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Article

Stability of Bi-Additive Mappings and Bi-Jensen Mappings

1
Humanitas College, Kyung Hee University, Yongin 17104, Korea
2
Department of Mathematics Education, College of Education, Mokwon University, Daejeon 35349, Korea
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Symmetry 2021, 13(7), 1180; https://doi.org/10.3390/sym13071180
Submission received: 21 May 2021 / Revised: 26 June 2021 / Accepted: 28 June 2021 / Published: 30 June 2021
(This article belongs to the Special Issue Advance in Functional Equations)

Abstract

:
Symmetry is repetitive self-similarity. We proved the stability problem by replicating the well-known Cauchy equation and the well-known Jensen equation into two variables. In this paper, we proved the Hyers-Ulam stability of the bi-additive functional equation f ( x + y , z + w ) = f ( x , z ) + f ( y , w ) and the bi-Jensen functional equation 4 f x + y 2 , z + w 2 = f ( x , z ) + f ( x , w ) + f ( y , z ) + f ( y , w ) .

1. Introduction

A functional equation is stable if there is a function that exactly satisfies the given equation in the vicinity of a function that approximately satisfies it. Any approximate solution can actually be an exact solution. In Cauchy’s equation f ( x + y ) = f ( x ) + f ( y ) we can deal with a class of approximate solutions defined by the functional inequality introduced by Rassias.
f ( x + y ) f ( x ) f ( y ) ε ( x p + y p ) .
It turns out that for p 1 each solution of the above inequality can be approximated by an additive function A in such a way that the inequality
f ( x ) A ( x ) k ε x p .
holds, with a suitable k, on the whole domain (for p = 0 it coincides with the classical Hyers–Ulam result).
Let us say X and Y are vector spaces. The mapping h : X Y is called an additional mapping (respectively, an affine mapping) if h satisfies the Cauchy functional equation h ( x + y ) = h ( x ) + h ( y ) (respectively, the Jensen functional equation 2 h x + y 2 = h ( x ) + h ( y ) ). T. Aoki [1] and Th. M. Rassias [2,3] extended Hyers-Ulam stability taking into account the variables for the Cauchy equation. S.-M. Jung [4] got the result of the Jensen equation. It was also generalized as a functional case by P. Găvruta [5] and S.-M. Jung [6] and Y.-H. Lee and K.-W. Jun [7].
The following functional Equations (1) and (3) are functional equations those combine the existing well-known the Cauchy equation and the Jensen equation.
f ( x + y , z + w ) = f ( x , z ) + f ( y , w ) .
The authors [8] introduce the system of equations
2 f ( x + y 2 , z ) = f ( x , z ) + f ( y , z ) , 2 f x , y + z 2 = f ( x , y ) + f ( x , z ) .
and the bi-Jensen functional equation
4 f x + y 2 , z + w 2 = f ( x , z ) + f ( x , w ) + f ( y , z ) + f ( y , w ) .
We made the above functional equations with a symmetrical structure. Symmetry is repetitive self-similarity. The solution of (2) is coincide with the solution of (3). The solution of (1) is of the form A 1 ( x ) + A 2 ( y ) , where A 1 and A 2 are additive mappings. The solution of (2) is of the form A 1 ( x ) + A 2 ( y ) + f ( 0 , 0 ) , where A 1 and A 2 are additive mappings. The solution of (2) contains the solution of (1). The difference of the solutions (1) and (2) is merely a constant, that is, the solutions (1) and (2) are similar.
Jun, Jung, and Lee [9] obtained the stability on a bi-Jensen functional equation in Banach spaces. Additionally, the authors [10] proved the stability on a Cauchy-Jensen functional equation Banach spaces.
In this paper, we investigate the generalized Hyers-Ulam stability of (1) in Banach spaces and 2-Banach spaces. We proved the Hyers-Ulam stability of (2) and (3) in quasi-Banach spaces.

2. Solution and Stability of a Bi-Additive Functional Equation

In the following theorem, we find out the general solution of the bi-additive functional Equation (1).
Theorem 1.
A mapping f : X × X Y satisfies ( 1 ) if and only if there exist two additive mappings A 1 , A 2 : X × X Y such that
f ( x , y ) = A 1 ( x ) + A 2 ( y )
for all x , y X .
Proof. 
We first assume that f is a solution of (1). Define A 1 , A 2 : X Y by A 1 ( x ) : = f ( x , 0 ) and A 2 ( x ) : = f ( 0 , x ) for all x X . One can easily verify that A 1 , A 2 are additive. Letting y = z = 0 in (1), we get
f ( x , w ) = f ( x , 0 ) + f ( 0 , w ) = A 1 ( x ) + A 2 ( w )
for all x , w X .
Conversely, we assume that there is two additive mappings A 1 , A 2 : X × X Y , such that f ( x , y ) = A 1 ( x ) + A 2 ( y ) for all x , y X . Since A 1 , A 2 are additive, we gain
f ( x + y , z + w ) = A 1 ( x + y ) + A 2 ( z + w ) = A 1 ( x ) + A 1 ( y ) + A 2 ( z ) + A 2 ( w ) = A 1 ( x ) + A 2 ( z ) + A 1 ( y ) + A 2 ( w ) = f ( x , z ) + f ( y , w )
for all x , y , z , w X . □
From now on, let X and Y be a normed linear space and a Banach space, respectively.
Theorem 2.
Let 0 < p < 1 , ε > 0 , δ 0 and f : X × X Y be a mapping such that
f ( x + y , z + w ) f ( x , z ) f ( y , w ) ε + δ ( x p + y p + z p + w p )
for all x , y , z , w X . Then there is unique bi-additive mapping F : X × X Y , such that
f ( x , y ) F ( x , y ) ε + 2 δ 2 2 p ( x p + y p )
for all x , y X . The mapping F is given by F ( x , y ) : = lim j 1 2 j f ( 2 j x , 2 j y ) for all x , y X .
Proof. 
Putting y = x and w = z in (4), we have
f ( x , z ) 1 2 f ( 2 x , 2 z ) ε 2 + δ ( x p + z p )
for all x , z X . Thus, we obtain
1 2 j f ( 2 j x , 2 j z ) 1 2 j + 1 f ( 2 j + 1 x , 2 j + 1 z ) ε 2 j + 1 + 2 j ( p 1 ) δ ( x p + z p )
for all x , z X and all j. Replacing z by y in the above inequality, we see that
1 2 j f ( 2 j x , 2 j y ) 1 2 j + 1 f ( 2 j + 1 x , 2 j + 1 y ) ε 2 j + 1 + 2 j ( p 1 ) δ ( x p + y p )
for all x , y X and all j. For given integers l , m ( 0 l < m ) , we get
1 2 l f ( 2 l x , 2 l y ) 1 2 m f ( 2 m x , 2 m y ) j = l m 1 ε 2 j + 1 + 2 j ( p 1 ) δ ( x p + y p )
for all x , y X . By (6), the sequence { 1 2 j f ( 2 j x , 2 j y ) } is a Cauchy sequence for all x , y X . Since Y is complete, the sequence { 1 2 j f ( 2 j x , 2 j y ) } converges for all x , y X . Define F : X × X Y by
F ( x , y ) : = lim j 1 2 j f ( 2 j x , 2 j y )
for all x , y X . By (4), we have
1 2 j f 2 j ( x + y ) , 2 j ( z + w ) f ( 2 j x , 2 j z ) f ( 2 j y , 2 j w ) ε 2 j + 2 j ( p 1 ) δ ( x p + y p + z p + w p )
for all x , y , z , w X and all j N . Letting j in the above inequality, we see that F satisfies (1). Setting l = 0 and taking m in (6), one can obtain the inequality (5). If G : X × X Y is another 2-variable additive mapping satisfying (5), we obtain
F ( x , y ) G ( x , y ) = 1 2 n F ( 2 n x , 2 n y ) G ( 2 n x , 2 n y ) 1 2 n F ( 2 n x , 2 n y ) f ( 2 n x , 2 n y ) + 1 2 n f ( 2 n x , 2 n y ) G ( 2 n x , 2 n y ) 1 2 n 1 ε + 2 n p + 1 2 2 p δ ( x p + y p ) 0 a s n
for all x , y X . Hence the mapping F is the unique bi-additive mapping, as desired. □
Corollary 1.
Let f : X × X Y be a mapping such that
f ( x + y , z + w ) f ( x , z ) f ( y , w ) ε
for all x , y , z , w X . Then, there exists a unique mapping F : X × X Y satisfying ( 1 ) , such that
f ( x , y ) F ( x , y ) ε 2
for all x , y X .
Proof. 
If we insert δ = 0 in Theorem 2, we obtain ε as an estimate of the difference between the exact and the approximate solution of the considered equation. □
In the case p > 2 in Theorem 2, one can also obtain the similar result.
We explain some definitions [11,12] on 2-Banach spaces.
Definition 1.
Let X be a vector space over R with dimension greater than 1 and · , · : X 2 R be a function. Then we say ( X , · , · ) is a linear 2-normed space if
(a) x , y = 0 if and only if x and y are linearly dependent;
(b) x , y = y , x ;
(c) α x , y = | α | x , y ;
(d) x , y + z x , y + x , z
for all α R and x , y , z X . In this case, the function · , · is called a 2-norm on X .
Definition 2.
Let X be linear 2-normed space and { x n } a sequence in X . The sequence { x n } is said to convergent in X if there is an x X , such that
lim n x n x , y = 0
for all y X . In this case, we say that a sequence { x n } converges to x, simply denoted by lim n x n = x .
Definition 3.
Let X be linear 2-normed space and { x n } a sequence in X is called a Cauchy sequence if for any ε > 0 , there exists N N such that for all m , n N , x m x n , y < ε for all y X . For convenience, we will write lim m , n x n x m , y = 0 for a Cauchy sequence { x n } . A 2-Banach space is defined to be a linear 2-normed space in which every Cauchy sequence is convergent.
In the following lemma, we get some primitive properties in linear 2-normed spaces that will be used to prove our stability results.
Lemma 1
([13]). Let ( X , · , · ) be a linear 2-normed space and x X .
(a) If x , y = 0 for all y X , then x = 0 .
(b) | x , z y , z | x y , z for all x , y , z X .
(c) If a sequence { x n } is convergent in X , then lim n x n , y = lim n x n , y for all y X .
In the rest of this section, let X be a normed space and Y a 2-Banach space.
Theorem 3.
Let p ( 0 , 1 ) , ε > 0 , δ , η 0 and let f : X × X Y be a surjective mapping such that
f ( x + y , z + w ) f ( x , z ) f ( y , w ) , f ( u , v ) ε + δ ( x p + y p + z p + w p ) + η ( u + v )
for all x , y , z , w , u , v X . Then there exists a unique mapping F : X × X Y satisfying ( 1 ) , such that
f ( x , y ) F ( x , y ) , f ( u , v ) ε 2 + 2 δ 2 2 p ( x p + y p ) + η 2 ( u + v )
for all x , y , u , v X .
Proof. 
Letting y = x and w = z in (7), we have
f ( x , z ) 1 2 f ( 2 x , 2 z ) , f ( u , v ) ε 2 + δ ( x p + z p ) + η 2 ( u + v )
for all x , z , u , v X . Thus, we obtain
1 2 j f ( 2 j x , 2 j z ) 1 2 j + 1 f ( 2 j + 1 x , 2 j + 1 z ) , f ( u , v ) ε 2 j + 1 + 2 j ( p 1 ) δ ( x p + z p ) + η 2 j + 1 ( u + v )
for all x , z , u , v X and all j. Replacing z by y in the above inequality, we see that
1 2 j f ( 2 j x , 2 j y ) 1 2 j + 1 f ( 2 j + 1 x , 2 j + 1 y ) , f ( u , v ) ε 2 j + 1 + 2 j ( p 1 ) δ ( x p + y p ) + η 2 j + 1 ( u + v )
for all x , y , u , v X and all j. For given integers l , m ( 0 l < m ) , we get
1 2 l f ( 2 l x , 2 l y ) 1 2 m f ( 2 m x , 2 m y ) , f ( u , v ) j = l m 1 ε 2 j + 1 + 2 j ( p 1 ) δ ( x p + z p ) + η 2 j + 1 ( u + v )
for all x , y , u , v X . By (9), the sequence { 1 2 j f ( 2 j x , 2 j y ) } is a Cauchy sequence for all x , y X . Since Y is complete, the sequence { 1 2 j f ( 2 j x , 2 j y ) } converges for all x , y X . Define F : X × X Y by F ( x , y ) : = lim j 1 2 j f ( 2 j x , 2 j y ) for all x , y X . By (7), we have
1 2 j f 2 j ( x + y ) , 2 j ( z + w ) 1 2 j f ( 2 j x , 2 j z ) 1 2 j f ( 2 j y , 2 j w ) , f ( u , v ) 1 2 j ε + 2 j p δ ( x p + y p + z p + w p ) + η ( u + v )
for all x , y , z , w , u , v X and all j. Letting j , we see that F satisfies (1). Setting l = 0 and taking m in (9), one can obtain the inequality (8). If G : X × X Y is another mapping satisfying (1) and (8), we obtain
F ( x , y ) G ( x , y ) , f ( u , v ) = 1 2 n F ( 2 n x , 2 n y ) G ( 2 n x , 2 n y ) , f ( u , v ) 1 2 n F ( 2 n x , 2 n y ) f ( 2 n x , 2 n y ) , f ( u , v ) + 1 2 n f ( 2 n x , 2 n y ) G ( 2 n x , 2 n y ) , f ( u , v ) 2 2 n ε 2 + 2 n p + 1 δ 2 2 p ( x p + y p ) + η ( u + v ) 2 0 a s n
for all x , y , u , v X . Hence the mapping F is the unique mapping satisfying (1), as desired. □
Corollary 2.
Let f : X × X Y be a mapping, such that
f ( x + y , z + w ) f ( x , z ) f ( y , w ) , f ( u , v ) ε
for all x , y , z , w , u , v X . Then there exists a unique mapping F : X × X Y satisfying ( 1 ) such that
f ( x , y ) F ( x , y ) , f ( u , v ) ε 2
for all x , y , u , v X .
Proof. 
Taking δ = η = 0 in Theorem 3, we have the desired result. □
In the case p > 2 in Theorem 3, one can also obtain the similar result.

3. Solution and Stability of a Bi-Jensen Functional Equation

In [14,15], one can find the concept of quasi-Banach spaces.
Definition 4.
Let X be a real linear space. A quasi-norm is real-valued function on X satisfying the following:
(i) x 0 for all x X and x = 0 if, and only if, x = 0 .
(ii) λ x = | λ | x for all λ R and all x X .
(iii) There is a constant K 1 , such that x + y K ( x + y ) for all x , y X .
The pair ( X , · ) is called a quasi-normed space if · is a quasi-norm on X . The smallest possible K is called the modulus of concavity of · . A quasi-Banach space is a complete quasi-normed space. A quasi-norm · is called a p-norm ( 0 < p 1 ) if
x + y p x p + y p
for all x , y X . In this case, a quasi-Banach space is called a p-Banach space.
A quasi-norm gives rise to a linear topology on X, namely the least linear topology for which the unit ball B = { x X : x 1 } is a neighborhood of zero. This topology is locally bounded, that is, it has a bounded neighborhood of zero. Actually, every locally bounded topology arises in this way.
From now on, assume that X is a quasi-normed space with quasi-norm · X and that Y is a p-Banach space with p-norm · Y . Let K be the modulus of concavity of · Y .
Let φ : X × X × X [ 0 , ) and ψ : X × X × X [ 0 , ) be two functions such that
lim n 1 3 n φ ( 3 n x , 3 n y , z ) = 0 , lim n 1 3 n ψ ( 3 n x , y , z ) = 0
and
lim n 1 3 n φ ( x , y , 3 n z ) = 0 , lim n 1 3 n ψ ( x , 3 n y , 3 n z ) = 0
for all x , y , z X , and
M ( x , y , z ) : = j = 0 1 3 p j φ ( 3 j x , 3 j y , z ) p <
and
N ( z , x , y ) : = j = 0 1 3 p j ψ ( z , 3 j x , 3 j y ) p <
for all x , z X and all y { x , 3 x } .
We will use the following lemma in order to prove Theorem 4.
Lemma 2
([16]). Let 0 p 1 and let x 1 , x 2 , , x n be non-negative real numbers. Then
j = 1 n x j p j = 1 n x j p .
Theorem 4.
Let 0 < p 1 and suppose that a mapping f : X × X Y satisfies the inequalities
2 f x + y 2 , z f ( x , z ) f ( y , z ) Y φ ( x , y , z ) ,
2 f x , y + z 2 f ( x , y ) f ( x , z ) Y ψ ( x , y , z )
for all x , y , z X . Then there exists a unique additive-Jensen mapping J 1 : X × X Y satisfying
f ( x , y ) f ( 0 , y ) J 1 ( x , y ) Y K 3 [ M ( x , x , y ) + M ( x , 3 x , y ) ] 1 p
for all x , y X . There exists a unique Jensen-additive mapping J 2 : X × X Y satisfying
f ( x , y ) f ( x , 0 ) J 2 ( x , y ) Y K 3 [ N ( x , y , y ) + N ( x , y , 3 y ) ] 1 p
for all x , y X .
Proof. 
Let g ( x , y ) : = f ( x , y ) f ( 0 , y ) for all x , y X . Then g ( 0 , y ) = 0 for all y X . Letting y by x in (14), we get
g ( x , z ) + g ( x , z ) Y φ ( x , x , z )
for all x , z X . Replacing x by x and y by 3 x in (14), we have
2 g ( x , z ) g ( x , z ) g ( 3 x , z ) Y φ ( x , 3 x , z )
for all x , z X . By two above inequalities and replacing z by y, we get
3 g ( x , y ) g ( 3 x , y ) Y K [ φ ( x , x , y ) + φ ( x , 3 x , y ) ]
for all x , y X . Thus we have
1 3 j g ( 3 j x , y ) 1 3 j + 1 g ( 3 j + 1 x , y ) Y K 3 j + 1 [ φ ( 3 j x , 3 j x , y ) + φ ( 3 j x , 3 j + 1 x , y ) ]
for all x , y X and all j. For given integer l , m ( 0 l < m ) , by Lemma 2, we get
1 3 l g ( 3 l x , y ) 1 3 m g ( 3 m x , y ) Y p K 3 p j = l m 1 1 3 p j φ ( 3 j x , 3 j x , y ) p + φ ( 3 j x , 3 j + 1 x , y ) p
for all x , y X . By (12), the sequence { 1 3 j g ( 3 j x , y ) } is a Cauchy sequence for all x , y X . Since Y is complete, the sequence { 1 3 j g ( 3 j x , y ) } converges for all x , y X . Define J 1 : X × X Y by
J 1 ( x , y ) : = lim j 1 3 j g ( 3 j x , y )
for all x , y X . Putting l = 0 and taking m in (18), one can obtain the inequality (16). From the definition of J 1 , we get
3 j J 1 ( x , y ) = J 1 ( 3 j x , y ) and J 1 ( 0 , y ) = 0
for all x , y X and all j. By (14), (16) and (19), we gain
2 J 1 ( 2 x , y ) 4 J 1 ( x , y ) Y = 2 J 1 ( 2 x , y ) J 1 ( 3 x , y ) J 1 ( x , y ) Y 3 j 2 J 1 ( 3 j · 2 x , y ) J 1 ( 3 j · 3 x , y ) J 1 ( 3 j x , y ) Y 3 j 2 J 1 ( 3 j · 2 x , y ) 2 f ( 3 j · 2 x , y ) Y + J 1 ( 3 j · 3 x , y ) f ( 3 j · 3 x , y ) Y + 3 j J 1 ( 3 j x , y ) f ( 3 j x , y ) Y + 3 j 2 f 3 j ( 3 x + x ) 2 , y f ( 3 j · 3 x , y ) f ( 3 j x , y ) Y 2 · 3 j 1 K [ M ( 3 j · 2 x , 3 j ( 2 x ) , y ) + M ( 3 j · ( 2 x ) , 3 j + 1 · 2 x , y ) ] 1 p + 3 j 1 K [ M ( 3 j · 3 x , 3 j ( 3 x ) , y ) + M ( 3 j · ( 3 x ) , 3 j + 1 · 3 x , y ) ] 1 p + 3 j 1 K [ M ( 3 j x , 3 j ( x ) , y ) + M ( 3 j · ( x ) , 3 j + 1 x , y ) ] 1 p + 3 j φ ( 3 j x , 3 j + 1 x , y )
for all x , y X and all j. From this and (19), we obtain
2 J 1 ( x , y ) = J 1 ( 2 x , y )
for all x , y X . From (12) and (14),
2 J 1 x + y 2 , z J 1 ( x , z ) J 1 ( y , z ) Y = lim j 3 j 2 J 1 3 j x + 3 j y 2 , z J 1 ( 3 j x , z ) J 1 ( 3 j y , z ) Y lim j 3 j φ ( 3 j x , 3 j y , z ) = 0
for all x , y , z X . From (20) and the above inequality,
J 1 ( x + y , z ) = 2 J 1 x + y 2 , z = J 1 ( x , z ) + J 1 ( y , z )
for all x , y , z X . Hence
J 1 ( x + y , z ) = J 1 ( x , z ) + J 1 ( y , z )
for all x , y , z X . That is, J 1 is an additive mapping with respect to the first variable. By (15), we get
2 3 j g 3 j x , y + z 2 + 1 3 j g ( 3 j x , y ) 1 3 j g ( 3 j x , z ) Y 1 3 j ψ ( 3 j x , y , z )
for all x , y , z X and all j. Letting j in the above inequality and using (10), J 1 is a Jensen mapping with respect to the second variable. To prove the uniqueness of J 1 , let S 1 be another additive-Jensen mapping satisfying (16). Then we obtain
2 S 1 ( 2 x , y ) 4 S 1 ( x , y ) Y p = 2 S 1 ( 2 x , y ) S 1 ( 3 x , y ) S 1 ( x , y ) Y p = 3 j p 2 S 1 ( 2 · 3 j x , y ) S 1 ( 3 · 3 j x , y ) S 1 ( 3 j x , y ) } } Y p 3 j p 2 S 1 ( 2 · 3 j x , y ) 2 g ( 2 · 3 j x , y ) Y p + 3 j p S 1 ( 3 · 3 j x , y ) g ( 3 · 3 j x , y ) Y p + 3 j p S 1 ( 3 j x , y ) g ( 3 j x , y ) Y p + 3 j p 2 g 3 j · 3 x + x 2 , y g ( 3 · 3 j x , y ) g ( 3 j x , y ) Y p
for all x , y X and all j. It follows from (16), we have
J 1 ( x , y ) S 1 ( x , y ) Y p = 1 3 j J 1 ( 3 j x , y ) 1 3 j S 1 ( 3 j x , y ) Y p 1 3 j J 1 ( 3 j x , y ) 1 3 j f ( 3 j x , y ) + 1 3 j f ( 0 , y ) Y p + 1 3 j f ( 3 j x , y ) 1 3 j f ( 0 , y ) 1 3 j S 1 ( 3 j x , y ) Y p 2 K p 3 p ( j + 1 ) [ M ( 3 j x , 3 j x , y ) + M ( 3 j x , 3 j + 1 x , y ) ]
for all x , y X and all j. Taking j in the above inequality and using (12), we get J 1 = S 1 .
Define J 2 : X × X Y by
J 2 ( x , y ) : = lim j 1 3 j f ( x , 3 j y )
for all x , y X . By the same method in the above arguments, J 2 is a unique Jensen-additive mapping satisfying (17). □
Corollary 3.
Let 0 < p 1 and ε , δ > 0 be fixed. Suppose that a mapping f : X × X Y satisfies the inequalities
2 f x + y 2 , z f ( x , z ) f ( y , z ) Y ε , 2 f x , y + z 2 f ( x , y ) f ( x , z ) Y δ
for all x , y , z X . Then there exists a unique additive-Jensen mapping J 1 : X × X Y satisfying
f ( x , y ) f ( 0 , y ) J 1 ( x , y ) Y K ε 2 3 p 1 1 p
for all x , y X . There exists a unique Jensen-additive mapping J 2 : X × X Y satisfying
f ( x , y ) f ( x , 0 ) J 2 ( x , y ) Y K δ 2 3 p 1 1 p
for all x , y X .
Proof. 
Let φ ( x , y , z ) : = ε and ψ ( x , y , z ) : = δ for all x , y , z X . By Theorem 4, we have an additive-Jensen mapping J 1 and a Jensen-additive mapping J 2 , as desired. □
From now on, let χ : X × X × X × X [ 0 , ) be a function such that
lim n 1 4 n χ ( 2 n x , 2 n y , 2 n z , 2 n w ) = 0
and
L ( x , y , z , w ) : = j = 0 1 4 p j χ ( 2 j x , 2 j y , 2 j z , 2 j w ) p <
for all x , y , z , w X .
Theorem 5.
Let 0 < p 1 and suppose that a mapping f : X × X Y satisfies f ( x , 0 ) = 0 and the inequality
4 f x + y 2 , z + w 2 f ( x , z ) f ( x , w ) f ( y , z ) f ( y , w ) Y χ ( x , y , z , w )
for all x , y , z , w X . Then the limit F ( x , y ) : = lim j 1 4 j f ( 2 j x , 2 j y ) exists for all x , y X and the mapping F : X × X Y is the unique bi-Jensen mapping satisfying
f ( x , y ) f ( 0 , y ) F ( x , y ) Y χ ˜ ( x , y ) 1 p ,
where
χ ˜ ( x , y ) = j = 0 1 4 p ( j + 1 ) χ ( 2 j + 1 x , 0 , 2 j + 1 y , 0 ) p + χ ( 0 , 0 , 2 j + 1 y , 0 ) p
for all x , y X .
Proof. 
Replacing x by 2 j + 1 x and putting y = 0 , z = 2 j + 1 y , w = 0 in (23), we gain
1 4 j f ( 2 j x , 2 j y ) 1 4 j + 1 f ( 2 j + 1 x , 2 j + 1 y ) 1 4 j + 1 f ( 0 , 2 j + 1 y ) Y 1 4 j + 1 χ ( 2 j + 1 x , 0 , 2 j + 1 y , 0 )
for all x , y X and all j. Letting x = 0 in (25), we get
1 4 j f ( 0 , 2 j y ) 2 4 j + 1 f ( 0 , 2 j + 1 y ) Y 1 4 j + 1 χ ( 0 , 0 , 2 j + 1 y , 0 )
for all y X and all j. By (25) and(26), we have
1 4 j f ( 2 j x , 2 j y ) f ( 0 , 2 j y ) 1 4 j + 1 f ( 2 j + 1 x , 2 j + 1 y ) f ( 0 , 2 j + 1 y ) Y p 1 4 p ( j + 1 ) χ ( 2 j + 1 x , 0 , 2 j + 1 y , 0 ) p + χ ( 0 , 0 , 2 j + 1 y , 0 ) p
for all x , y X and all j. Thus we have
1 4 l f ( 2 l x , 2 l y ) f ( 0 , 2 l y ) 1 4 m f ( 2 m x , 2 m y ) f ( 0 , 2 m y ) Y p j = l m 1 1 4 p ( j + 1 ) χ ( 2 j + 1 x , 0 , 2 j + 1 y , 0 ) p + χ ( 0 , 0 , 2 j + 1 y , 0 ) p
for all integers l , m ( 0 l < m ) and all x , y X . By (22), the sequence { 1 4 j [ f ( 2 j x , 2 j y ) f ( 0 , 2 j y ) ] } is a Cauchy sequence for all x , y X . Since Y is complete, the sequence { 1 4 j [ f ( 2 j x , 2 j y ) f ( 0 , 2 j y ) ] } converges for all x , y X . So one can define the mapping F : X × X Y by
F ( x , y ) : = lim n 1 4 n [ f ( 2 n x , 2 n y ) f ( 0 , 2 n y ) ]
for all x , y X . Letting l = 0 and taking the limit m in (28), we get (24). Now, we show that F is a bi-Jensen mapping.
On the other hand it follows from (22), (23) and (29) that
4 F x + y 2 , z + w 2 F ( x , z ) F ( x , w ) F ( y , z ) F ( y , w ) Y p = lim n 1 4 p n 4 f 2 n x + 2 n y 2 , 2 n z + 2 n w 2 f ( 2 n x , 2 n z ) f ( 2 n x , 2 n w ) f ( 2 n y , 2 n z ) f ( 2 n y , 2 n w ) 4 f 0 , 2 n z + 2 n w 2 + 2 f ( 0 , 2 n z ) + 2 f ( 0 , 2 n w ) Y p = lim n 1 4 p n [ χ ( 2 n x , 2 n y , 2 n z , 2 n w ) p + χ ( 0 , 0 , 2 n z , 2 n w ) p ] = 0
for all x , y , z , w X . Hence the mapping F satisfies (3).
To prove the uniqueness of F, let G : X Y be another bi-Jensen mapping satisfying (24). It follows from (22) that
lim n 1 4 p n L ( 2 n x , 2 n y , 2 n z , 2 n w ) = lim n j = n 1 4 p j χ ( 2 j x , 2 j y , 2 j z , 2 j w ) p = 0
for all x , y , z , w X . Hence lim n 1 4 p n χ ˜ ( 2 n x , 2 n y ) = 0 for all x , y X . It follows from (21), (27) and (29) the above equality that
F ( 2 x , 2 y ) 4 F ( x , y ) Y = lim n 1 4 n f ( 2 n + 1 x , 2 n + 1 y ) f ( 0 , 2 n + 1 y ) 1 4 n 1 f ( 2 n x , 2 n y ) + f ( 0 , 2 n y ) Y = 4 lim n 1 4 n [ f ( 2 n x , 2 n y ) f ( 0 , 2 n y ) ] 1 4 n + 1 [ f ( 2 n + 1 x , 2 n + 1 y ) f ( 0 , 2 n + 1 y ) ] Y 4 lim n 1 4 p ( n + 1 ) χ ( 2 n + 1 x , 0 , 2 n + 1 y , 0 ) p + χ ( 0 , 0 , 2 n + 1 y , 0 ) p = 0
for all x , y X . So F ( 2 x , 2 y ) = 4 F ( x , y ) for all x , y X . Thus it follows from (24) and (29) that
F ( x , y ) G ( x , y ) Y p = lim n 1 4 p n f ( 2 n x , 2 n y ) f ( 0 , 2 n y ) G ( 2 n x , 2 n y ) Y p lim n 1 4 p n χ ˜ ( 2 n x , 2 n y ) = 0
for all x , y X . So F = G . □
Corollary 4.
Let 0 < p 1 and ε > 0 be fixed. Suppose that a mapping f : X × X Y satisfies the inequalities
4 f x + y 2 , z + w 2 f ( x , z ) f ( x , w ) f ( y , z ) f ( y , w ) Y ε
for all x , y , z , w X . Then there exists a unique bi-Jensen mapping F : X × X Y satisfying
f ( x , y ) f ( 0 , y ) F ( x , y ) Y ε 2 4 p 1 1 p
for all x , y X .
Proof. 
Taking χ ( x , y , z , w ) : = ε for all x , y , z , w X in Theorem 5, we obtain χ ˜ ( x , y ) = 2 ε p 4 p 1 for all x , y X . Thus we obtain the estimate value χ ˜ ( x , y ) 1 p = ε 2 4 p 1 1 p for all x , y X . □

Author Contributions

Conceptualization, J.-H.B. and W.-G.P.; methodology, J.-H.B. and W.-G.P.; validation, J.-H.B. and W.-G.P.; investigation, J.-H.B. and W.-G.P.; writing—original draft preparation, J.-H.B. and W.-G.P.; writing—review and editing, J.-H.B. and W.-G.P.; project administration, J.-H.B. and W.-G.P.; funding acquisition, J.-H.B. and W.-G.P. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Bae, J.-H.; Park, W.-G. Stability of Bi-Additive Mappings and Bi-Jensen Mappings. Symmetry 2021, 13, 1180. https://doi.org/10.3390/sym13071180

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Bae J-H, Park W-G. Stability of Bi-Additive Mappings and Bi-Jensen Mappings. Symmetry. 2021; 13(7):1180. https://doi.org/10.3390/sym13071180

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Bae, Jae-Hyeong, and Won-Gil Park. 2021. "Stability of Bi-Additive Mappings and Bi-Jensen Mappings" Symmetry 13, no. 7: 1180. https://doi.org/10.3390/sym13071180

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Bae, J. -H., & Park, W. -G. (2021). Stability of Bi-Additive Mappings and Bi-Jensen Mappings. Symmetry, 13(7), 1180. https://doi.org/10.3390/sym13071180

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