Hyers-Ulam stability of Davison functional equation on restricted domains

Choonkil Park; Mohammad Amin Tareeghee; Abbas Najati; Batool Noori

doi:10.1515/dema-2024-0039

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Hyers-Ulam stability of Davison functional equation on restricted domains

Choonkil Park , Mohammad Amin Tareeghee , Abbas Najati and Batool Noori

Published/Copyright: September 26, 2024

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From the journal Demonstratio Mathematica Volume 57 Issue 1

Abstract

In this article, we study the Hyers-Ulam stability of Davison functional equation

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

on some unbounded restricted domains. Using the obtained results, we study an interesting asymptotic behavior of Davison functions. We also investigate the Hyers-Ulam stability of Davison functional equation and its generalized form given by

f ( x y ) + g ( x + y ) = h ( x y + x ) + k ( y ) ,

for x , y ∈ R ⩾ 0 = { t ∈ R : t ⩾ 0 } .

Keywords: Hyers-Ulam stability; Davison functional equation; Davison function; asymptotic behavior

MSC 2010: 39B82; 39B62

1 Introduction and preliminaries

During the 17th International Symposium on functional equations, the following functional equation

(1) f ( x y ) + f ( x + y ) = f ( x y + x ) + f ( y ) ,

was introduced by Davison [1]. He specifically asked about the general solution of (1) when the domain and range of f are assumed to be (commutative) fields. Benz [2] proved that if f : R → R is a continuous solution of (1), then f is of the form f ( x ) = a x + b , where a and b are arbitrary constants. The general solution (without any regularity assumptions on f : R → R ) of the Davison functional equation (1) was obtained by Girgensohn and Lajkó [3]. They proved that if f : R → R is a solution (1), then f has the form f ( x ) = A ( x ) + b , where A : R → R is an additive function and b ∈ R is an arbitrary constant. Moreover, they obtained the general solution of the generalized version of (1) as follows.

Theorem 1.1

[3] The functions f , g , h , k : R → R satisfy

(2) f ( x y ) + g ( x + y ) = h ( x y + x ) + k ( y ) , x , y ∈ R ,

if and only if they have the form

f ( x ) = A ( x ) + b 1 , g ( x ) = A ( x ) + b 2 , h ( x ) = A ( x ) + b 3 , k ( x ) = A ( x ) + b 4 ,

where A : R → R is additive and b 1 , b 2 , b 3 , b 4 are real constants with b 1 + b 2 = b 3 + b 4 .

It should be noted that the aforementioned results can be extended to functions f : K → G , where K is a commutative field of characteristic different from 2 and 3, and G is an abelian group.

Girgensohn and Lajkó [3] also presented the general solution of (1) and (2) for x , y ∈ ( 0 , + ∞ ) .

Davison [4] solved the functional equation (1) in two cases for functions with values in an abelian group G . First f : N → G , and second f : Z → G , where N and Z are, respectively, the set of natural numbers and the set of integer numbers. Indeed, he proved that every solution of (1) is a linear combination (in the codomain) of four functions.

In 1940, Ulam [5] posed the following problem concerning the stability of group homomorphisms: Let ( G 1 , * ) be a group, ( G 2 , ∘ ) a metric group with a metric d ( . , . ) and ε > 0 . Find a δ > 0 such that if f : G 1 → G 2 fulfills d ( f ( x * y ) , f ( x ) ∘ f ( y ) ) ⩽ δ for all x , y ∈ G 1 , then there exists a homomorphism h : G 1 → G 2 such that d ( f ( x ) , h ( x ) ) ⩽ ε for all x ∈ G 1 . This problem was first solved by Hyers [6] for the case where G 1 and G 2 are Banach spaces. The Hyers-Ulam stability problem for various functional equations have been investigated by numerous mathematicians. For more information on this area we refer the reader to [7–9].

The Hyers-Ulam stability of the Davison functional equation was first treated by Jung and Sahoo [10]. They investigated the Hyers-Ulam stability of Davison functional equation (1) by following an idea of Girgensohn and Lajkó [3] (see also [11]).

Theorem 1.2

[10] If the function f : R → R satisfies the inequality

∣ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ∣ ⩽ ε , x , y ∈ R ,

for some ε ⩾ 0 , then there exists an additive function A : R → R such that

∣ f ( x ) − A ( x ) − f ( 0 ) ∣ ⩽ 12 ε , x ∈ R .

It was shown in [12,13] that the estimate 12 ε in Theorem 1.2 can be improved to 9 ε .

Kim [14] studied the Hyers-Ulam stability of two generalized Davison functional equations, namely,

(3) f ( x y ) + f ( x + y ) = g ( x y + x ) + g ( y )

and

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

for all x , y ∈ X and f , g : X → Y , where X is a normed algebra with a unit element and Y is a Banach space. The Hyers-Ulam stability of (3) has been investigated in [12].

It will also be interesting to study the Hyers-Ulam stability of Davison functional equations (1) and (2) on restricted domains. In this article, we prove the Hyers-Ulam stability of the Davison functional equation on unbounded restricted domains by following ideas of [3] and [12]. We apply the obtained results to the study of an interesting asymptotic behavior of Davison functions.

2 Hyers-Ulam stability

Throughout this article, we denote by A a normed algebra with the unit element 1 and by ℬ a Banach space.

Theorem 2.1

Let f : A → ℬ be a function which for ε ⩾ 0 and d > 0 satisfies

(4) ‖ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ‖ ⩽ ε , ‖ x ‖ ⩾ d , y ∈ A .

Then there is a unique additive function φ : A → ℬ , such that

∥ φ ( x ) − f ( x ) + f ( 0 ) ∥ ⩽ 28 ε , x ≠ − 1 ; ‖ φ ( − 1 ) − f ( − 1 ) + f ( 0 ) ‖ ⩽ 19 ε .

Proof

Replace y by y + 1 in (4) to obtain

(5) ‖ f ( x y + x ) + f ( x + y + 1 ) − f ( x y + 2 x ) − f ( y + 1 ) ‖ ⩽ ε ,

for all x , y ∈ A with ‖ x ‖ ⩾ d . Thus, it follows from (4) and (5) that

(6) ‖ f ( x y ) + f ( x + y ) + f ( x + y + 1 ) − f ( y ) − f ( x y + 2 x ) − f ( y + 1 ) ‖ ⩽ ‖ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ‖ + ‖ f ( x y + x ) + f ( x + y + 1 ) − f ( x y + 2 x ) − f ( y + 1 ) ‖ ⩽ 2 ε ,

for all x , y ∈ A with ‖ x ‖ ⩾ d . Replacing y by 4 y in (6), we obtain

‖ f ( 4 x y ) + f ( x + 4 y ) + f ( x + 4 y + 1 ) − f ( 4 y ) − f ( 4 x y + 2 x ) − f ( 4 y + 1 ) ‖ ⩽ 2 ε ,

for all x , y ∈ A with ‖ x ‖ ⩾ d . From (4) and the last inequality, we obtain

‖ f ( x + 4 y ) + f ( x + 4 y + 1 ) − f ( 2 x + 2 y ) − f ( 4 y ) − f ( 4 y + 1 ) + f ( 2 y ) ‖ ⩽ ‖ f ( 4 x y ) + f ( x + 4 y ) + f ( x + 4 y + 1 ) − f ( 4 y ) − f ( 4 x y + 2 x ) − f ( 4 y + 1 ) ‖ + ‖ f ( 4 x y ) + f ( 2 x + 2 y ) − f ( 4 x y + 2 x ) − f ( 2 y ) ‖ ⩽ 3 ε ,

for all x , y ∈ A with ‖ x ‖ ⩾ d . If we replace x by x − y in the above inequality and then substitute 3 x for x in the resultant inequality, we obtain

(7) ‖ f ( 3 x + 3 y ) + f ( 3 x + 3 y + 1 ) − f ( 6 x ) − f ( 4 y ) − f ( 4 y + 1 ) + f ( 2 y ) ‖ ⩽ 3 ε ,

for all x , y ∈ A with ‖ 3 x − y ‖ ⩾ d . Letting y = 0 and y = − x separately in (7), we obtain

(8) ‖ f ( 3 x ) + f ( 3 x + 1 ) − f ( 6 x ) − f ( 1 ) ‖ ⩽ 3 ε ,

(9) ‖ f ( 0 ) + f ( 1 ) − f ( 6 x ) − f ( − 4 x ) − f ( − 4 x + 1 ) + f ( − 2 x ) ‖ ⩽ 3 ε ,

for all x ∈ A with ‖ x ‖ ⩾ d . If we replace x , y by 2 x , − x , respectively in (7), we have

(10) ‖ − f ( 3 x ) − f ( 3 x + 1 ) + f ( 12 x ) + f ( − 4 x ) + f ( − 4 x + 1 ) − f ( − 2 x ) ‖ ⩽ 3 ε ,

for all x ∈ A with ‖ x ‖ ⩾ d . Adding (8), (9), and (10), we infer that

‖ f ( 12 x ) − 2 f ( 6 x ) + f ( 0 ) ‖ ⩽ 9 ε , ‖ x ‖ ⩾ d .

Let g ( x ) ≔ f ( 6 x ) − f ( 0 ) . Then the last inequality means

‖ g ( 2 x ) − 2 g ( x ) ‖ ⩽ 9 ε , ‖ x ‖ ⩾ d .

It is easy to see that

(11) g ( 2 n + 1 x ) 2 n + 1 − g ( 2 m x ) 2 m ⩽ ∑ k = m n 9 ε 2 k + 1 , ‖ x ‖ ⩾ d .

This implies that g ( 2 n x ) 2 n n is a Cauchy sequence for ‖ x ‖ ⩾ d . It is easy to show that g ( 2 n x ) 2 n n is a Cauchy sequence for each x ∈ A and thus converges. Therefore, we can define

T : A → ℬ , T ( x ) = lim n g ( 2 n x ) 2 n .

Obviously,

(12) T ( 0 ) = 0 , T ( 2 x ) = 2 T ( x ) , T ( x ) = lim n f ( 2 n ⋅ 6 x ) 2 n , x ∈ A .

By (8), we obtain

(13) T ( x ) = lim n f ( 2 n ⋅ 6 x + 1 ) 2 n , x ∈ A .

Using (7), (12), and (13), we obtain

T ( 3 x + 3 y ) = T ( 3 x ) + 3 T ( y ) , x , y ∈ A , y ≠ 3 x .

Letting x = 0 in the last equation and using T ( 0 ) = 0 , we obtain T ( 3 y ) = 3 T ( y ) for all y ∈ A . So, T is additive. Setting m = 0 and taking the limit on both sides of (11) as n → ∞ , we obtain

‖ T ( x ) − f ( 6 x ) + f ( 0 ) ‖ ⩽ 9 ε , ‖ x ‖ ⩾ d .

Since T is additive, this inequality implies

(14) ‖ φ ( x ) − f ( x ) + f ( 0 ) ‖ ⩽ 9 ε , ‖ x ‖ ⩾ 6 d ,

where φ ( x ) = 1 6 T ( x ) . Obviously, (14) holds for x = 0 . Let y ∈ A and y ≠ − 1 . We can choose x ∈ A such that min { ‖ x ‖ , ‖ x y ‖ , ‖ x + y ‖ , ‖ x y + x ‖ } ⩾ 6 d . By (14), we have

‖ φ ( x y ) − f ( x y ) + f ( 0 ) ‖ ⩽ 9 ε , ‖ φ ( x + y ) − f ( x + y ) + f ( 0 ) ‖ ⩽ 9 ε , ‖ f ( x y + x ) − φ ( x y + x ) − f ( 0 ) ‖ ⩽ 9 ε .

Adding these inequalities and (4), we obtain

‖ φ ( x y ) + φ ( x + y ) − φ ( x y + x ) − f ( y ) + f ( 0 ) ‖ ⩽ 28 ε .

Since φ is additive, the last inequality yields

‖ φ ( y ) − f ( y ) + f ( 0 ) ‖ ⩽ 28 ε .

In the case y = − 1 , we choose x ∈ A such that ‖ x ‖ ⩾ 6 d + ‖ 1 ‖ . By (4) and (14), we have

‖ f ( − x ) + f ( x − 1 ) − f ( 0 ) − f ( − 1 ) ‖ ⩽ ε , ‖ φ ( x − 1 ) − f ( x − 1 ) + f ( 0 ) ‖ ⩽ 9 ε , ‖ φ ( − x ) − f ( − x ) + f ( 0 ) ‖ ⩽ 9 ε .

Adding these inequalities and using the additivity of φ , we obtain

‖ φ ( − 1 ) − f ( − 1 ) + f ( 0 ) ‖ ⩽ 19 ε .

The uniqueness of φ is obvious. Hence, the proof is complete.□

With a proof similar to the one presented in Theorem 2.1, we obtain the following result.

Corollary 2.2

Let Y be a linear space and f : A → Y a function satisfying

f ( x y ) + f ( x + y ) = f ( x y + x ) + f ( y ) , ‖ x ‖ ⩾ d .

Then f is affine, that is, f − f ( 0 ) is additive on A .

Remark 2.3

Since

{ ( x , y ) ∈ A × A : ‖ x ‖ ⩾ 2 d } ⊆ { ( x , y ) ∈ A × A : ‖ x ‖ + ‖ y ‖ ⩾ 2 d } ⊆ { ( x , y ) ∈ A × A : max { ‖ x ‖ , ‖ y ‖ } ⩾ d } ,

the above results are valid if the condition ‖ x ‖ ⩾ d is replaced by ‖ x ‖ + ‖ y ‖ ⩾ d or max { ‖ x ‖ , ‖ y ‖ } ⩾ d .

Corollary 2.4

Let Y be a normed linear space. A function f : A → Y is affine on A if and only if one of the following conditions hold:

lim ‖ x ‖ → ∞ sup y ∈ A ‖ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ‖ = 0 ;
lim ‖ x ‖ + ‖ y ‖ → ∞ [ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ] = 0 ;
lim max { ‖ x ‖ , ‖ y ‖ } → ∞ [ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ] = 0 .

Proof

It is clear that an affine function f : A → Y fulfills ( i ) , ( i i ) , ( i i i ) . Obviously, ( i i ) ⇒ ( i ) and ( i i i ) ⇒ ( i ) . Let f satisfy ( i ) and ε > 0 be arbitrary. Then there exists d > 0 such that

‖ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ‖ ⩽ ε , ‖ x ‖ ⩾ d , y ∈ A .

Let Y ˜ be the completion of Y . By Theorem 2.1, there exists an additive function φ : A → Y ˜ satisfying

‖ f ( x ) − f ( 0 ) − φ ( x ) ‖ ⩽ 28 ε , x ∈ A .

Let g ( x ) = f ( x ) − f ( 0 ) . Then

‖ g ( x + y ) − g ( x ) − g ( y ) ‖ ⩽ ‖ g ( x + y ) − φ ( x + y ) ‖ + ‖ g ( x ) − φ ( x ) ‖ + ‖ g ( y ) − φ ( y ) ‖ ⩽ 84 ε , x , y ∈ A .

Since ε > 0 is arbitrary, we obtain g is additive. This means f is affine.□

Corollary 2.5

Let f : R → R be a function given by

f ( x ) ≔ 1 1 + x 2 , x ∈ R .

It is clear that

lim ∣ y ∣ → ∞ [ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ] = 0 ,

but f is not affine.

Note that, as a consequence of Corollary 2.4, we obtain the result that every function f : R → R is either affine or satisfies the condition

sup x , y ∈ R ∣ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ∣ ∣ x ∣ p = ∞ ,

for all p > 0 .

3 Stability on R ⩾ 0

The aim of this section is to present the Hyers-Ulam stability of Davison functional equation and its generalized form (2) for x , y ∈ R ⩾ 0 = { t ∈ R : t ⩾ 0 } .

Theorem 3.1

Let ε ⩾ 0 and f : [ 0 , + ∞ ) → ℬ be a function satisfying

(15) ‖ f ( x y ) + f ( x + y ) − f ( x y + x ) − f ( y ) ‖ ⩽ ε , x , y ⩾ 0 .

Then there is an affine function ψ : R → ℬ such that

‖ ψ ( x ) − f ( x ) ‖ ⩽ 55 ε , x > 0 .

Proof

Replacing y by y + 1 in (15), we obtain

‖ f ( x y + x ) + f ( x + y + 1 ) − f ( x y + 2 x ) − f ( y + 1 ) ‖ ⩽ ε , x , y ⩾ 0 .

Adding this inequality and (15), we obtain

‖ f ( x y ) + f ( x + y ) + f ( x + y + 1 ) − f ( y ) − f ( x y + 2 x ) − f ( y + 1 ) ‖ ⩽ 2 ε , x , y ⩾ 0 .

Now, replacing x by x 2 and y by 2 y , we obtain

f ( x y ) + f x 2 + 2 y + f x 2 + 2 y + 1 − f ( 2 y ) − f ( x y + x ) − f ( 2 y + 1 ) ⩽ 2 ε , x , y ⩾ 0 .

By this inequality and (15), one obtains

(16) f x 2 + 2 y + f x 2 + 2 y + 1 − f ( x + y ) − f ( 2 y ) − f ( 2 y + 1 ) + f ( y ) ⩽ 3 ε ,

for all x , y ⩾ 0 . Replacing x by x − y 3 and y by y 3 in (16), we obtain

f x + y 2 + f x + y 2 + 1 − f 2 3 y − f 2 3 y + 1 + f 1 3 y − f ( x ) ⩽ 3 ε ,

for all x ⩾ y 3 ⩾ 0 . The last inequality can be written as follows:

(17) ‖ G ( x + y ) − f ( x ) − H ( y ) ‖ ⩽ 3 ε , x ⩾ y 3 ⩾ 0 ,

where

G ( t ) ≔ f t 2 + f t 2 + 1 , H ( t ) ≔ f 2 3 t + f 2 3 t + 1 − f 1 3 t .

Letting y = x and y = 0 separately in (17), we obtain

(18) ‖ f ( x + 1 ) − H ( x ) ‖ ⩽ 3 ε , x ⩾ 0 ,

(19) ‖ G ( x ) − f ( x ) − f ( 1 ) ‖ ⩽ 3 ε , x ⩾ 0 .

By (19), we obtain

(20) ‖ G ( x + y ) − f ( x + y ) − f ( 1 ) ‖ ⩽ 3 ε , x , y ⩾ 0 .

It follows from (17), (18), and (20) that

(21) ‖ f ( x + y ) − f ( x ) − f ( y + 1 ) + f ( 1 ) ‖ ⩽ 9 ε , x ⩾ y 3 ⩾ 0 .

Putting y = 1 in (21), we obtain

(22) ‖ f ( x + 1 ) − f ( x ) − f ( 2 ) + f ( 1 ) ‖ ⩽ 9 ε , x ⩾ 1 3 .

It follows from (21) and (22) that

(23) ‖ f ( x + y ) − f ( x ) − f ( y ) − f ( 2 ) + 2 f ( 1 ) ‖ ⩽ 18 ε , x ⩾ y 3 , y ⩾ 1 3 .

Letting y = x in (23), we obtain

‖ f ( 2 x ) − 2 f ( x ) + c ‖ ⩽ 18 ε , x ⩾ 1 3 ,

where c = 2 f ( 1 ) − f ( 2 ) . Then

(24) f ( 2 n + 1 x ) 2 n + 1 − f ( 2 m x ) 2 m + ∑ k = m n c 2 k + 1 ⩽ ∑ k = m n 9 ε 2 k , x ⩾ 1 3 .

Then { f ( 2 n x ) 2 n } n is a Cauchy sequence for all x ⩾ 0 . Define

A : [ 0 , + ∞ ) → ℬ , A ( x ) ≔ lim n f ( 2 n x ) 2 n .

It follows from (19) and the definition of G that

A ( x ) ≔ lim n G ( 2 n x ) 2 n = lim n f ( 2 n x + 1 ) 2 n , x ⩾ 0 .

Hence, (18) yields

A ( x ) ≔ lim n H ( 2 n x ) 2 n , x ⩾ 0 .

Therefore, (17) implies

A ( x + y ) = A ( x ) + A ( y ) , ( x , y ) ∈ D ≔ ( x , y ) ∈ R 2 : x > y 3 > 0 .

Since D is an open connected set in R 2 , by the extension theorem of Rimán [15] there is an additive function φ : R → ℬ and a constant a ∈ R such that A ( x ) = φ ( x ) + a for all x > 0 . Putting m = 0 and letting n → ∞ in (24), it follows

(25) ‖ φ ( x ) − f ( x ) + a + 2 f ( 1 ) − f ( 2 ) ‖ ⩽ 18 ε , x ⩾ 1 3 .

Let y > 0 and choose x > 0 such that min { x y , x + y , x y + x } > 1 3 . By (25), we have

‖ φ ( x y ) − f ( x y ) + a + 2 f ( 1 ) − f ( 2 ) ‖ ⩽ 18 ε , ‖ φ ( x + y ) − f ( x + y ) + a + 2 f ( 1 ) − f ( 2 ) ‖ ⩽ 18 ε , ‖ f ( x y + x ) − φ ( x y + x ) − a − 2 f ( 1 ) + f ( 2 ) ‖ ⩽ 18 ε .

Adding these inequalities and (15), we obtain

‖ φ ( x y ) + φ ( x + y ) − φ ( x y + x ) − f ( y ) + a + 2 f ( 1 ) − f ( 2 ) ‖ ⩽ 55 ε .

Since φ is additive, we infer

‖ φ ( y ) − f ( y ) + a + 2 f ( 1 ) − f ( 2 ) ‖ ⩽ 55 ε .

This completes the proof.□

Theorem 3.2

Let ε ⩾ 0 and f , g , h , k : [ 0 , + ∞ ) → ℬ be functions satisfying

(26) ‖ f ( x y ) + g ( x + y ) − h ( x y + x ) − k ( y ) ‖ ⩽ ε , x , y ⩾ 0 .

Then there is an affine function φ : R → ℬ such that

(27) ‖ φ ( x ) − g ( x ) ‖ ⩽ 330 ε , x > 0 ;

(28) ‖ φ ( x ) − h ( x ) + f ( 0 ) − k ( 0 ) ‖ ⩽ 331 ε , x > 0 ;

(29) ‖ φ ( x ) − k ( x ) + f ( 0 ) − h ( 0 ) ‖ ⩽ 331 ε , x > 0 ;

(30) ‖ φ ( x ) − f ( x ) + 2 f ( 0 ) − h ( 0 ) − k ( 0 ) ‖ ⩽ 333 ε , x > 0 .

Proof

Letting x = 0 and y = 0 separately in (26), we obtain

(31) ‖ f ( 0 ) + g ( x ) − h ( x ) − k ( 0 ) ‖ ⩽ ε , x ⩾ 0 ,

(32) ‖ f ( 0 ) + g ( y ) − h ( 0 ) − k ( y ) ‖ ⩽ ε , y ⩾ 0 .

It follows from (31) that

(33) ‖ f ( 0 ) + g ( x y + x ) − h ( x y + x ) − k ( 0 ) ‖ ⩽ ε , x , y ⩾ 0 .

By (26), (32), and (33), one obtains

(34) ‖ f ( x y ) + g ( x + y ) − g ( x y + x ) − g ( y ) − 2 f ( 0 ) + h ( 0 ) + k ( 0 ) ‖ ⩽ 3 ε , x , y ⩾ 0 .

Setting x = 1 in (34), we have

(35) ‖ f ( y ) − g ( y ) − 2 f ( 0 ) + h ( 0 ) + k ( 0 ) ‖ ⩽ 3 ε , y ⩾ 0 .

Then

(36) ‖ g ( x y ) − f ( x y ) + 2 f ( 0 ) − h ( 0 ) − k ( 0 ) ‖ ⩽ 3 ε , x , y ⩾ 0 .

Adding (34) and (36), we obtain

‖ g ( x y ) + g ( x + y ) − g ( x y + x ) − g ( y ) ‖ ⩽ 6 ε , x , y ⩾ 0 .

By Theorem 3.1, there is an affine function φ : R → ℬ such that

‖ φ ( x ) − g ( x ) ‖ ⩽ 330 ε , x > 0 .

This inequality with (31), (32), and (35) yield (28), (29), and (30).□

Corollary 3.3

Let ε ⩾ 0 and f , g , h , k : [ 0 , + ∞ ) → ℬ be functions satisfying

f ( x y ) + g ( x + y ) − h ( x y + x ) − k ( y ) = 0 , x , y ⩾ 0 .

Then f , g , h , k are affine functions on ( 0 , + ∞ ) . Moreover,

h ( x ) = g ( x ) + f ( 0 ) − k ( 0 ) , k ( x ) = g ( x ) + f ( 0 ) − h ( 0 ) , f ( x ) = g ( x ) + f ( 0 ) − g ( 0 ) , x > 0 .

4 Conclusions

The functional equation f ( x y ) + f ( x + y ) = f ( x y + x ) + f ( y ) is known as the Davison functional equation. We treated the Hyers-Ulam stability of the Davison functional equation in two cases. The first is for functions defined on a normed unitary algebra with values in a Banach space and the functional inequality associated with Davison equation is valid only on a restricted unbounded domain. The second one is for the case of functions defined on interval [ 0 , + ∞ ) with values in a Banach space. In the second case, we considered also a generalization form of Davison functional equation. Finally, as an application of the obtained results, the asymptotic behavior of Davison functions has been investigated. Investigation and solving Davison’s functional equation and its generalization in other certain restricted domains can be future research prospects.

Acknowledgments

The authors would like to thank the reviewers for their valuable suggestions and comments.

Funding information: The authors state no funding involved.
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: The authors declare that they have no competing interests.
Data availability statement: Not applicable.

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Received: 2022-08-05

Revised: 2023-10-28

Accepted: 2024-06-07

Published Online: 2024-09-26

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2024-0039

Keywords for this article

Hyers-Ulam stability; Davison functional equation; Davison function; asymptotic behavior

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