On the stability of pexider functional equation in non-archimedean spaces

In this paper, the Hyers-Ulam stability of the Pexider functional equation

in a non-Archimedean space is investigated, where σ is an involution in the domain of the given mapping f.

MSC 2010:26E30, 39B52, 39B72, 46S10

The stability problem for functional equations first was planed in 1940 by Ulam [1]:

Let G₁ be group and G₂ be a metric group with the metric d(·,·). Does, for any ε > 0, there exists δ > 0 such that, for any mapping f : G₁ → G₂ which satisfies d(f(xy), f(x)f(y)) ≤ δ for all x, y ∈ G₁, there exists a homomorphism h : G₁ → G₂ so that, for any x ∈ G₁, we have d(f (x), h(x)) ≤ ε?

In 1941, Hyers [2] answered to the Ulam's question when G₁ and G₂ are Banach spaces. Subsequently, the result of Hyers was generalized by Aoki [3] for additive mappings and Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The paper of Rassias [4] has provided a lot of influences in the development of the Hyers-Ulam-Rassias stability of functional equations (for more details, see [5] where a discussion on definitions of the Hyers-Ulam stability is provided by Moszner, also [6-12]).

In this paper, we give a modification of the approach of Belaid et al. [13] in non-Archimedean spaces. Recently, Ciepliński [14] studied and proved stability of multi-additive mappings in non-Archimedean normed spaces, also see [15-22].

Definition 1.1. The function | · | : K → ℝ is called a non-Archimedean valuation or absolute value over the field K if it satisfies following conditions: for any a, b ∈ K,

(1) |a| ≥ 0;

(2) |a| = 0 if and only if a = 0;

(3) |ab| = |a| |b|

(4) |a + b| ≤ max{|a|, |b|};

(5) there exists a member a₀ ∈ K such that |a₀| ≠ 0, 1.

A field K with a non-Archimedean valuation is called a non-Archimedean field.

Corollary 1.2. |-1| = |1| = 1 and so, for any a ∈ K, we have |-a| = |a|. Also, if |a| < |b| for any a, b ∈ K, then |a + b| = |b|.

In a non-Archimedean field, the triangle inequality is satisfied and so a metric is defined. But an interesting inequality changes the usual Archimedean sense of the absolute value. For any n ∈ ℕ, we have |n · 1| ≤ ℝ. Thus, for any a ∈ K, n ∈ ℕ and nonzero divisor k ∈ ℤ of n, the following inequalities hold:

Definition 1.3. Let V be a vector space over a non-Archimedean field K. A non-Archimedean norm over V is a function || · || : V → R satisfying the following conditions: for any α ∈ K and u, v ∈ V,

(1) ||u|| = 0 if and only if u = 0;

(2) ||αu|| = |α| ||u||;

(3) ||u + v|| ≤ max{||u||, ||v||}.

Since 0 = ||0|| = ||v - v|| ≤ max{||v||, ||-v||} = ||v|| for any v ∈ V, we have ||v|| ≥ 0. Any vector space V with a non-Archimedean norm || · || : V → ℝ is called a non-Archimedean space. If the metric d(u, v) = ||u - v|| is induced by a non-Archimedean norm || · || : V → ℝ on a vector space V which is complete, then (V, || · ||) is called a complete non-Archimedean space.

Proposition 1.4. ([23]) A sequence in a non-Archimedean space is a Cauchy sequence if and only if the sequence converges to zero.

Since any non-Archimedean norm satisfies the triangle inequality, any non-Archimedean norm is a continuous function from its domain to real numbers.

Proposition 1.5. Let V be a normed space and E be a non-Archimedean space. Let f : V → E be a function, continuous at 0 ∈ V such that, for any × ∈ V, f(2x) = 2f(x) (for example, additive functions). Then, f = 0.

Proof. Since f(0) = 0, for any ε > 0, there exists δ > 0 that, for any x ∈ V with ||x|| ≤ δ,

and, for any x ∈ V, there exists n ∈ ℕ that and hence

Since this inequality holds for all ε > 0, it follows that, for any x ∈ V, f(x) = 0. This completes the proof.

The preceding fact is a special case of a general result for non-Archimedean spaces, that is, every continuous function from a connected space to a non-Archimedean space is constant. This is a consequence of totally disconnectedness of every non-Archimedean space (see [23]).

Throughout this section, we assume that V₁ is a normed space and V₂ is a complete non-Archimedean space. Let σ : V₁ → V₁ be a continuous involution (i.e., σ (x + y) = σ (x) + σ (y) and σ (σ (x)) = x) and φ : V₁ × V₁ → ℝ be a function with

and define a function ϕ : V₁ × V₁ → ℝ by

which easily implies

Theorem 2.1. Suppose that φ satisfies the condition 2.1 and let ϕ is defined by Equation 2.2. If f : V₁ → V₂ satisfies the inequality

for all x, y ∈ V₁, then there exists a unique solution q : V₁ → V₂ of the functional equation

such that

for all x ∈ V₁.

Proof. Replacing x and y in Equation 2.4 with and , respectively, we obtain

Replacing x and y in Equation 2.4 with and , respectively, we obtain

Also, replacing both of x, y in Equation 2.4 with , we get

and so, for any n ∈ ℕ, we get

Similarly, replacing both of x, y in Equation 2.4 with , we get

Replacing x in Equation 2.7 with , we obtain

for all x ∈ V₁ and so, by assumption Equation 2.1,

Thus, f(0) = 0 and the inequality Equation 2.10 reduces to

and so,

For any n ∈ ℕ, define

and

Then,

for all x, y ∈ V₁.

From Equations (2.9) and (2.11), we get

and so Proposition 1.4 and the hypothesis Equation 2.1 imply that is a Cauchy sequence. Since V₂ is complete, the sequence converges to a point of V₂ which defines a mapping q : V₁ → V₂.

Now, we prove

for all n ∈ ℕ. Since Equation 2.7 implies

Assume that ||f(x) -q_n(x)|| ≤ ϕ_n(x, x) holds for some n ∈ ℕ. Then, we have

Therefore, by induction on n, Equation 2.13 follows from Equation 2.12. Taking the limit of both sides of Equation 2.13, we prove that q satisfies Equation 2.6.

For any n ∈ ℕ and x, y ∈ V₁, we have

and so, by the continuity of non-Archimedean norm and taking the limit of both sides of the above inequality, we get

Thus, q is a solution of the Equation 2.5 which satisfies Equation 2.6.

Then, by replacing x, y with in Equation 2.5, we obtain the following identities: for any solution g : V₁ → V₂ of the Equation (2.5),

and

By induction on n, one can show that

and

for all n ∈ ℕ.

Now, suppose that q' : V₁ → V₂ is another solution of 2.5 that satisfies the Equation 2.6. It follows from Equations 2.14 to 2.16 that

Therefore, since

we have q(x) = q'(x) for all x ∈ V₁. This completes the proof.

In the proof of the next theorem, we need a result concerning the Cauchy functional equation

which has been established in [20].

Theorem 2.2. ([20]) Suppose that φ(x, y) satisfies the condition 2.1 and, for a mapping f : V₁ → V₂,

for all x, y ∈ V₁. Then, there exists a unique solution q : V₁ → V₂ of the Equation 2.17 such that

for all x ∈ V₁, where

for all x, y ∈ V₁

In this section, we assume that V₁ is a normed space and V₂ is a complete non-Archimedean space. For any mapping f : V₁ → V₂, we define two mappings F^e and F^o as follows:

and also define F(x) = f(x) -f(0). Then, we have obviously

Theorem 3.1. Let σ : V₁ → V₁ be a continuous involution and the mappings f_i : V₁ → V₂ for i = 1, 2, 3, 4 and δ > 0, satisfy

for all x, y ∈ V₁, then there exists a unique solution q : V₁ → V₂ of the Equation 2.5 and a mapping v : V₁ → V₂ which satisfies

for all x, y ∈ V₁ and exists two additive mappings such that for i = 1, 2 and, for all x ∈ V₁,

Proof. It follows from (3.2) that

and so, for all x, y ∈ V₁,

then,

Similarly, we have

for all x, y ∈ V₁.

Now, first by putting y = 0 in Equation 3.7 and applying Equation 3.2 and second by putting x = 0 in Equation 3.7 and applying Equation 3.2 once again, we obtain

for all x, y ∈ V₁ and so these inequalities with Equation 3.7 imply

Replacing y with σ(y) in Equation 3.11, we get

It follows from Equations 3.1, 3.11 and 3.12 that

By Theorem 2.1 of [24], there exists a unique solution q : V₁ → V₂ of the functional Equation 2.5 such that

for all x ∈ V₁.

As a result of the inequalities Equations 3.11 and 3.12, we have

It is easily seen that the mapping v : V₁ → V₂ defined by

is a solution of the functional equation

for all x, y ∈ V₁.

Replacing both of x, y in Equation 3.14 with , We get

for all x ∈ V₁. Now, Equations 3.13 and 3.15 imply

and

Similarly, it follows from the inequalities Equations 3.7, 3.10 and 3.13 that

Since Equation 3.8 implies

for all x, y ∈ V₁, we have

for all x ∈ V₁. Now, from Equations 3.8 and 3.20, we obtain

and so, by interchanging role of x, y in the preceding inequality,

for all x, y ∈ V₁. Since y + σ (x) = σ (x + σ (y), it follows from Equations 3.1, 3.24 and 3.25 that

By Theorem 2.2, there exists a unique additive mapping such that

Since

for all x ∈ V₁, we deduce for all x ∈ V₁.

By a similar deduction, Equations 3.8 and 3.21 imply that there exists a unique additive mapping such that

Moreover, we have for all x ∈ V₁. Thus, by Equations 3.16, 3.22, 3.27 and 3.28, we obtain

This proves Equation 3.3. Similarly, one can prove Equations 3.4 to 3.6.

The authors declare that they have no competing interests.

All authors carried out the proof. All authors conceived of the study, and participated in its design and coordination. All authors read and approved the final manuscript.

The authors would like to thank the referee and area editor Professor Ondrĕj Došlý for giving useful suggestions and comments for the improvement of this paper.

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