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Additive functional inequalities in Banach spaces

Abstract

In this paper, we prove the Hyers-Ulam stability of the following function inequalities:

in Banach spaces.

MSC:39B62, 39B52, 46B25.

1 Introduction and preliminaries

The stability problem of functional equations originated from the question of Ulam [1] in 1940 concerning the stability of group homomorphisms. Let ( G 1 ,) be a group and let ( G 2 ,) be a metric group with the metric d(,). Given ϵ>0, does there exist a δ 0 such that if a mapping h: G 1 G 2 satisfies the inequality d(h(xy),h(x)h(y))<δ for all x,y G 1 , then there exists a homomorphism H: G 1 G 2 with d(h(x),H(x))<ϵ for all x G 1 ? In other words, under what condition does there exist a homomorphism near an approximate homomorphism? The concept of stability for a functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. In 1941, Hyers [2] gave the first affirmative answer to the question of Ulam for Banach spaces. Let f:E E be a mapping between Banach spaces such that

f ( x + y ) f ( x ) f ( y ) δ

for all x,yE and for some δ>0. Then there exists a unique additive mapping T:E E such that

f ( x ) T ( x ) δ

for all xE. Moreover, if f(tx) is continuous in tR for each fixed xE, then T is -linear. In 1978, Th.M. Rassias [3] proved the following theorem.

Theorem 1.1 Let f:E E be a mapping from a normed vector space E into a Banach space E subject to the inequality

f ( x + y ) f ( x ) f ( y ) ϵ ( x p + y p )
(1.1)

for all x,yE, where ϵ and p are constants with ϵ>0 and p<1. Then there exists a unique additive mapping T:E E such that

f ( x ) T ( x ) 2 ϵ 2 2 p x p
(1.2)

for all xE. If p<0, then inequality (1.1) holds for all x,y0, and (1.2) for x0. Also, if the function tf(tx) from into E is continuous in tR for each fixed xE, then T is -linear.

In 1991, Gajda [4] answered the question for the case p>1, which was raised by Th.M. Rassias. On the other hand, J.M. Rassias [5] generalized the Hyers-Ulam stability result by presenting a weaker condition controlled by a product of different powers of norms.

Theorem 1.2 [6, 7]

If it is assumed that there exist constants Θ0 and p 1 , p 2 R such that p= p 1 + p 2 1, and f:E E is a mapping from a norm space E into a Banach space E such that the inequality

f ( x + y ) f ( x ) f ( y ) ϵ x p 1 y p 2

holds for all x,yE, then there exists a unique additive mapping T:E E such that

f ( x ) T ( x ) Θ 2 2 p x p

for all xE. If, in addition, for every xE, f(tx) is continuous in tR for each fixed xE, then T is -linear.

More generalizations and applications of the Hyers-Ulam stability to a number of functional equations and mappings can be found in [811].

In [12], Park et al. investigated the following inequalities:

in Banach spaces. Recently, Cho et al. [13] investigated the following functional inequality:

f ( x ) + f ( y ) + f ( z ) K f ( x + y + z K ) ( 0 < | K | < | 3 | )

in non-Archimedean Banach spaces. Lu and Park [14] investigated the following functional inequality:

i = 1 N f ( x i ) K f ( i = 1 N ( x i ) K ) ( 0 < | K | N )

in Fréchet spaces.

In this paper, we investigate the following functional inequalities:

(1.3)
(1.4)

and prove the Hyers-Ulam stability of functional inequalities (1.3) and (1.4) in Banach spaces.

Throughout this paper, assume that X is a normed vector space and that (Y,) is a Banach space.

2 Hyers-Ulam stability of functional inequality (1.3)

Throughout this section, assume that K is a real number with 0<|K|<3.

Proposition 2.1 Let f:XY be a mapping such that

f ( x ) + f ( y ) + f ( z ) K f ( x + y + z K )
(2.1)

for all x,y,zX. Then the mapping f:XY is additive.

Proof Letting x=y=z=0 in (2.1), we get

3 f ( 0 ) K f ( 0 ) .

So, f(0)=0.

Letting z=0 and y=x in (2.1), we get

f ( x ) + f ( x ) K f ( 0 ) =0

for all xX. So, f(x)=f(x) for all xX.

Letting z=xy in (2.1), we get

f ( x ) + f ( y ) f ( x + y ) = f ( x ) + f ( y ) + f ( x y ) K f ( 0 ) =0

for all x,yX. Thus,

f(x+y)=f(x)+f(y)

for all x,yX, as desired. □

Theorem 2.2 Assume that a mapping f:XY satisfies the inequality

f ( x ) + f ( y ) + f ( z ) K f ( x + y + z K ) +ϕ(x,y,z),
(2.2)

where ϕ: X 3 [0,) satisfies

ϕ ˜ (x,y,z):= j = 1 2 j ϕ ( x 2 j , y 2 j , z 2 j ) <
(2.3)

for all x,y,zX. Then there exists a unique additive mapping A:XY such that

A ( x ) f ( x ) 1 2 ϕ ˜ (x,x,2x)+ ϕ ˜ (x,x,0)
(2.4)

for all xX.

Proof It follows from (2.3) that ϕ(0,0,0)=0. Letting x=y=z=0 in (2.2), we get 3f(0)Kf(0)+ϕ(0,0,0)=Kf(0). So, f(0)=0.

Letting y=x, z=2x in (2.2), we get

2 f ( x ) + f ( 2 x ) ϕ(x,x,2x)

for all xX. So,

2 f ( x 2 ) + f ( x ) ϕ ( x 2 , x 2 , x )
(2.5)

for all xX.

Letting y=x and z=0 in (2.2), we get

f ( x ) + f ( x ) ϕ(x,x,0)
(2.6)

for all xX. It follows from (2.5) and (2.6) that

for all nonnegative integers m and l with m>l and all xX. It means that the sequence { 2 n f( x 2 n )} is a Cauchy sequence for all xX. Since Y is complete, the sequence { 2 n f( x 2 n )} converges. We define the mapping A:XY by A(x)= lim n 2 n f( x 2 n ) for all xX. Moreover, letting l=0 and passing the limit m, we get (2.4).

Next, we show that A(x) is an additive mapping.

A ( x ) + A ( x ) = lim n 2 n f ( x 2 n ) + f ( x 2 n ) lim n 1 2 2 n + 1 ϕ ( 2 x 2 n + 1 , 2 x 2 n + 1 , 0 ) = 0

and so A(x)=A(x) for all xX.

A ( x ) + A ( y ) A ( x + y ) = A ( x ) + A ( y ) + A ( x y ) = lim n 2 n f ( x 2 n ) + f ( y 2 n ) + f ( x y 2 n ) lim n 1 2 2 n + 1 ϕ ( 2 x 2 n + 1 , 2 y 2 n + 1 , 2 ( x + y ) 2 n + 1 ) = 0

for all x,yX. Thus, the mapping A:XY is additive.

Now, we prove the uniqueness of A. Assume that T:XY is another additive mapping satisfying (2.4). Then we obtain

A ( x ) T ( x ) = lim n 2 n A ( x 2 n ) T ( x 2 n ) lim n 2 n [ A ( x 2 n ) f ( x 2 n ) + T ( x 2 n ) f ( x 2 n ) ] lim n [ ϕ ˜ ( x 2 n , x 2 n , 2 x 2 n ) + 2 ϕ ˜ ( x 2 n , x 2 n , 0 ) ] = 0

for all xX. Then we can conclude that A(x)=T(x) for all xX. This completes the proof. □

Corollary 2.3 Let p and θ be positive real numbers with p>1. Let f:XY be a mapping satisfying

f ( x ) + f ( y ) + f ( z ) K f ( x + y + z K ) +θ ( x p + y p + z p )

for all x,y,zX. Then there exists a unique additive mapping A:XY such that

f ( x ) A ( x ) 2 p + 6 2 p 2 θ x p

for all xX.

3 Hyers-Ulam stability of functional inequality (1.4)

Throughout this section, assume that K is a real number with 0<K2.

Proposition 3.1 Let f:XY be a mapping such that

f ( x ) + f ( y ) + K f ( z ) K f ( x + y K + z )
(3.1)

for all x,y,zX. Then the mapping f:XY is additive.

Proof Letting x=y=z=0 in (3.1), we get

( K + 2 ) f ( 0 ) K f ( 0 ) .

So, f(0)=0.

Letting z=0 and y=x in (3.1), we get

f ( x ) + f ( x ) K f ( 0 ) =0

for all xX. So, f(x)=f(x) for all xX.

Letting z= x y K in (3.1), we get

f ( x ) + f ( y ) K f ( x + y K ) = f ( x ) + f ( y ) + K f ( x y K ) K f ( 0 ) =0

for all x,yX. Thus,

Kf ( x + y K ) =f(x)+f(y)
(3.2)

for all x,yX. Letting y=0 in (3.2), we get Kf( x K )=f(x) for all xX. So,

f(x+y)=Kf ( x + y K ) =f(x)+f(y)

for all x,yX, as desired. □

Theorem 3.2 Let K be a positive real number with K<2. Assume that a mapping f:XY satisfies the inequality

f ( x ) + f ( y ) + K f ( z ) K f ( x + y K + z ) +ϕ(x,y,z),
(3.3)

where ϕ: X 3 [0,) satisfies

ϕ ˜ (x,y,z):= j = 1 ( 2 K ) j ϕ ( ( K 2 ) j x , ( K 2 ) j y , ( K 2 ) j z ) <
(3.4)

for all x,y,zX. Then there exists a unique additive mapping A:XY such that

A ( x ) f ( x ) 1 2 ϕ ˜ ( x , x , 2 K x ) + ϕ ˜ (x,x,0)
(3.5)

for all xX.

Proof It follows from (3.4) that ϕ(0,0,0)=0. Letting x=y=z=0 in (3.3), we get (K+2)f(0)Kf(0)+ϕ(0,0,0)=Kf(0). So, f(0)=0.

Letting y=x, z=0 in (3.3), we get

f ( x ) + f ( x ) ϕ(x,x,0)
(3.6)

for all xX. Letting x=y=Kx, z=2x in (3.3), we obtain

f ( K x ) + f ( K x ) + f ( 2 x ) ϕ(Kx,Kx,2x)

for all xX. So,

2 K f ( K 2 x ) + f ( x ) 1 K ϕ ( K x 2 , K x 2 , x )
(3.7)

for all xX. It follows from (3.6) and (3.7) that

for all nonnegative integers m and l with m>l and all xX. It means that the sequence { ( 2 K ) n f( ( K 2 ) n x)} is a Cauchy sequence for all xX. Since Y is complete, the sequence { ( 2 K ) n f( ( K 2 ) n x)} converges. So, we may define the mapping A:XY by A(x)= lim n ( ( 2 K ) n f( ( K 2 ) n x)) for all xX.

Moreover, by letting l=0 and passing the limit m, we get (3.5).

Next, we claim that A(x) is an additive mapping. It follows from (3.6) that

A ( x ) + A ( x ) = lim n ( 2 K ) n f ( ( K 2 ) n x ) + f ( ( K 2 ) n x ) lim n ( 2 K ) n ϕ ( ( K 2 ) n x , ( K 2 ) n x , 0 ) = lim n K 2 ( 2 K ) n + 1 ϕ ( ( K 2 ) n + 1 ( 2 K x ) , ( K 2 ) n + 1 ( 2 K x ) , 0 ) = 0

and so A(x)=A(x) for all xX.

It follows from (3.3) that

f ( K x ) K f ( x ) = f ( K x ) + f ( 0 ) + K f ( x ) ϕ(Kx,0,x)

for all xX. Hence,

for all x,yX. So, the mapping A:XY is an additive mapping.

Now, we show the uniqueness of A. Assume that T:XY is another additive mapping satisfying (3.5). Then we get

for all xX. Thus, we may conclude that A(x)=T(x) for all xX. This proves the uniqueness of A. So, the mapping A:XY is a unique additive mapping satisfying (3.5). □

Corollary 3.3 Let p, θ and K be positive real numbers with p>1 and K<2. Let f:XY be a mapping satisfying

f ( x ) + f ( y ) + K f ( z ) K f ( x + y K + z ) +θ ( x p + y p + z p )
(3.8)

for all x,y,zX. Then there exists a unique additive mapping A:XY such that

f ( x ) A ( x ) 1 K ( 2 K ) p + 6 K ( 2 K ) p 2 K θ x p

for all xX.

Theorem 3.4 Let K be a real number with K>2. Assume that a mapping f:XY satisfies inequality (3.3), where ϕ: X 3 [0,) satisfies

ϕ ˜ (x,y,z):= j = 0 ( K 2 ) j ϕ ( ( 2 K ) j x , ( 2 K ) j y , ( 2 K ) j z ) <
(3.9)

for all x,y,zX. Then there exists a unique additive mapping A:XY such that

A ( x ) f ( x ) 1 2 ϕ ˜ ( x , x , 2 K x ) + K 2 ϕ ˜ ( 2 K x , 2 K x , 0 )
(3.10)

for all xX.

Proof It follows from (3.9) that ϕ(0,0,0)=0. Letting x=y=z=0 in (3.3), we get (K+2)f(0)Kf(0)+ϕ(0,0,0)=Kf(0). So, f(0)=0.

Replacing x by 2 K x in (3.7), we get

f ( x ) + K 2 f ( 2 K x ) 1 2 ϕ ( x , x , 2 K x )
(3.11)

for all xX. It follows from (3.6) and (3.11) that

for all nonnegative integers m and l with m>l and all xX. It means that the sequence { ( K 2 ) n f( ( 2 K ) n x)} is a Cauchy sequence for all xX. Since Y is complete, the sequence { ( K 2 ) n f( ( 2 K ) n x)} converges. So, we may define the mapping A:XY by A(x)= lim n ( ( K 2 ) n f( ( 2 K ) n x)) for all xX.

Moreover, by letting l=0 and passing the limit m, we get (3.10).

The rest of the proof is similar to the proof of Theorem 3.2. □

Corollary 3.5 Let p, θ and K be positive real numbers with p>1 and K>2. Let f:XY be a mapping satisfying (3.8). Then there exists a unique additive mapping A:XY such that

f ( x ) A ( x ) 1 K ( 2 K ) p + 6 K 2 K ( 2 K ) p θ x p

for all xX.

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Correspondence to Choonkil Park.

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Lu, G., Park, C. Additive functional inequalities in Banach spaces. J Inequal Appl 2012, 294 (2012). https://doi.org/10.1186/1029-242X-2012-294

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Keywords

  • Hyers-Ulam stability
  • additive functional inequality
  • additive mapping