# Stability of Exponential Functional Equations with Involutions

Authors: Jaeyoung Chung; Soon-Yeong Chung
Journal: Journal of Function Spaces
 (2014)

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## Abstract

LetS be a commutative semigroup if not otherwise specified and f:S→ℝ. In this paper we consider the stability of exponential functional equations |f(x+σ(y))-g(x)f(y)|≤ϕ(x) or ϕ(y), |f(x+σ(y))-f(x)g(y)|≤ϕ(x) or ϕ(y) for all x,y∈S and where σ:S→S is an involution. As main results we investigate bounded and unbounded functions satisfying the above inequalities. As consequences of our results we obtain the Ulam-Hyers stability of functional equations (Chung and Chang (in press); Chávez and Sahoo (2011); Houston and Sahoo (2008); Jung and Bae (2003)) and a generalized result of Albert and Baker (1982).


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## Body

## 1. Introduction

Throughout this paper we denote byℝ, ℝ+, ℂ, ℝn the set of real numbers, nonnegative real numbers, complex numbers, and the n-dimensional Euclidean space, respectively, and ϕ:S→ℝ+, ϵ≥0. A function m:S→ℂ is calledexponential provided that m(x+y)=m(x)m(y) for all x,y∈S and σ:S→S is calledan involution provided that σ(x+y)=σ(x)+σ(y) and σ(σ(x))=x for all x,y∈S. In [1], Baker proved the stability of the exponential functional equation: let f:S→ℂ satisfy the exponential functional inequality
(1)|f(x+y)-f(x)f(y)|≤ϵ
for all x,y∈S. Then, either f is a bounded function satisfying
(2)|f(x)|≤1+1+4ϵ2
for all x∈S, or f is an unbounded exponential function (see also Baker et al. [2]). In particular, if S=V, where V is a vector space over the field ℚ of rational numbers and f:V→ℝ is a bounded function satisfying (1) for 0≤ϵ<1/4, then f satisfies either
(3)1+1-4ϵ2≤f(x)≤1+1+4ϵ2
for all x∈V, or else
(4)-ϵ≤f(x)≤1-1-4ϵ2
for all x∈V (see Albert and Baker [3]). In [4–6], some functional equations arising from number theory or product of matrices are introduced. The equations therein can be reduced to the functional equations with involution σ:S→S:
(5)f(x+σ(y))-f(x)f(y)=0
for all x,y∈S (see Section 4). As we see in Section 4, the equations in [4–6] can be reduced to the equation of the form (5) when S=ℂ with operation of multiplication and σ(z)=z¯ for all z∈ℂ or σ(z)=z for all z∈ℂ and the equations in [6] are reduced to those forms (5) in the complex numbers with σ(z)=z¯ and in the set of quaternions with σ(q)=q*, where q* denotes the conjugate of the quaternion q.As main results of the paper, we consider, in Section2, the Ulam-Hyers stability of exponential functional equations with involution σ:S→S:
(6)|f(x+σ(y))-g(x)f(y)|≤ϕ(x),|f(x+σ(y))-f(x)g(y)|≤ϕ(y),|f(x+σ(y))-g(x)f(y)|≤ϕ(y),|f(x+σ(y))-f(x)g(y)|≤ϕ(x)
for all x,y∈S. In Section 3, as a consequence of our main results we obtain the behaviors of bounded functions f satisfying each of the inequalities
(7)|f(x+σ(y))-f(x)f(y)|≤ϕ(x),|f(x+σ(y))-f(x)f(y)|≤ϕ(y)
for all x,y∈S. Also, as a direct consequence of the stability of (7) we obtain a generalized version of the result of Albert and Baker [3]. Our method of proof also works for investigation of bounded solutions of some other functional inequalities (see [7–9]). We refer to [7–12] for the exponential functional equations, inequalities, and related results.In Section4, as an application of our result, we obtain the stability of the functional equations:
(8)f(x1,y1)f(x2,y2)=f(x1x2+y1y2,x1y2-x2y1)
for all x1,y1,x2,y2∈ℝ, and
(9)f(x1,y1,u1,v1)f(x2,y2,u2,v2)=f(x1x2+y1y2+u1u2+v1v2,x1y2-y1x2+u1v2-v1u2,x1u2-y1v2-u1x2+v1y2,x1v2+y1u2-u1y2-v1x2)
for all x1,y1,u1,v1,x2,y2,u2,v2∈ℝ, where f:ℝ2→ℝ and g:ℝ4→ℝ. As stated in [6], (8) and (9) arise from a well-known theorem in number theory.Reducing (8) and (9) to those in the complex numbers and the quaternions, we obtain the stability of (8) and (9); that is, we investigate bounded and unbounded functions f satisfying each of the following functional inequalities:
(10)|f(x1,y1)f(x2,y2)-f(x1x2+y1y2,x1y2-x2y1)|≤ϕ(x2,y2),(11)|f(x1,y1)f(x2,y2)-f(x1x2+y1y2,x1y2-x2y1)|≤ϕ(x1,y1)
for all x1,y1,x2,y2∈ℝ and for some ϕ:ℝ2→ℝ+, and we investigate bounded and unbounded functions g satisfying each of the following functional inequalities:
(12)|g(x1,y1,u1,v1)g(x2,y2,u2,v2)-g(x1x2+y1y2+u1u2+v1v2,x1y2-y1x2+u1v2-v1u2,x1u2-y1v2-u1x2+v1y2,x1v2+y1u2-u1y2-v1x2)|≤ψ(x2,y2,u2,v2),(13)|g(x1,y1,u1,v1)g(x2,y2,u2,v2)-g(x1x2+y1y2+u1u2+v1v2,x1y2-y1x2+u1v2-v1u2,x1u2-y1v2-u1x2+v1y2,x1v2+y1u2-u1y2-v1x2)|≤ψ(x1,y1,u1,v1)
for all x1,y1,u1,v1,x2,y2,u2,v2∈ℝ and for some ψ:ℝ4→ℝ+. We also refer the reader to [5] for another proof of finding general solutions of (8) and refer to [13] for the stability of inequalities (10)~(13).


## 2. Stability of (6)

Throughout this section we assume thatS is a commutative semigroup if not otherwise specified. An exponential function m:S→ℂ is called σ-exponential if m satisfies m(σ(x))=m(x) for all x∈S. We denote σ-exponential functions m by mσ. It is easy to see that if S is uniquely 2-divisible (i.e., for each x∈S there exists a unique y∈S such that 2y=x; we write y=x/2), then mσ is a σ-exponential function if and only if
(14)mσ(x)=m(x+σ(x)2),x∈S
for some exponential function m:S→ℂ. Throughout this section we denote by σ:S→S an involution. In the following we exclude the trivial cases when f=0 or g=0.Theorem 1.
Letf,g:S→ℂ satisfy
(15)|f(x+σ(y))-g(x)f(y)|≤ϕ(x)
for all x,y∈S. If S is a group and f is bounded, then
(16)|1-|g(x)||≤ϕ(x)Mf
for all x∈S, where Mf=sup⁡x∈S|f(x)|. If f is unbounded, then there exists a σ-exponential function mσ:S→ℂ such that
(17)g(x)=g(0)mσ(x),|f(x)-f(0)g(0)mσ(x)|≤ϕ(x)
for all x∈S, where g(0)=±1. In particular, if ϕ(x)=ϵ≥0 for all x∈S, then we have
(18)g(x)=mσ(x),f(x)=f(0)mσ(x)
for all x∈S.Proof.
First, we assume thatf is bounded (in this case we assume that S is a group). Using the triangle inequality with (15) we have
(19)|g(x)f(y)|≤|f(x+σ(y))|+ϕ(x)≤Mf+ϕ(x)
for all x,y∈S. Taking the supremum of the left-hand side of (19) with respect to y we have
(20)|g(x)|Mf≤Mf+ϕ(x)
for all x∈S, which implies
(21)Mf(|g(x)|-1)≤ϕ(x)
for all x∈S. Replacing y by σ(-x+y) in (15) and using the triangle inequality with the result we have
(22)|f(y)|≤|g(x)f(σ(-x+y))|≤|g(x)|Mf+ϕ(x)
for all x,y∈S. Taking the supremum of the left-hand side of (22) with respect to y we have
(23)Mf≤|g(x)|Mf+ϕ(x)
for all x∈S, which implies
(24)Mf(1-|g(x)|)≤ϕ(x)
for all x∈S. From (21) and (24) we get (16). Now, we assume that f is unbounded. Putting x=0 in (15) we have
(25)|f(σ(y))-g(0)f(y)|≤ϕ(0)
for all y∈S. Now, using the triangle inequality, (15), and (25) we have
(26)|f(z)||g(y)g(x)-g(0)g(x+σ(y))|≤|g(y)g(x)f(z)-g(y)f(x+σ(z))|+|g(y)f(x+σ(z))-f(y+σ(x)+z)|+|f(y+σ(x)+z)-g(0)f(x+σ(y)+σ(z))|+|g(0)f(x+σ(y)+σ(z))-g(0)g(x+σ(y))f(z)|≤|g(y)|ϕ(x)+ϕ(y)+ϕ(0)+|g(0)|ϕ(x+σ(y))
for all x,y,z∈S. Since f is unbounded, it follows from (26) that
(27)g(x)g(y)=g(0)g(x+σ(y))
for all x,y∈S. Dividing (27) by g(0)2 we have
(28)g0(x)g0(y)=g0(x+σ(y))
for all x,y∈S, where g0(x)=g(x)/g(0). Thus, g0 is σ-exponential and g(x)=g(0)mσ(x) for some σ-exponential mσ. Replacing y by σ(y) in (25) and multiplying g(0) in the result we have
(29)|g(0)f(y)-g(0)2f(σ(y))|≤|g(0)|ϕ(0)
for all y∈S. Using the triangle inequality with (25) and (29) we have
(30)|1-g(0)2||f(σ(y))|≤(1+|g(0)|)ϕ(0)
for all y∈S. Since f is unbounded, we have g(0)2=1. Putting y=0 in (15) we have
(31)|f(x)-f(0)g(x)|≤ϕ(x)
for all x∈S. Thus, we get (17). For the particular case when ϕ(x)=ϵ for all x∈S, replacing (x,y) by (σ(x),y) and (x,y) by (σ(y),x) in (15) and using triangle inequality with the resulting two inequalities, we have
(32)|f(x)g(σ(y))-f(y)g(σ(x))|≤2ϵ
for all x,y∈S, which is
(33)|f(x)g(y)-f(y)g(x)|≤2ϵ
for all x,y∈S, since g(x)=g(σ(x)) for all x∈S. Dividing both sides of (33) by |g(y)| we have
(34)|f(x)-α(y)g(x)|≤2ϵ|g(y)|
for all y∈J:={y:g(y)≠0}, where α(y)=f(y)/g(y). Since f is unbounded, we have α(y)≠0 for all y∈J and hence g is unbounded. Replacing y by y1∈J and y by y2∈J in (34) we have two inequalities. Using the triangle inequality with these two resulting inequalities we have
(35)|α(y1)-α(y2)||g(x)|≤2ϵ(1|g(y1)|+1|g(y2)|)
for all x∈S. Since g is unbounded, from (35) we have α(y1)=α(y2) and hence α(y):=α is independent of y∈J. From (34) we have
(36)|f(x)-αg(x)|≤2ϵ|g(y)|
for all x∈S and y∈J. Since g is unbounded we have
(37)f(x)=αg(x)=αg(0)mσ(x)
for all x∈S. Putting (37) in (15) we have
(38)|αg(0)(1-g(0))||mσ(x)mσ(y)|≤ϵ
for all x,y∈S. Since α≠0, g(0)≠0, and mσ is unbounded, we have g(0)=1. Thus, we get (18). This completes the proof.Remark 2.
We have no idea if the caseg(x)=-mσ(x) in (17) occurs or not. For example, it can be verified that if there exists a sequence xn∈S, n=1,2,3,…, such that (1+ϕ(xn))/|g(xn)|→0 as n→∞, the case g(x)=-mσ(x) in (17) does not occur; in particular (g,f) satisfies (18).Remark 3.
In general, the inequality (16) cannot be replaced by the weaker inequality:
(39)|1-g(x)|≤ϵMf
for all x∈S. Indeed, let ϕ(x)=ϵ>0, σ(y)=y, f(x)=ϵsinx, g(x)=cos⁡x for all x,y∈S. Then we have
(40)|f(x+y)-f(x)g(y)|=|ϵcos⁡xsiny|≤ϵ
for all x,y∈S. However, the inequality (39) fails since
(41)|1-g(π)|=|1-cos⁡π|=2>1=ϵMf.As a direct consequence of Theorem1 we have the following.Corollary 4.
Letf:S→ℂ satisfy
(42)|f(x+σ(y))-f(x)f(y)|≤ϕ(x)
for all x,y∈S. If S is a group and f is bounded, then
(43)|f(y)||1-|f(x)||≤ϕ(x)
for all x,y∈S. If f is unbounded, then f is σ-exponential.Proof.
The inequality (16) implies (43). Now, by Theorem 1, we have
(44)f(x)=f(0)mσ(x),x∈S
for some σ-exponential function mσ. Putting (44) to (42) and letting x=0 we have
(45)|f(0)(1-f(0))||mσ(y)|≤ϕ(0)
for all y∈S. Since f is unbounded, from (45) we have f(0)=1. This completes the proof.Theorem 5.
Letf,g:S→ℂ satisfy
(46)|f(x+σ(y))-f(x)g(y)|≤ϕ(y)
for all x,y∈S. If S is a group and f is bounded, then
(47)|1-|g(x)||≤ϕ(x)Mf
for all x∈S, where Mf=sup⁡x∈S|f(x)|. If f is unbounded, then there exists an exponential function m:S→ℂ such that
(48)g(x)=m(x),|f(x)-f(0)m(σ(x))|≤ϕ(σ(x))
for all x∈S. In particular, if ϕ(x)=ϵ≥0 for all x∈S, then we have
(49)g(x)=m(x),f(x)=f(0)m(σ(x))
for all x∈S.Proof.
Using the same method as in the proof of Theorem1, we can show that if f is bounded and S is a group, then f satisfies (47). Assume that f is unbounded. Using the triangle inequality and (46) we have
(50)|f(z)||g(x)g(y)-g(y+x)|≤|f(z)g(x)g(y)-f(z+σ(x))g(y)|+|f(z+σ(x))g(y)-f(z+σ(x)+σ(y))|+|f(z+σ(y+x))-f(z)g(y+x)|≤ϕ(x)|g(y)|+ϕ(y)+ϕ(x+y)
for all x,y,z∈S. Since f is unbounded, it follows from (50) that
(51)g(x+y)=g(x)g(y)
for all x,y∈S. Therefore, g is an exponential function: say g=m. Putting x=0 and replacing y by σ(x) in (46) we get (48). Assume that ϕ(y)=ϵ for all y∈S. Replacing (x,y) by (x,σ(y)) and (x,y) by (y,σ(x)) in (46) using triangle inequality with the resulting two inequalities we have
(52)|f(x)g(σ(y))-f(y)g(σ(x))|≤2ϵ
for all x,y∈S. Now, using the same method as in the proof of Theorem 1 (after the inequality (33)) we can show that
(53)f(x)=αg(σ(x))
for all x∈S. Since g is an exponential function, we have g(σ(0))=g(0)=1 and hence α=f(0). This completes the proof.As a direct consequence of Theorem5 we have the following.Corollary 6.
Letf:S→ℂ satisfy
(54)|f(x+σ(y))-f(x)f(y)|≤ϕ(y)
for all x,y∈S. If S is a group and f is bounded, then
(55)|f(y)||1-|f(x)||≤ϕ(x)
for all x,y∈S. If f is unbounded, then f is σ-exponential.Proof.
Iff is bounded, the inequality (55) follows from (47). If f is unbounded, then, by Theorem 5, f is exponential. Thus, from (54) we have
(56)|f(x)||f(σ(y))-f(y)|≤ϕ(y)
for all x,y∈S. Since f is unbounded, it follows from (56) that f(y)=f(σ(y)) for all y∈S. This completes the proof.Theorem 7.
Letf,g:S→ℂ satisfy
(57)|f(x+σ(y))-g(x)f(y)|≤ϕ(y)
for all x,y∈S. If f(0)=0, then
(58)|f(x)|≤ϕ(0),|g(x)|≤inf⁡y∈Sϕ(y)+ϕ(0)|f(y)|
for all x∈S. If f(0)≠0, S is a group, and g is bounded, then
(59)||f(0)|-|f(x)||≤ϕ(x)+ϕ(0)Mg
for all x∈S, where Mg=sup⁡x∈S|g(x)|. If f(0)≠0 and g is unbounded, then there exists an unbounded σ-exponential function mσ:S→ℂ such that
(60)f(x)=f(0)mσ(x),|g(x)-mσ(x)|≤ϕ(0)|f(0)|
for all x∈S.Proof.
Puttingy=0 in (57) we have
(61)|f(x)-f(0)g(x)|≤ϕ(0)
for all x∈S. If f(0)=0, then we have
(62)|f(x)|≤ϕ(0)
for all x∈S. Using the triangle inequality with (57) and using (62) we have
(63)|g(x)f(y)|≤ϕ(y)+ϕ(0)
for all x,y∈S. Dividing (63) by |f(y)| and taking infimum of the right-hand side of the result we get
(64)|g(x)|≤inf⁡y∈Sϕ(y)+ϕ(0)|f(y)|
for all x∈S, which gives (58). If f(0)≠0, using the triangle inequality with (57), using (61), and dividing the result by |f(0)|, we have
(65)|g(x+σ(y))-g(x)f0(y)|≤ϕ0(y)
for all x,y∈S, where f0(y)=f(y)/f(0), ϕ0(y)=(ϕ(y)+ϕ(0))/|f(0)|. If g is bounded and S is a group, then by Theorem 5, f0 satisfies
(66)|1-|f0(x)||≤ϕ0(x)Mg
for all x∈S, where Mg=sup⁡x∈S|g(x)|. Multiplying (66) by |f(0)| we have
(67)||f(0)|-|f(x)||≤ϕ(x)+ϕ(0)Mg
for all x∈S. If g is unbounded, then by Theorem 5, f0 is exponential function; say f0=m. Putting y=0 in (57) and dividing the result by |f(0)| we have
(68)|g(x)-m(x)|≤ϕ(0)|f(0)|
for all x∈S. Multiplying (68) by |f(y)|,
(69)|g(x)f(y)-m(x)f(y)|≤ϕ(0)|f(y)||f(0)|
for all x,y∈S. Using the triangle inequality with (57) and (69) we have
(70)|f(0)m(x)||m(σ(y))-m(y)|=|f(x+σ(y))-m(x)f(y)|≤|f(x+σ(y))-g(x)f(y)|+|g(x)f(y)-m(x)f(y)|≤ϕ(y)+ϕ(0)|f(y)||f(0)|
for all x,y∈S. Since m is unbounded, it follows from (70) that m(σ(y))=m(y) for all y∈S. This completes the proof.Theorem 8.
Letf,g:S→ℂ satisfy
(71)|f(x+σ(y))-f(x)g(y)|≤ϕ(x)
for all x,y∈S. If f(0)=0, then
(72)|f(x)|≤ϕ(0),|g(x)|≤inf⁡y∈Sϕ(y)+ϕ(0)|f(y)|
for all x∈S. If f(0)≠0, S is a group, and g is bounded, then
(73)||f(0)|-|f(x)||≤ϕ(x)+ϕ(0)Mg
for all x∈S, where Mg=sup⁡x∈S|g(x)|. If f(0)≠0 and g is unbounded, then there exists an unbounded exponential m:S→ℂ such that
(74)f(x)=f(0)m(x),|g(x)-m(σ(x))|≤ϕ(0)|f(0)|
for all x∈S.Proof.
Puttingx=0 in (71) we have
(75)|f(σ(y))-f(0)g(y)|≤ϕ(0)
for all y∈S. If f(0)=0, then replacing y by σ(y) in (75) we have
(76)|f(y)|≤ϕ(0)
for all y∈S. Using the triangle inequality with (71) and using (76) we have
(77)|f(x)g(y)|≤ϕ(x)+ϕ(0)
for all x,y∈S. Dividing (77) by |f(x)| and taking infimum of the right-hand side of the result we have
(78)|g(y)|≤inf⁡x∈Sϕ(x)+ϕ(0)|f(x)|
for all x∈S, which gives (72). If f(0)≠0, using the triangle inequality with (71) and (75), replacing (x,y) by (y,x), and dividing the result by |f(0)|, we have
(79)|g(x+σ(y))-g(x)f0(y)|≤ϕ0(y)
for all x,y∈S, where f0(y)=f(y)/f(0), ϕ0(y)=(ϕ(y)+ϕ(0))/|f(0)|. If g is bounded and S is a group, then by Theorem 5 we get (73). If g is unbounded, then by Theorem 5, f0 is exponential; say f0=m. Putting x=0 in (71), replacing y by σ(x) in the result, and dividing the result by |f(0)| we get (74). This completes the proof.Finally, we consider the functional inequality with three unknown functionsf,g,h:S→ℂ:
(80)|f(x+σ(y))-g(x)h(y)|≤ϵ
for all x,y∈S. The inequality (80) is no more than the inequality
(81)|f(x+y)-g(x)h(y)|≤ϵ
for all x,y∈S, since, if we replace y by σ(y) in (80) and h by h∘σ, inequality (80) is reduced to (81).In the following we exclude the trivial cases wheng=0 or h=0.Theorem 9.
Letf,g,h:S→ℂ satisfy the inequality (81). If g or h is unbounded, then there exists an unbounded exponential function m:S→ℂ such that
(82)g(x)=g(0)m(x),h(x)=h(0)m(x),|f(x)-g(0)h(0)m(x)|≤ϵ
for all x∈S. If S is a group and g or h is bounded, then
(83)||h(x)|-|h(0)||≤2ϵMg,||g(x)|-|g(0)||≤2ϵMh,||f(x)|-|g(0)h(0)||≤3ϵ
for all x∈S, where Mg=sup⁡x∈S|g(x)|, Mh=sup⁡x∈S|h(x)|.Proof.
Replacing(x,y) by (y,x) in (81) and using the triangle inequality with the result and (81) we have
(84)|g(x)h(y)-h(x)g(y)|≤2ϵ
for all x,y∈S. Since we exclude the trivial cases when g=0 or h=0, it follows from (84) that g is bounded if and only if h is bounded. If h(0)=0, putting y=0 in (81), and if h(0)=0, putting y=0 in (81), we have |f(x)|≤ϵ for all x∈S. Using the triangle inequality with (81) we have
(85)|g(x)h(y)|≤|f(x+y)|+ϵ≤2ϵ
for all x,y∈S. Thus, we have
(86)MgMh≤2ϵ.
Assume that g or h is unbounded. From (86) we have g(0)≠0, h(0)≠0. Putting y=0 and replacing x by x+y in (81) we have
(87)|f(x+y)-g(x+y)h(0)|≤ϵ
for all x,y∈S. Using the triangle inequality with (81) and (87) and dividing the result by |h(0)| we have
(88)|g(x+y)-g(x)h0(y)|≤2ϵ|h(0)|
for all x,y∈S, where h0(x)=h(x)/h(0). Replacing y by σ(y) in (87) and using Theorem 5 we get
(89)h0(x)=m(x),g(x)=g(0)m(x),∀x∈S
for some exponential function m. Putting (89) in (81) we obtain (82). Now, we assume that g or h is bounded. If h(0)≠0, then using Theorem 5 with (88) we have
(90)||h(x)|-|h(0)||≤2ϵMg
for all x∈S. If g(0)=0, from (86) we have
(91)|g(x)|≤2ϵMh
for all x∈S. Thus, h satisfies (90) for both cases h(0)=0 and h(0)≠0. Changing the role of g and h we get
(92)||g(x)|-|g(0)||≤2ϵMh
for all x∈S. Putting y=0 in (81) and using the triangle inequality we have
(93)||f(x)|-|h(0)g(x)||≤ϵ
for all x∈S. Multiplying |h(0)| in (92) we have
(94)||h(0)g(x)|-|g(0)h(0)||≤2ϵ|h(0)|Mh≤2ϵ
for all x∈S. From (92) and (93), using the triangle inequality we have
(95)||f(x)|-|g(0)h(0)||≤3ϵ
for all x∈S. This completes the proof.


## 3. Bounded Solutions

Throughout this section, we assume thatG is a group and σ:S→S an involution. We describe bounded functions satisfying following functional inequalities:
(96)|f(x+σ(y))-f(x)f(y)|≤ϕ(x),(97)|f(x+σ(y))-f(x)f(y)|≤ϕ(y)
for all x,y∈G and for some ϕ:G→ℝ+.Theorem 10.
Letf:G→ℂ be bounded functions satisfying (96) or (97). Then f satisfies
(98)|f(x)|≤12(1+1+4ϕ(x))
for all x∈G. Let K:={x∈G:ϕ(x)<1/4}. Then, either f satisfies
(99)12(1+1-4ϕ(x))≤|f(x)|≤12(1+1+4ϕ(x))
for all x∈K, or else
(100)|f(x)|≤12(1-1-4ϕ(x))
for all x∈K.Proof.
By Corollaries4 and 6, every bounded solution of (96) and (97) satisfies
(101)|f(y)(1-|f(x)|)|≤ϕ(x)
for all x,y∈G. Letting y=x in (101) we have
(102)-ϕ(x)≤|f(x)|(|f(x)|-1)≤ϕ(x)
for all x∈G. Solving the inequality (102) we get (98). Also from (102), for each x∈K, we have
(103)12(1+1-4ϕ(x))≤|f(x)|≤12(1+1+4ϕ(x)),
or
(104)|f(x)|≤12(1-1-4ϕ(x)).
Assume that there exist x1,x2∈K such that
(105)|f(x1)|≤12(1-1-4ϕ(x1)),|f(x2)|≥12(1+1-4ϕ(x2)).
Putting y=x2, x=x1 in (101) we have the contradiction
(106)14<|f(x2)(1-|f(x1)|)|≤ϕ(x2)<14.
Thus, f satisfies (103) for all x∈K or (104) for all x∈K. This completes the proof.Defineσ:ℝ2→ℝ2 by σ(x,y)=(y,x) for all x,y∈ℝ. Then it is easy to see that every σ-exponential mσ:ℝ2→ℂ has the form mσ(x,y)=m(x+y), x,y∈ℝ for some exponential m:ℝ→ℂ. Thus, by Corollary 4 and Theorem 10, we have the following.Example 11.
Letf:ℝ2→ℂ satisfy
(107)|f(x+v,y+u)-f(x,y)f(u,v)|≤14(u2+v2)
for all x,y,u,v∈ℝ. Then, if f is unbounded function, then f has the form
(108)f(x,y)=m(x+y)
for all x,y∈ℝ, where m:ℝ→ℂ is an exponential function; if f is bounded, then f satisfies
(109)|f(x,y)|≤12(1+1+1(x2+y2))
for all x,y∈ℝ. Let K={(x,y):x2+y2>1}. Then we have
(110)12(1+1-1(x2+y2))≤|f(x,y)|≤12(1+1+1(x2+y2))
for all x,y∈K, or
(111)|f(x,y)|≤12(1-1-1(x2+y2))
for all x,y∈K.Corollary 12.
Letϵ<1/4 and f:G→ℂ be a bounded function satisfying
(112)|f(x+σ(y))-f(x)f(y)|≤ϵ
for all x,y∈G. Then either f satisfies
(113)12(1+1-4ϵ)≤|f(x)|≤12(1+1+4ϵ)
for all x∈G, or else
(114)|f(x)|≤12(1-1-4ϵ)
for all x∈G.In particular, ifG is 2-divisible and σ(x)=x for all x∈G, then we obtain the following result which is a generalized version of the result of Albert and Baker [3].Corollary 13.
LetS be 2-divisible, ϵ<1/4, and let f:G→ℝ be bounded functions satisfying
(115)|f(x+y)-f(x)f(y)|≤ϵ
for all x,y∈G. Then either f satisfies
(116)12(1+1-4ϵ)≤f(x)≤12(1+1+4ϵ)
for all x∈G, or else
(117)-ϵ≤f(x)≤12(1-1-4ϵ)
for all x∈G.Proof.
Replacingy by x in (115) and using the triangle inequality with the result we have f(2x)≥f(x)2-ϵ≥-ϵ for all x∈G, which implies that f(x)≥-ϵ for all x∈G. Since ϵ<(1/2)(1-1-4ϵ), inequality (113) implies (116), and (114) implies (117). This completes the proof.


## 4. Applications

In this section we consider the stability of (8) and (9) which were dealt with in [13]. Let ℍ={a+bi+cj+dk∣a,b,c,d∈ℝ} be the quaternion group. Recall that i2=j2=k2=-1, ij=k, jk=i, ki=j, ji=-k, kj=-i, ik=-j, and the conjugate of q=a+bi+cj+dk∈ℍ is given by q*=a-bi-cj-dk. We denote ∥q∥=qq*=a2+b2+c2+d2. We first consider unbounded functions f:ℝ2→ℝ satisfying (10) and (11) and unbounded functions g:ℝ4→ℝ satisfying (12) and (13).Theorem 14.
Letf:ℝ2→ℝ be an unbounded function satisfying (10). Then f has the form
(118)f(x,y)=m(x2+y2)
for all x,y∈ℝ.Proof.
LetF(x+yi)=f(x,y) and Φ(x+yi)=ϕ(x,y) for all x,y∈ℝ. Then the functional inequality (10) is converted to
(119)|F(z¯w)-F(z)F(w)|≤Φ(w)
for all z=x2+iy2, w=x1+iy1∈ℂ. Letting σ(z)=z¯ and using Corollary 4 when S=〈ℂ,·〉 we have F(z)=mσ(z) for all z∈ℂ. In view of (14), mσ(z) is written in the form mσ(z)=m(|z|) for all z∈ℂ. Thus, we get (118). This completes the proof.Using Corollary6 we obtain the following.Theorem 15.
Letf:ℝ2→ℝ be an unbounded function satisfying (11). Then f has the form
(120)f(x,y)=m(x2+y2)
for all x,y∈ℝ.Theorem 16.
Letg:ℝ4→ℝ be an unbounded function satisfying (12). Then g has the form
(121)g(x,y,u,v)=m(x2+y2+u2+v2)
for all x,y,u,v∈ℝ.Proof.
LetF(x+yi+uj+vk)=g(x,y,u,v) and Ψ(x+yi+uj+vk)=ψ(x,y,u,v) for all x,y,u,v∈ℝ. Then the functional inequality (12) is reduced to
(122)|F(q)F(p)-F(qp*)|≤Ψ(q)
for all p,q∈ℍ. Letting σ(q)=q* and using Corollary 4 when S=〈ℍ,·〉 we have that F(q)=mσ(q) for all q∈ℍ. In view of (14), mσ(q) is written in the form mσ(q)=m(∥q∥) for all q∈ℍ. Thus, we get (121). This completes the proof.Using Corollary6 we obtain the following.Theorem 17.
Letg:ℝ4→ℝ be an unbounded function satisfying (13). Then g has the form
(123)g(x,y,u,v)=m(x2+y2+u2+v2)
for all x,y,u,v∈ℝ.Using the results in Section3 we can also obtain bounded functions f:ℝ2→ℝ satisfying the inequalities (10), (11) and bounded functions g:ℝ4→ℝ satisfying (12) and (13) as in [13].



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