 2 ( p + q + r ) j 2 2 j θ x p + q + r w
for all x X and all w Y and all integers j 0 . For all integers l, m with 0 l < m , we get
1 2 l f ( 2 l x ) 1 2 m f ( 2 m x ) , w j = l m 1 2 ( p + q + r ) j 2 2 j θ x p + q + r w
for all x X and all w Y . So, we get
lim l 1 2 l f ( 2 l x ) 1 2 m f ( 2 m x ) , w = 0
for all x X and all w Y . Thus the sequence { 1 2 j f ( 2 j x ) } is a Cauchy sequence in for each x X . Since is a 2-Banach space, the sequence { 1 2 j f ( 2 j x ) } converges for each x X . So, one can define the mapping A : X Y by
A ( x ) : = lim j 1 2 j f ( 2 j x ) = lim j 1 2 j f ( 2 j x )
for all x X . That is,
lim j 1 2 j f ( 2 j x ) A ( x ) , w = lim j 1 2 j f ( 2 j x ) A ( x ) , w = 0

for all x X and all w Y .

The further part of the proof is similar to the proof of Theorem2.2. □

Theorem 3.3 Let θ [ 0 , ) , p , q , r ( 0 , ) with p + q + r > 1 , and let f : X Y be a mapping satisfying (3.2). Then there is a unique additivemapping A : X Y such that
f ( x ) A ( x ) , w θ 2 p + q + r 2 x p + q + r w

for all x X and all w Y .

Proof It follows from (3.3) that
f ( x ) 2 f ( x 2 ) , w 1 2 p + q + r θ x p + q + r w
(3.4)
for all x X and all w Y . Replacing x by x 2 j in (3.4) and multiplying by 2 j , we obtain
2 j f ( x 2 j ) 2 j + 1 f ( x 2 j + 1 ) , w 2 j 2 ( p + q + r ) ( j + 1 ) θ x p + q + r w
for all x X and all w Y and all integers j 0 . For all integers l, m with 0 l < m , we get
2 l f ( x 2 l ) 2 m f ( x 2 m ) , w j = l m 1 2 j 2 ( p + q + r ) ( j + 1 ) θ x p + q + r w
for all x X and all w Y . So, we get
lim l 2 l f ( x 2 l ) 2 m f ( x 2 m ) , w = 0
for all x X and all w Y . Thus the sequence { 2 j f ( x 2 j ) } is a Cauchy sequence in for each x X . Since is a 2-Banach space, the sequence { 2 j f ( x 2 j ) } converges for each x X . So, one can define the mapping A : X Y by
A ( x ) : = lim j 2 j f ( x 2 j )
for all x X . That is,
lim j 2 j f ( x 2 j ) A ( x ) , w = 0

for all x X and all w Y .

The further part of the proof is similar to the proof of Theorem2.2. □

Now we prove the superstability of the Jensen functional equation in 2-Banachspaces.

Theorem 3.4 Let θ [ 0 , ) , p , q , r , t ( 0 , ) with t 1 , and let f : X Y be a mapping satisfying
f ( x ) + f ( y ) + 2 f ( z ) , w 2 f ( x + y 2 + z ) , w + θ x p y q z r w t
(3.5)

for all x , y , z X and all w Y . Then f : X Y is an additive mapping.

Proof Replacing w by sw in (3.5) for s R { 0 } , we get
f ( x ) + f ( y ) + 2 f ( z ) , s w 2 f ( x + y 2 + z ) , s w + θ x p y q z r s w t
and so
f ( x ) + f ( y ) + 2 f ( z ) , w 2 f ( x + y 2 + z ) , w + θ x p y q z r w t | s | t | s |

for all x , y , z X , all w Y and all s R { 0 } .

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

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Research Institute for Natural Sciences, Hanyang University

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© Park; licensee Springer. 2013

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