Additive functional inequalities in Banach spaces

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 ( x ⋅ y ) , 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

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

for all x ∈ E. Moreover, if f ( t x ) is continuous in t ∈ R for each fixed x ∈ E, then T is ℝ-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

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

for all x ∈ E. If p < 0, then inequality (1.1) holds for all x , y ≠ 0, and (1.2) for x ≠ 0. Also, if the function t ↦ f ( t x ) from ℝ into E ′ is continuous in t ∈ R for each fixed x ∈ E, then T is ℝ-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 Θ ≥ 0 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

holds for all x , y ∈ E, then there exists a unique additive mapping T : E → E ′ such that

for all x ∈ E. If, in addition, for every x ∈ E, f ( t x ) is continuous in t ∈ R for each fixed x ∈ E, then T is ℝ-linear.

More generalizations and applications of the Hyers-Ulam stability to a number of functional equations and mappings can be found in [8‐11].

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

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

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

in Fréchet spaces.

In this paper, we investigate the following functional inequalities:

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.

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

Proposition 2.1 Let f : X → Y be a mapping such that

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

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

So, f ( 0 ) = 0.

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

for all x ∈ X. So, f ( − x ) = − f ( x ) for all x ∈ X.

Letting z = − x − y in (2.1), we get

for all x , y ∈ X. Thus,

for all x , y ∈ X, as desired. □

Theorem 2.2 Assume that a mapping f : X → Y satisfies the inequality

where ϕ : X 3 → [ 0 , ∞ ) satisfies

for all x , y , z ∈ X. Then there exists a unique additive mapping A : X → Y such that

for all x ∈ X.

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

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

for all x ∈ X. So,

for all x ∈ X.

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

for all x ∈ X. It follows from (2.5) and (2.6) that

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

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

and so A ( − x ) = − A ( x ) for all x ∈ X.

for all x , y ∈ X. Thus, the mapping A : X → Y is additive.

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

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

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

for all x , y , z ∈ X. Then there exists a unique additive mapping A : X → Y such that

for all x ∈ X.

Throughout this section, assume that K is a real number with 0 < K ≠ 2.

Proposition 3.1 Let f : X → Y be a mapping such that

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

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

So, f ( 0 ) = 0.

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

for all x ∈ X. So, f ( − x ) = − f ( x ) for all x ∈ X.

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

for all x , y ∈ X. Thus,

for all x , y ∈ X. Letting y = 0 in (3.2), we get K f ( x K ) = f ( x ) for all x ∈ X. So,

for all x , y ∈ X, as desired. □

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

where ϕ : X 3 → [ 0 , ∞ ) satisfies

for all x , y , z ∈ X. Then there exists a unique additive mapping A : X → Y such that

for all x ∈ X.

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 ) ∥ ≤ ∥ K f ( 0 ) ∥ + ϕ ( 0 , 0 , 0 ) = ∥ K f ( 0 ) ∥. So, f ( 0 ) = 0.

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

for all x ∈ X. Letting x = y = K x, z = − 2 x in (3.3), we obtain

for all x ∈ X. So,

for all x ∈ X. It follows from (3.6) and (3.7) that

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

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

and so A ( − x ) = − A ( x ) for all x ∈ X.

It follows from (3.3) that

for all x ∈ X. Hence,

for all x , y ∈ X. So, the mapping A : X → Y is an additive mapping.

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

for all x ∈ X. Thus, we may conclude that A ( x ) = T ( x ) for all x ∈ X. This proves the uniqueness of A. So, the mapping A : X → Y 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 : X → Y be a mapping satisfying

for all x , y , z ∈ X. Then there exists a unique additive mapping A : X → Y such that

for all x ∈ X.

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

for all x , y , z ∈ X. Then there exists a unique additive mapping A : X → Y such that

for all x ∈ X.

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 ) ∥ ≤ ∥ K f ( 0 ) ∥ + ϕ ( 0 , 0 , 0 ) = ∥ K f ( 0 ) ∥. So, f ( 0 ) = 0.

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

for all x ∈ X. It follows from (3.6) and (3.11) that

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

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 : X → Y be a mapping satisfying (3.8). Then there exists a unique additive mapping A : X → Y such that

for all x ∈ X.