# Stability of Jensen-Type Functional Equations on Restricted Domains in a Group and Their Asymptotic Behaviors

Authors: Jae-Young Chung; Dohan Kim; John Michael Rassias
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We consider the Hyers-Ulam stability problems for the Jensen-type functional equations in general restricted domains. The main purpose of this paper is to find the restricted domains for which the functional inequality satisfied in those domains extends to the inequality for whole domain. As consequences of the results we obtain asymptotic behavior of the equations.


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## Body

## 1. Introduction

The Hyers-Ulam stability problems of functional equations was originated by Ulam in 1960 when he proposed the following question [1].Letfbe a mapping from a groupG1to a metric groupG2with metricd(·,·)such that(1.1)d(f(xy),f(x)f(y))≤ε.Then does there exist a group homomorphismhandδϵ>0such that(1.2)d(f(x),h(x))≤δϵfor allx∈G1?One of the first assertions to be obtained is the following result, essentially due to Hyers [2], that gives an answer for the question of Ulam.Theorem 1.1.
Suppose thatS is an additive semigroup,  Y is a Banach space,  ϵ≥0, and  f:S→Y satisfies the inequality
(1.3)‖f(x+y)-f(x)-f(y)‖≤ϵ
for all x,y∈S. Then there exists a unique function A:S→Y satisfying
(1.4)A(x+y)=A(x)+A(y)
for which
(1.5)‖f(x)-A(x)‖≤ϵ
for all x∈S.We call the functions satisfying (1.4) additive functions. Perhaps Aoki in 1950 was the first author treating the generalized version of Hyers’ theorem [3]. Generalizing Hyers’ result he proved that if a mapping f:X→Y between two Banach spaces satisfies (1.6)‖f(x+y)-f(x)-f(y)‖≤Φ(x,y)forx,y∈X
with Φ(x,y)=ϵ(∥x∥p+∥y∥p)(ϵ≥0,0≤p<1), then there exists a unique additive function A:X→Y such that ∥f(x)-A(x)∥≤2ϵ∥x∥p/(2-2p) for all x∈X. In 1951 Bourgin [4, 5] stated that if Φ is symmetric in ∥x∥ and ∥y∥ with ∑j=1∞Φ(2jx,2jx)/2j<∞ for each x∈X, then there exists a unique additive function A:X→Y such that ∥f(x)-A(x)∥≤∑j=1∞Φ(2jx,2jx)/2j for all x∈X. Unfortunately, there was no use of these results until 1978 when Rassias [6] dealt with the inequality of Aoki [3]. Following Rassias’ result, a great number of papers on the subject have been published concerning numerous functional equations in various directions [6–15]. Among the results, stability problem in a restricted domain was investigated by Skof, who proved the stability problem of inequality (1.3) in a restricted domain [16, 17]. Developing this result, Jung, Rassias, and M. J. Rassias considered the stability problems in restricted domains for some functional equations including the Jensen functional equation [9] and Jensen-type functional equations [13]. We also refer the reader to [18–27] for some related results on Hyers-Ulam stabilities in restricted conditions. The results can be summarized as follows. Let X and Y be a real normed space and a real Banach space, respectively. For fixed d≥0, if f:X→Y satisfies the functional inequalities (such as that of Cauchy, quadratic, Jensen, and Jensen type) for all x,y∈X with ∥x∥+∥y∥≥d (which is the case where the inequalities are given by two indeterminate variables x and y), the inequalities hold for all x,y∈X. Following the approach in [28] we consider the Jensen-type equation in various restricted domains in an Abelian group. As applications, we obtain the stability problems for the above equations in more general restricted domains than that of the form {(x,y)∈X:∥x∥+∥y∥≥d}, which generalizes and refines the stability theorems in [13]. As an application we obtain asymptotic behaviors of the equations.


## 2. Stability of Jensen-Type Functional Equations

Throughout this section, we denote byG, X, and Y, an Abelian group, a real normed space, and a Banach space, respectively. In this section we consider the Hyers-Ulam stability of the Jensen and Jensen-type functional inequalities for the functions f:G→Y(2.1)‖f(x+y)+f(x-y)-2f(x)‖≤ϵ,(2.2)‖f(x+y)-f(x-y)-2f(y)‖≤ϵ
in restricted domains U⊂G×G.Inequalities (2.1) and (2.2) were previously treated by J. M. Rassias and M. J. Rassias [13], who proved the Hyers-Ulam stability of the inequalities in the restricted domain U={(x,y):∥x∥+∥y∥≥d},d≥0, for the functions f:X→Y:Theorem 2.1.
Letd≥0 and  ϵ>0 be fixed. Suppose that  f:X→Y satisfies the inequality
(2.3)‖f(x+y)+f(x-y)-2f(x)‖≤ϵ
for all x,y∈X, with ∥x∥+∥y∥≥d. Then there exists a unique additive function A:X→Y such that
(2.4)‖f(x)-A(x)-f(0)‖≤52ϵ
for all x∈X.Theorem 2.2.
Letd≥0 and  ϵ>0 be fixed. Suppose that  f:X→Y satisfies the inequality
(2.5)‖f(x+y)-f(x-y)-2f(y)‖≤ϵ
for all x,y∈X, with ∥x∥+∥y∥≥d and
(2.6)‖f(x)+f(-x)‖≤3ϵ
for all x∈X, with ∥x∥≥d. Then there exists a unique additive function A:X→Y such that
(2.7)‖f(x)-A(x)‖≤332ϵ
for all x∈X.We use the following usual notations. We denote byG×G={(a1,a2):a1,a2∈G} the product group; that is, for a=(a1,a2),b=(b1,b2)∈G×G, we define a+b=(a1+b1,a2+b2),a-b=(a1-b1,a2-b2). For a subset H of G×G and a,b∈G×G, we define a+H={a+h:h∈H}.For givenx,y∈G we denote by Px,y,Qx,y the subsets of points of the forms (not necessarily distinct) in G×G, respectively, (2.8)Px,y={(0,0),(x,-x),(y,y),(x+y,-x+y)},Qx,y={(-x,x),(y,y),(-x+y,x+y)}.
The set Px,y can be viewed as the vertices of rectangles in G×G, and Qx,y can be viewed as a subset of the vertices of rectangles in G×G.Definition 2.3.
LetU⊂G×G. One introduces the following conditions(J1)and(J2)onU.For anyx,y∈G, there exists az∈Gsuch that(2.9)(J1)(0,z)+Px,y={(0,z),(x,-x+z),(y,y+z),(x+y,-x+y+z)}⊂U,(J2)(z,0)+Qx,y={(-x+z,x),(y+z,y),(-x+y+z,x+y)}⊂U,
respectively.The sets(0,z)+Px,y,(z,0)+Qx,y can be understood as the translations of Px,y and Qx,y by (0,z) and (z,0), respectively.There are many interesting examples of the setsU satisfying some of the conditions (J1) and (J2). We start with some trivial examples.Example 2.4.
LetG be a real normed space. For d≥0,x0,y0∈G, let
(2.10)U={(x,y)∈G×G:k‖x‖+s‖y‖≥d},V={(x,y)∈G×G:‖kx+sy‖≥d}.
Then U satisfies (J1) if s>0, (J2) if k>0 and V satisfies (J1) if s≠0, (J2) if k≠0.Example 2.5.
LetG be a real inner product space. For d≥0,x0,y0∈G(2.11)U={(x,y)∈G×G:〈x0,x〉+〈y0,y〉≥d}.
Then U satisfies (J1) if y0≠0, (J2) if x0≠0.Example 2.6.
LetG be the group of nonsingular square matrices with the operation of matrix multiplication. For k,s∈ℝ,δ,d≥0, let
(2.12)U={(P1,P2)∈G×G:|det⁡P1|k|det⁡P2|s≤δ},V={(P1,P2)∈G×G:|det⁡P1|k|det⁡P2|s≥d}.
Then both U and V satisfy (J1) if s≠0, (J2) if k≠0.In the following one can see that ifPx,y and Qx,y are replaced by arbitrary subsets of four points (not necessarily distinct) in G×G, respectively, then the conditions become stronger; that is, there are subsets U1 and U2 which satisfy the conditions (J1) and (J2), respectively, but U1 and U2 fail to fulfill the following conditions (2.13) and (2.14), respectively. For any subset {X1,X2,X3,X4} of points (not necessarily distinct) in G×G, there exists a z∈G such that (2.13)(0,z)+{X1,X2,X3,X4}⊂U1,(2.14)(z,0)+{X1,X2,X3,X4}⊂U2,
respectively.Now we give examples ofU1 and U2 which satisfy (J1) and (J2), respectively, but not (2.13) and (2.14), respectively.Example 2.7.
LetG=ℤ be the group of integers. Enumerate
(2.15)Z×Z={(a1,b1),(a2,b2),…,(an,bn),…}⊂R2
such that
(2.16)|a1|+|b1|≤|a2|+|b2|≤⋯≤|an|+|bn|≤⋯,
and let Pn={(0,0),(an,-an),(bn,bn),(an+bn,-an+bn)},n=1,2,…. Then it is easy to see that U=⋃n=1∞((0,2n)+Pn) satisfies the condition (J1). Now let P={(p1,q1),(p2,q2)}⊂ℤ×ℤ with |q2-q1|≤|p2-p1|,p1p2>0. Then (0,z)+P is not contained in U for all z∈ℤ. Indeed, for any choices of (xn,yn)∈Pn+(0,2n),n=1,2,…, we have ym-yn>|xm-xn| for all m>n,m,n=1,2,…. Thus, if (0,z)+P⊂U for some z∈ℤ, then P+(0,z)⊂(0,2n)+Pn for some n∈ℕ. Thus, it follows from the condition q2-q1≤|p2-p1| that the line segment joining the points of P+(-z,z) intersects the line x=0 in ℝ2, which contradicts the condition p1p2>0. Similarly, let Qn={(-an,an),(bn,bn),(-an+bn,an+bn)}. Then it is easy to see that U=⋃n=1∞((2n,0)+Qn) satisfies the condition (J2) but not (2.14).Theorem 2.8.
LetU⊂G×G satisfy the condition  (J1) and  ϵ≥0. Suppose that  f:G→Y satisfies  (2.1) for all  (x,y)∈U. Then there exists an additive function  A:G→Y such that
(2.17)‖f(x)-A(x)-f(0)‖≤2ϵ
for all x∈G.Proof.
For givenx,y∈G, choose a z∈G such that (0,z)+Px,y⊂U. Replacing x by x+y, y by -x+y+z; x by x, y by -x+z; x by y, y by y+z; x by 0, y by z in (2.1), respectively, we have
(2.18)‖f(2y+z)+f(2x-z)-2f(x+y)‖≤ϵ,‖f(z)+f(2x-z)-2f(x)‖≤ϵ,‖f(2y+z)+f(-z)-2f(y)‖≤ϵ,‖f(z)+f(-z)-2f(0)‖≤ϵ.
From (2.18), using the triangle inequality and dividing the result by 2, we have
(2.19)|f(x+y)-f(x)-f(y)+f(0)|≤2ϵ
for all x,y∈G. From (2.19), using Theorem 1.1, we get the result.Letd≥0,s∈ℝ, and let U={(x,y):∥x∥+s∥y∥≥d}. Then U satisfies the condition (J1). Thus, as a direct consequence of Theorem 2.8, we obtain the following (cf. Theorem 2.1).Corollary 2.9.
Letd≥0,s∈ℝ. Suppose that  f:X→Y satisfies inequality  (2.1) for all  x,y∈X, with  ∥x∥+s∥y∥≥d. Then there exists a unique additive function  A:X→Y such that
(2.20)‖f(x)-A(x)-f(0)‖≤2ϵ
for all x∈X.Theorem 2.10.
LetU⊂G×G satisfy the condition  (J2) and  ϵ≥0. Suppose that  f:G→Y satisfies  (2.2) for all  (x,y)∈U. Then there exists a unique additive function  A:G→Y such that
(2.21)‖f(x)-A(x)‖≤32ϵ
for all x∈G.Proof.
For givenx,y∈G, choose a z∈G such that (z,0)+Qx,y⊂U. Replacing x by -x+y+z, y by x+y; x by -x+z, y by x; x by y+z, y by y in (2.2), respectively, we have
(2.22)‖f(2y+z)-f(-2x+z)-2f(x+y)‖≤ϵ,‖f(z)-f(-2x+z)-2f(x)‖≤ϵ,‖f(2y+z)-f(z)-2f(y)‖≤ϵ.
From (2.22), using the triangle inequality and dividing the result by 2, we have
(2.23)|f(x+y)-f(x)-f(y)|≤32ϵ.
Now by Theorem 1.1, we get the result.Letd≥0,k∈ℝ, and let U={(x,y):k∥x∥+∥y∥≥d}. Then U satisfies the condition (J2). Thus, as a direct consequence of Theorem 2.10, we generalize and refine Theorem 2.2 as follows.Corollary 2.11.
Letd≥0,k∈ℝ. Suppose that  f:X→Y satisfies inequality  (2.2) for all  x,y, with  k∥x∥+∥y∥≥d. Then there exists a unique additive function  A:X→Y such that
(2.24)‖f(x)-A(x)‖≤32ϵ
for all x∈X.Remark 2.12.
Corollary2.11 refines Theorem 2.2 in both the bounds and the condition (2.6).Now we discuss other possible restricted domains. We assume thatG is a 2-divisible Abelian group. For given x,y∈G, we denote by Rx,y,Sx,y⊂G×G, (2.25)Rx,y={(x,x),(x,y),(x-y2,x-y2),(x-y2,-x+y2)},Sx,y={(x,x),(y,x),(x-y2,x-y2),(-x+y2,x-y2)}.
One can see that Rx,y and Sx,y consist of the vertices of parallelograms in G×G, respectively.Definition 2.13.
LetU⊂G×G. One introduces the following conditions(J3),(J4)onU.For anyx,y∈G, there exists az∈Gsuch that(2.26)(J3)(z,-z)+Rx,y={(x+z,x-z),(x+z,y-z),(x-y2+z,x-y2-z),(x-y2+z,-x+y2-z)}⊂U,(J4)(z,-z)+Sx,y={(x+z,x-z),(y+z,x-z),(x-y2+z,x-y2-z),(-x+y2+z,x-y2-z)}⊂U,
respectively.Example 2.14.
LetG be a real normed space. For k,s,d∈ℝ, let
(2.27)U={(x,y)∈G×G:k‖x‖+s‖y‖≥d},V={(x,y)∈G×G:‖kx+sy‖≥d}.
Then U satisfies (J3) and (J4) if k+s>0, and V satisfies (J3) and (J4) if k≠s.Example 2.15.
LetG be a real inner product space. For d≥0,x0,y0∈G,
(2.28)U={(x,y)∈G×G:〈x0,x〉+〈y0,y〉≥d}.
Then U satisfies (J3),(J4) if x0≠y0.Example 2.16.
LetG be the group of nonsingular square matrices with the operation of matrix multiplication. For k,s∈ℝ,δ,d≥0, let
(2.29)U={(P1,P2)∈G×G:|det⁡P1|k|det⁡P2|s≤δ},V={(P1,P2)∈G×G:|det⁡P1|k|det⁡P2|s≥d}.
Then U and V satisfy both (J3) and (J4) if k≠s.From now on, we assume thatG is a 2-divisible Abelian group.Theorem 2.17.
LetU⊂G×G satisfy the condition  (J3) and  ϵ≥0. Suppose that  f:G→Y satisfies  (2.1) for all  (x,y)∈U. Then there exists a unique additive function  A:G→Y such that
(2.30)‖f(x)-A(x)-f(0)‖≤4ϵ
for all x∈G.Proof.
For givenx,y∈G, choose a z∈G such that (z,-z)+Rx,y⊂U. Replacing x by x+z, y by x-z; x by x+z, y by y-z; x by (x-y)/2+z, y by (x-y)/2-z; x by (x-y)/2+z, y by (-x+y)/2-z in (2.1), respectively, we have
(2.31)‖f(2x)+f(2z)-2f(x+z)‖≤ϵ,‖f(x+y)+f(x-y+2z)-2f(x+z)‖≤ϵ,‖f(x-y)+f(2z)-2f(x-y2+z)‖≤ϵ,‖f(0)+f(x-y+2z)-2f(x-y2+z)‖≤ϵ.
From (2.31), using the triangle inequality, we have
(2.32)|f(2x)-f(x+y)-f(x-y)+f(0)|≤4ϵ
for all x,y∈G. Replacing x by (x+y/2), y by (x-y/2) in (2.32), we have
(2.33)|f(x+y)-f(x)-f(y)+f(0)|≤4ϵ
for all x,y∈G. From (2.33), using Theorem 1.1, we get the result.Letd≥0,k,s∈ℝ with k+s>0, and let U={(x,y):k∥x∥+s∥y∥≥d}. Then U satisfies the conditions (J3) and (J4). Thus, as a direct consequence of Theorem 2.17 we generalize Theorem 2.1 as follows.Corollary 2.18.
Letd≥0,k,s∈ℝ with  k+s>0. Suppose that  f:X→Y satisfies the inequality  (2.1) for all  x,y, with  k∥x∥+s∥y∥≥d. Then there exists a unique additive function  A:X→Y such that
(2.34)‖f(x)-A(x)-f(0)‖≤4ϵ
for all x∈X.Theorem 2.19.
LetU⊂G×G satisfy the condition  (J4) and  ϵ≥0. Suppose that  f:G→Y satisfies  (2.2) for all  (x,y)∈U. Then there exists a unique additive function  A:G→Y such that
(2.35)‖f(x)-A(x)‖≤4ϵ
for all x∈G.Proof.
For givenx,y∈G, choose a z∈G such that (z,-z)+Sx,y⊂U. Replacing x by x+z, y by x-z; x by y+z, y by x-z; x by (x-y)/2+z, y by (x-y)/2-z; x by (-x+y)/2+z, y by (x-y)/2-z in (2.2), respectively, we have
(2.36)‖f(2x)-f(2z)-2f(x-z)‖≤ϵ,‖f(x+y)-f(-x+y+2z)-2f(x-z)‖≤ϵ,‖f(x-y)-f(2z)-2f(x-y2+z)‖≤ϵ,‖f(0)-f(-x+y+2z)-2f(x-y2+z)‖≤ϵ.
From (2.36), using the triangle inequality, we have
(2.37)|f(2x)-f(x+y)-f(x-y)+f(0)|≤4ϵ
for all x,y∈G. Replacing x by (x+y)/2, y by (x-y)/2 in (2.37) and using Theorem 1.1, we get the result.As a direct consequence of Theorem2.19, we have the following.Corollary 2.20.
Letd≥0,k,s∈ℝ with  k+s>0. Suppose that  f:X→Y satisfies the inequality  (2.2) for all  x,y, with  k∥x∥+s∥y∥≥d. Then there exists a unique additive function  A:X→Y such that
(2.38)‖f(x)-A(x)‖≤4ϵ
for all x∈X.


## 3. Asymptotic Behavior of the Equations

In this section we discuss asymptotic behaviors of the equations which gives refined versions of the results in [13].Using Theorems2.8 and 2.17, we have the following (cf. [13]).Theorem 3.1.
LetU satisfy  (J1) or  (J3). Suppose that  f:X→Y satisfies the asymptotic condition
(3.1)‖f(x+y)+f(x-y)-2f(x)‖⟶0
as ∥x∥+∥y∥→∞,(x,y)∈U. Then there exists a unique additive function A:X→Y such that
(3.2)f(x)=A(x)+f(0)
for all x∈X.Proof.
By the condition (3.1), for each n∈ℕ, there exists dn>0 such that
(3.3)‖f(x+y)+f(x-y)-2f(x)‖≤1n
for all (x,y)∈U with ∥x∥+∥y∥≥dn. Let U0=U∩{(x,y):∥x∥+∥y∥≥dn}. Then U0 satisfies both the conditions (J1) and (J3). By Theorems 2.8 and 2.17, there exists a unique additive function An:X→Y such that
(3.4)‖f(x)-An(x)-f(0)‖≤2nor4n
for all x∈X. Putting n=m in (3.4) and using the triangle inequality, we have
(3.5)‖An(x)-Am(x)‖≤8
for all x∈X. Using the additivity of An,Am, we have An=Am for all n,m∈ℕ. Letting n→∞ in (3.4), we get the result.Corollary 3.2.
Letk,s∈ℝ satisfy one of the conditions:  s>0,k+s>0. Suppose that  f:X→Y satisfies the condition
(3.6)‖f(x+y)+f(x-y)-2f(x)‖⟶0
as k∥x∥+s∥y∥→∞. Then there exists a unique additive function A:X→Y such that
(3.7)f(x)=A(x)+f(0)
for all x∈X.Using Theorems2.10 and 2.19, we have the following (cf. [13]).Theorem 3.3.
LetU satisfy  (J2) or  (J4). Suppose that  f:X→Y satisfies the condition
(3.8)‖f(x+y)-f(x-y)-2f(y)‖⟶0
as ∥x∥+∥y∥→∞,(x,y)∈U. Then f is an additive function.Corollary 3.4.
Letk,s∈ℝ satisfy one of the conditions:  k>0,k+s>0. Suppose that  f:X→Y satisfies the condition
(3.9)‖f(x+y)-f(x-y)-2f(y)‖⟶0
as k∥x∥+s∥y∥→∞. Then f is an additive function.



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