# Ulam’s Type Stability of Involutional-Exponential Functional Equations

Authors: Jaeyoung Chung
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

LetS be a commutative semigroup, f
,
g
:
S
→
C and σ
:
S
→
S an involution. In this paper we consider the stability of involution-exponential functional equations f
x
+
σ
y
-
g
x
f
y
≤
ϕ
x
resp
.,
ϕ
y
,
|
f
(
x
+
σ
y
)
-
f
(
x
)
g
(
y
)
|
≤
ϕ
(
x
)
[
resp
.,
ϕ
(
y
)
] for all x
,
y
∈
S, where ϕ
:
S
→
R
+ satisfies the growth condition: there exists C
>
1 such that lim
k
→
∞
C
-
k
ϕ
(
k
x
)
=
0 for each x
∈
S. We also consider the stability of L
∞-version |
f
(
x
+
σ
y
)
-
f
(
x
)
f
(
y
)
|
L
∞
(
R
2
n
)
≤
ϵ
, where f
:
R
n
→
C is a locally integrable function.


---

## Body

## 1. Introduction

Throughout this paper we denote byS, R
,
R
+
,
C
,
R
n, a commutative semigroup with an identity element, the set of real numbers, nonnegative real numbers, complex numbers, and the n-dimensional Euclidean space, respectively, and R
0
n
=
R
n
∖
{
0
}
,
C
0
=
C
∖
{
0
}, 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. An exponential function m
:
S
→
C is called σ
-exponential if m satisfies m
(
σ
x
)
=
m
(
x
) for all x
∈
S and denote by m
σ a σ-exponential function.In [1], the following functional inequalities with involution are investigated:
(1)
|
f
(
x
+
σ
y
)
-
g
(
x
)
f
(
y
)
|
≤
ϕ
(
x
)
[
resp
.
ϕ
(
y
)
]
,
∀
x
,
y
∈
S
,
(2)
|
f
(
x
+
σ
y
)
-
f
(
x
)
g
(
y
)
|
≤
ϕ
(
x
)
[
resp
.
ϕ
(
y
)
]
,
∀
x
,
y
∈
S
.
As a result, all unbounded functions f
,
g satisfying the inequalities (1) and (2) are exactly described only when ϕ is a constant function while only one of unbounded functions f
,
g satisfying each of (1) and (2) is exactly described when ϕ is an arbitrary unbounded function.In this paper we investigate the functional inequalities (1) and (2) by imposing some growth conditions on ϕ, f, or g. First, we introduce the condition on h
:
S
→
C:
(3)
inf
⁡
x
∈
S
1
+
ϕ
(
x
)
|
h
(
x
)
|
=
0
,
where h will stand for f and g.Secondly, we introduce the condition onϕ; there exists C
>
1 such that
(4)
lim
⁡
k
→
∞
C
-
k
ϕ
(
k
x
)
=
0
for all x
∈
S.As a result, we completely determinef and g satisfying each of the inequalities (1) and (2): if g satisfies (3) [resp., f satisfies (3)] or ϕ satisfies (4), then (
f
,
g
) satisfying (1) [resp., (2)] are of the form
(5)
f
(
x
)
=
f
(
0
)
m
σ
(
x
)
,
g
(
x
)
=
m
σ
(
x
)
×
[
resp
.
f
(
x
)
=
f
(
0
)
m
(
x
)
,
g
(
x
)
=
m
(
σ
x
)
]
for all x
∈
S, where m
σ is a σ-exponential function and m is an exponential function.As an application of our result, we determine all unbounded functionsf
,
g
:
R
0
2
→
R satisfying the functional inequalities
(6)
|
f
(
u
x
+
v
y
,
u
y
-
v
x
)
-
g
(
x
,
y
)
f
(
u
,
v
)
|
≤
ϕ
(
x
,
y
)
[
ϕ
(
u
,
v
)
]
,
(7)
|
f
(
u
x
+
v
y
,
u
y
-
v
x
)
-
f
(
x
,
y
)
g
(
u
,
v
)
|
≤
ϕ
(
x
,
y
)
[
ϕ
(
u
,
v
)
]
for all (
x
,
y
)
,
(
u
,
v
)
∈
R
0
2, where f
,
g satisfy (3) or ϕ
:
R
0
2
→
R
+ satisfies (4) (see [2–5] for related equations) and determine all unbounded functions f
,
g
:
R
0
4
→
R satisfying the functional inequalities
(8)
|
f
(
x
1
,
y
1
,
u
1
,
v
1
)
g
(
x
2
,
y
2
,
u
2
,
v
2
)
-
f
(
x
1
x
2
+
y
1
y
2
+
u
1
u
2
+
v
1
v
2
,
x
1
y
2
-
y
1
x
2
+
u
1
v
2
-
v
1
u
2
,
x
1
u
2
-
y
1
v
2
-
u
1
x
2
+
v
1
y
2
,
x
1
v
2
+
y
1
u
2
-
u
1
y
2
-
v
1
x
2
)
|
≤
ψ
(
x
1
,
y
1
,
u
1
,
v
1
)
[
ψ
(
x
2
,
y
2
,
u
2
,
v
2
)
]
,
for all x
1, y
1, u
1, v
1, x
2, y
2, u
2, and v
2
∈
R, where f
,
g satisfy (3) or ψ
:
R
0
4
→
R
+ satisfies (4) (see [2, 4] for related equations). Finally, we consider the stability of  L
∞-version
(9)
|
f
(
x
+
σ
y
)
-
f
(
x
)
f
(
y
)
|
L
∞
(
R
2
n
)
≤
ϵ
,
where f
:
R
n
→
C is a locally integrable function. As a result, we prove that every unbounded solution f (i.e., ∥
f
∥
L
∞
(
R
n
)
=
∞) of (9) satisfies
(10)
f
(
x
)
=
m
(
x
+
σ
x
2
)
for almost every x
∈
R
n, where m
:
R
n
→
C is an unbounded exponential function. Every bounded solution f (i.e.,∥
f
∥
L
∞
(
R
n
)
<
∞) satisfies
(11)
∥
f
∥
L
∞
(
R
n
)
≤
1
2
(
1
+
1
+
4
ϵ
)
;
If  ϵ
<
1
/
4, then f satisfies either
(12)
1
2
(
1
+
1
-
4
ϵ
)
≤
∥
f
∥
L
∞
(
R
n
)
≤
1
2
(
1
+
1
+
4
ϵ
)
or
(13)
∥
f
∥
L
∞
(
R
n
)
≤
1
2
(
1
-
1
-
4
ϵ
)
.We refer the reader to [1, 6–16] for related functional equations and their stabilities. We also refer the reader to [17–19] for some recent developments on the issues of stability and superstability for functional equations.


## 2. Stability of (1) and (2)

In this section we investigate unbounded functionsf
,
g satisfying (1) and (2) when some of f and g satisfy (3) or ϕ satisfies (4). For bounded solutions of (1) and (2) we refer the reader to [1].Lemma 1.
Assume thatm
:
S
→
C is an unbounded exponential function and ϕ
:
S
→
R
+ satisfies (4). Then m satisfies (3).Proof.
Sincem is unbounded, we can choose a x
0
∈
S such that |
m
(
x
0
)
|
≥
C, where C
>
1 is the constant in (4). Since ϕ satisfies (4) we have
(14)
lim
⁡
k
→
∞
1
+
ϕ
(
k
x
0
)
|
m
(
k
x
0
)
|
=
lim
⁡
k
→
∞
1
+
ϕ
(
k
x
0
)
|
m
(
x
0
)
|
k
≤
lim
⁡
k
→
∞
(
C
-
k
+
C
-
k
ϕ
(
k
x
0
)
)
=
0
.
This completes the proof.Theorem 2.
Letf
,
g
:
S
→
C be unbounded functions satisfying
(15)
|
f
(
x
+
σ
y
)
-
g
(
x
)
f
(
y
)
|
≤
ϕ
(
x
)
for all x
,
y
∈
S. Then g is a σ-exponential function. In particular if g satisfies (3) or ϕ satisfies (4), then there exists a σ-exponential function m
σ
:
S
→
C such that
(16)
f
(
x
)
=
f
(
0
)
m
σ
(
x
)
,
g
(
x
)
=
m
σ
(
x
)
for all x
∈
S.Proof.
Choosing a sequencey
n
∈
S
,
n
=
1,2
,
3
,
…, such that |
f
(
y
n
)
|
→
∞ as n
→
∞, putting y
=
y
n
,
n
=
1,2
,
3
,
…, in (15), dividing the result by |
f
(
y
n
)
|, and letting n
→
∞ we have
(17)
g
(
x
)
=
lim
⁡
n
→
∞
f
(
x
+
σ
y
n
)
f
(
y
n
)
for all x
∈
S. Putting x
=
0 in (15) we have
(18)
|
f
(
σ
y
)
-
g
(
0
)
f
(
y
)
|
≤
ϕ
(
0
)
for all y
∈
S. Multiplying both sides of (17) by g
(
y
) and using (15), (17), and (18) we have
(19)
g
(
y
)
g
(
x
)
=
lim
⁡
n
→
∞
g
(
y
)
f
(
x
+
σ
y
n
)
f
(
y
n
)
=
lim
⁡
n
→
∞
f
(
y
+
σ
x
+
y
n
)
f
(
y
n
)
=
lim
⁡
n
→
∞
g
(
0
)
f
(
x
+
σ
y
+
σ
y
n
)
f
(
y
n
)
=
g
(
0
)
g
(
x
+
σ
y
)
for all x
,
y
∈
S. Dividing (19) by g
(
0
)
2 we have
(20)
g
0
(
x
)
g
0
(
y
)
=
g
0
(
x
+
σ
y
)
for all x
,
y
∈
S, where g
0
(
x
)
=
g
(
x
)
/
g
(
0
). From (20) we have
(21)
g
(
x
)
=
g
(
0
)
m
σ
(
x
)
for some σ-exponential m
σ. If g satisfies (3) or ϕ satisfies (4), then, by Lemma 1, we can choose a sequence x
n
∈
S
,
n
=
1,2
,
3
,
…, such that (
1
+
ϕ
(
x
n
)
)
/
|
g
(
x
n
)
|
→
0 as n
→
∞. Putting x
=
x
n
,
n
=
1,2
,
3
,
…, in (15), dividing the result by |
g
(
x
n
)
|, and letting n
→
∞ we have
(22)
f
(
y
)
=
lim
⁡
n
→
∞
f
(
x
n
+
σ
y
)
g
(
x
n
)
for all y
∈
S. Multiplying both sides of (22) by g
(
x
) and using (15), (18), and (22) we have
(23)
g
(
x
)
f
(
y
)
=
lim
⁡
n
→
∞
g
(
x
)
f
(
x
n
+
σ
y
)
g
(
x
n
)
=
lim
⁡
n
→
∞
f
(
x
+
σ
x
n
+
y
)
g
(
x
n
)
=
lim
⁡
n
→
∞
g
(
0
)
f
(
σ
x
+
σ
y
+
x
n
)
g
(
x
n
)
=
g
(
0
)
f
(
x
+
y
)
for all x
,
y
∈
S. Putting y
=
0 in (23) and dividing the result by g
(
0
) we have
(24)
f
(
x
)
=
f
(
0
)
g
0
(
x
)
=
f
(
0
)
m
σ
(
x
)
for all x
∈
S. Putting x
=
0 in (15) and using (24) we have
(25)
|
f
(
0
)
(
1
-
g
(
0
)
)
|
|
m
σ
(
y
)
|
≤
ϕ
(
0
)
for all y
∈
S. Since m
σ is unbounded, from (25) we have g
(
0
)
=
1. Now, from (21) and (24) we get (16). This completes the proof.We denote byc
·
x the inner product of c
=
(
c
1
,
c
2
,
…
,
c
n
)
∈
C
n and x
=
(
x
1
,
x
2
,
…
,
x
n
)
∈
R
n which is defined as c
·
x
=
∑
j
=
1
n
‍
c
j
x
j, and R
c
=
(
R
c
1
,
…
,
R
c
n
), where R
c
j are the real parts of c
j
,
j
=
1,2
,
…
,
n. 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 2
y
=
x), then m
σ is σ-exponential if and only if
(26)
m
σ
(
x
)
=
m
(
x
+
σ
x
2
)
,
x
∈
S
for some exponential function m
:
S
→
C.Corollary 3.
LetP
(
x
)
,
x
∈
R
n, be a polynomial. Suppose that f
,
g
:
R
n
→
C are unbounded function satisfying
(27)
|
f
(
x
+
σ
y
)
-
g
(
x
)
f
(
y
)
|
≤
|
P
(
x
)
|
for all x
,
y
∈
R
n. Then there exists a σ-exponential function m
σ
:
R
n
→
C such that
(28)
f
(
x
)
=
f
(
0
)
m
σ
(
x
)
,
g
(
x
)
=
m
σ
(
x
)
for all x
∈
R
n. In particular if g is continuous, then there exists c
∈
C
n
,
R
(
c
+
c
σ
)
≠
0 such that
(29)
f
(
x
)
=
f
(
0
)
e
(
1
/
2
)
(
c
+
c
σ
)
·
x
,
g
(
x
)
=
e
(
1
/
2
)
(
c
+
c
σ
)
·
x
for all x
∈
R
n.Proof.
It is easy to see that|
P
(
x
)
| satisfies (4). Thus, by Theorem 2 we get (28). Assume that g is continuous. It is well known that every continuous exponential functional m
:
R
n
→
C is given by m
(
x
)
=
e
c
·
x for some c
∈
C
n. Thus, from (26) we have m
σ
(
x
)
=
e
(
1
/
2
)
(
c
+
c
σ
)
·
x for all x
∈
R
n, where c
σ denotes matrix multiplication. Thus, we get (29). This completes the proof.Remark 4.
Leta
,
b
∈
R
n be two nonzero vectors that are not parallel; that is, b
≠
r
a for all r
∈
R. Then, the hyperplane b
·
x
=
0 is not parallel to (
b
-
a
)
·
x
=
0 and hence there exists x
0
∈
R
n such that b
·
x
0
>
0 and (
b
-
a
)
·
x
0
>
0. If b
=
t
a for some t
∈
R, then there exists x
0
∈
R
n such that b
·
x
0
>
0 and (
b
-
a
)
·
x
0
>
0 if and only if t
>
1. Thus, if b
≠
t
a for all t
≤
1, then there exists x
0
∈
R
n such that b
·
x
0
>
0 and (
b
-
a
)
·
x
0
>
0.Corollary 5.
Letγ
∈
R
n be fixed. Suppose that f
,
g
:
R
n
→
C are unbounded continuous function satisfying
(30)
|
f
(
x
+
σ
y
)
-
g
(
x
)
f
(
y
)
|
≤
e
γ
·
x
for all x
,
y
∈
R
n. Then there exists c
∈
C
n
,
R
(
c
+
c
σ
)
≠
0 such that g
(
x
)
=
e
(
1
/
2
)
(
c
+
c
σ
)
·
x for all x
∈
R
n. If (
1
/
2
)
R
(
c
+
c
σ
)
≠
t
γ for all t
≤
1, then one has
(31)
f
(
x
)
=
f
(
0
)
e
(
1
/
2
)
(
c
+
c
σ
)
·
x
,
g
(
x
)
=
e
(
1
/
2
)
(
c
+
c
σ
)
·
x
for all x
∈
R
n.Proof.
Recall that every continuousσ-exponential functional m
σ
:
R
n
→
C is given by
(32)
m
σ
(
x
)
=
e
(
1
/
2
)
(
c
+
c
σ
)
·
x
for all x
∈
R
n, where c
σ denotes matrix multiplication. If (
1
/
2
)
R
(
c
+
c
σ
)
≠
t
γ for all t
≤
1, then by Remark 4 there exists x
0
∈
R
n such that
(33)
1
2
(
c
+
c
σ
)
·
x
0
>
0
,
1
2
(
c
+
c
σ
)
·
x
0
>
γ
·
x
0
.
From (32) and (33) we have
(34)
lim
⁡
k
→
∞
1
+
e
γ
·
k
x
0
g
(
k
x
0
)
=
1
+
e
γ
·
k
x
0
e
(
1
/
2
)
(
c
+
c
σ
)
·
k
x
0
=
0
,
which implies that g satisfies the condition (3). Thus, we get (31). This completes the proof.Theorem 6.
Letf
,
g
:
S
→
C be unbounded functions satisfying
(35)
|
f
(
x
+
σ
y
)
-
g
(
x
)
f
(
y
)
|
≤
ϕ
(
y
)
for all x
,
y
∈
S. Then there exists an unbounded σ-exponential function m
σ
:
S
→
C such that f
(
x
)
=
f
(
0
)
m
σ
(
x
) for all x
∈
S. In particular if f satisfies (3) or ϕ satisfies (4), then one has
(36)
f
(
x
)
=
f
(
0
)
m
σ
(
x
)
,
g
(
x
)
=
m
σ
(
x
)
for all x
∈
S.Proof.
Puttingy
=
0 in (35) we have
(37)
|
f
(
x
)
-
f
(
0
)
g
(
x
)
|
≤
ϕ
(
0
)
for all x
∈
S. Choose a sequence x
n
∈
S
,
n
=
1,2
,
3
,
…, such that |
g
(
x
n
)
|
→
∞ as n
→
∞. Putting x
=
x
n
,
n
=
1,2
,
3
,
…, in (35), dividing the result by |
g
(
x
n
)
|, letting n
→
∞, and using (37) we have
(38)
f
(
y
)
=
lim
⁡
n
→
∞
f
(
x
n
+
σ
y
)
g
(
x
n
)
=
lim
⁡
n
→
∞
f
(
0
)
g
(
x
n
+
σ
y
)
g
(
x
n
)
for all y
∈
S. Multiplying both sides of (38) by f
(
x
) and using (35) and (38) we have
(39)
f
(
y
)
f
(
x
)
=
lim
⁡
n
→
∞
f
(
0
)
g
(
x
n
+
σ
y
)
f
(
x
)
g
(
x
n
)
=
lim
⁡
n
→
∞
f
(
0
)
f
(
x
n
+
σ
y
+
σ
x
)
g
(
x
n
)
=
f
(
0
)
f
(
x
+
y
)
for all x
,
y
∈
S. From (39) we have f
0
(
x
)
∶
=
f
(
x
)
/
f
(
0
) is an exponential function, say m. Now, from (37) we can write
(40)
f
(
x
)
=
f
(
0
)
m
(
x
)
,
g
(
x
)
=
m
(
x
)
+
r
(
x
)
for all x
∈
S, where |
r
(
x
)
|
≤
ϕ
(
0
)
/
|
f
(
0
)
| for all x
∈
S. Putting (40) in (35) and using the triangle inequality we have
(41)
|
f
(
0
)
m
(
x
)
|
|
m
(
σ
y
)
-
m
(
y
)
|
≤
ϕ
(
y
)
+
r
(
x
)
|
f
(
y
)
|
≤
ϕ
(
y
)
+
ϕ
(
0
)
|
f
(
y
)
|
|
f
(
0
)
|
for all x
,
y
∈
S. Since m is unbounded, from (41) we have m
(
σ
y
)
=
m
(
y
) for all y
∈
S. Assume that f satisfies (3) or ϕ satisfies (4). Choose a sequence y
n
∈
S
,
n
=
1,2
,
3
,
…, such that (
1
+
ϕ
(
y
n
)
)
/
|
f
(
y
n
)
|
→
0 as n
→
0. Putting y
=
y
n
,
n
=
1,2
,
3
,
…, in (35), dividing the result by |
f
(
y
n
)
|, letting n
→
0, and using (37) we have
(42)
g
(
x
)
=
lim
⁡
n
→
∞
f
(
x
+
σ
y
n
)
f
(
y
n
)
=
lim
⁡
n
→
∞
f
(
0
)
g
(
x
+
σ
y
n
)
f
(
y
n
)
for all x
∈
S. Multiplying both sides of (42) by f
(
y
) and using (35) and (42) we have
(43)
g
(
x
)
f
(
y
)
=
lim
⁡
n
→
∞
f
(
0
)
g
(
x
+
σ
y
n
)
f
(
y
)
f
(
y
n
)
=
lim
⁡
n
→
∞
f
(
0
)
f
(
x
+
σ
y
+
σ
y
n
)
f
(
y
n
)
=
f
(
0
)
g
(
x
+
σ
y
)
for all x
,
y
∈
S. Putting x
=
0 in (43), replacing σ
y by x, and dividing the result by f
(
0
) we have
(44)
g
(
x
)
=
g
(
0
)
f
0
(
σ
x
)
=
g
(
0
)
m
σ
(
σ
x
)
=
g
(
0
)
m
σ
(
x
)
for all x
∈
S. Putting y
=
0 in (35) and using (40) and (44) we get g
(
0
)
=
1. This completes the proof.Using Theorem6 and applying the same method as in the proof of Corollary 3 we have the following.Corollary 7.
LetP
(
x
)
,
x
∈
R
n, be a polynomial. Suppose that f
,
g
:
R
n
→
C are unbounded function satisfying
(45)
|
f
(
x
+
σ
y
)
-
g
(
x
)
f
(
y
)
|
≤
|
P
(
y
)
|
for all x
,
y
∈
R
n. Then there exists a σ-exponential function m
σ
:
R
n
→
C such that
(46)
f
(
x
)
=
f
(
0
)
m
σ
(
x
)
,
g
(
x
)
=
m
σ
(
x
)
for all x
∈
R
n.Using Theorem6 and applying the same method as in the proof of Corollary 5 we have the following.Corollary 8.
Letγ
∈
R
n be fixed. Suppose that f
,
g
:
R
n
→
C are unbounded continuous function satisfying
(47)
|
f
(
x
+
σ
y
)
-
g
(
x
)
f
(
y
)
|
≤
e
γ
·
y
for all x
,
y
∈
R
n. Then there exists c
∈
C
n
,
R
(
c
+
c
σ
)
≠
0 such that f
(
x
)
=
f
(
0
)
e
(
1
/
2
)
(
c
+
c
σ
)
·
x for all x
∈
R
n. If (
1
/
2
)
R
(
c
+
c
σ
)
≠
t
γ for all t
≤
1, then one has
(48)
f
(
x
)
=
f
(
0
)
e
(
1
/
2
)
(
c
+
c
σ
)
·
x
,
g
(
x
)
=
e
(
1
/
2
)
(
c
+
c
σ
)
·
x
for all x
∈
R
n.Theorem 9.
Letf
,
g
:
S
→
C be unbounded functions satisfying
(49)
|
f
(
x
+
σ
y
)
-
f
(
x
)
g
(
y
)
|
≤
ϕ
(
x
)
for all x
,
y
∈
S. Then there exists an unbounded exponential m
:
S
→
C such that f
(
x
)
=
f
(
0
)
m
(
x
) for all x
∈
S. In particular if f satisfies (3) or ϕ satisfies (4), then one has
(50)
f
(
x
)
=
f
(
0
)
m
(
x
)
,
g
(
x
)
=
m
(
σ
x
)
for all x
∈
S.Proof.
Puttingx
=
0 in (49) we have
(51)
|
f
(
σ
y
)
-
f
(
0
)
g
(
y
)
|
≤
ϕ
(
0
)
for all y
∈
S. Choose a sequence y
n
∈
S
,
n
=
1,2
,
3
,
…, such that |
g
(
y
n
)
|
→
∞ as n
→
∞. Putting y
=
y
n
,
n
=
1,2
,
3
,
…, in (49), dividing the result by |
g
(
y
n
)
|, letting n
→
∞, and using (51) we have
(52)
f
(
x
)
=
lim
⁡
n
→
∞
f
(
x
+
σ
y
n
)
g
(
y
n
)
=
lim
⁡
n
→
∞
f
(
0
)
g
(
σ
x
+
y
n
)
g
(
y
n
)
for all x
∈
S. Multiplying both sides of (52) by f
(
y
) and using (49) and (52) we have
(53)
f
(
y
)
f
(
x
)
=
lim
⁡
n
→
∞
f
(
0
)
f
(
y
)
g
(
σ
x
+
y
n
)
g
(
y
n
)
=
lim
⁡
n
→
∞
f
(
0
)
f
(
y
+
x
+
σ
y
n
)
g
(
y
n
)
=
f
(
0
)
f
(
x
+
y
)
for all x
,
y
∈
S. From (53) we have f
0
(
x
)
:
=
f
(
x
)
/
f
(
0
) is an exponential function, say m. Assume that g satisfies (3) or ϕ satisfies (4). Choose a sequence x
n
∈
S
,
n
=
1,2
,
3
,
…, such that (
1
+
ϕ
(
x
n
)
)
/
|
f
(
x
n
)
|
→
0 as n
→
0. Putting x
=
x
n
,
n
=
1,2
,
3
,
…, in (49), dividing the result by |
f
(
x
n
)
|, letting n
→
0, and using (51) we have
(54)
g
(
y
)
=
lim
⁡
n
→
∞
f
(
x
n
+
σ
y
)
f
(
x
n
)
=
lim
⁡
n
→
∞
f
(
0
)
g
(
y
+
σ
x
n
)
f
(
x
n
)
for all y
∈
S. Multiplying both sides of (54) by f
(
x
) and using (49) and (54) we have
(55)
f
(
x
)
g
(
y
)
=
lim
⁡
n
→
∞
f
(
0
)
f
(
x
)
g
(
y
+
σ
x
n
)
f
(
x
n
)
=
lim
⁡
n
→
∞
f
(
0
)
f
(
x
+
σ
y
+
x
n
)
f
(
x
n
)
=
f
(
0
)
g
(
σ
x
+
y
)
for all x
,
y
∈
S. Putting y
=
0 in (55), replacing σ
x by x, and dividing the result by f
(
0
) we have
(56)
g
(
x
)
=
g
(
0
)
f
0
(
σ
x
)
=
g
(
0
)
m
(
σ
x
)
for all x
∈
S. Putting x
=
0 in (49) and using (56) we get g
(
0
)
=
1. This completes the proof.Using Theorem9 we have the following.Corollary 10.
LetP
(
x
)
,
x
∈
R
n, be a polynomial. Suppose that f
,
g
:
R
n
→
C are unbounded function satisfying
(57)
|
f
(
x
+
σ
y
)
-
f
(
x
)
g
(
y
)
|
≤
|
P
(
x
)
|
for all x
,
y
∈
R
n. Then there exists an exponential function m
:
R
n
→
C such that
(58)
f
(
x
)
=
f
(
0
)
m
(
x
)
,
g
(
x
)
=
m
(
σ
x
)
for all x
∈
R
n.Corollary 11.
Letγ
∈
R
n be fixed. Suppose that f
,
g
:
R
n
→
C are unbounded continuous function satisfying
(59)
|
f
(
x
+
σ
y
)
-
f
(
x
)
g
(
y
)
|
≤
e
γ
·
x
for all x
,
y
∈
R
n. Then there exists c
∈
C
n
,
R
c
≠
0 such that f
(
x
)
=
f
(
0
)
e
c
·
x for all x
∈
R
n. If R
c
≠
t
γ for all t
≤
1, then we have
(60)
f
(
x
)
=
f
(
0
)
e
c
·
x
,
g
(
x
)
=
e
c
σ
·
x
for all x
∈
R
n.Theorem 12.
Letf
,
g
:
S
→
C be unbounded functions satisfying
(61)
|
f
(
x
+
σ
y
)
-
f
(
x
)
g
(
y
)
|
≤
ϕ
(
y
)
for all x
,
y
∈
S. Then g is an exponential function. In particular, if g satisfies the condition (3) or ϕ satisfies (4), then there exists an unbounded exponential m
:
S
→
C such that
(62)
f
(
x
)
=
f
(
0
)
m
(
x
)
,
g
(
x
)
=
m
(
σ
x
)
for all x
∈
S.Proof.
Choose a sequencex
n
∈
S
,
n
=
1,2
,
3
,
…, such that |
f
(
x
n
)
|
→
∞ as n
→
∞. Putting x
=
x
n
,
n
=
1,2
,
3
,
…, in (61), dividing the result by |
f
(
x
n
)
|, and letting n
→
∞ we have
(63)
g
(
y
)
=
lim
⁡
n
→
∞
f
(
x
n
+
σ
y
)
f
(
x
n
)
for all y
∈
S. Multiplying both sides of (63) by g
(
x
) and using (61) and (63) we have
(64)
g
(
y
)
g
(
x
)
=
lim
⁡
n
→
∞
f
(
x
n
+
σ
y
)
g
(
x
)
f
(
x
n
)
=
lim
⁡
n
→
∞
f
(
x
n
+
σ
y
+
σ
x
)
f
(
x
n
)
=
g
(
x
+
y
)
for all x
,
y
∈
S. Therefore, g is an exponential function, say m. Assume that g satisfies (3) or ϕ satisfies (4). Choose a sequence y
n
∈
S
,
n
=
1,2
,
3
,
…, such that (
1
+
ϕ
(
y
n
)
)
/
|
g
(
y
n
)
|
→
0 as n
→
∞. Putting y
=
y
n
,
n
=
1,2
,
3
,
…, in (61), dividing the result by |
g
(
y
n
)
|, and letting n
→
∞ we have
(65)
f
(
x
)
=
lim
⁡
n
→
∞
f
(
x
+
σ
y
n
)
g
(
y
n
)
for all x
∈
S. Multiplying both sides of (65) by g
(
y
) and using (61) and (65) we have
(66)
f
(
x
)
g
(
y
)
=
lim
⁡
n
→
∞
f
(
x
+
σ
y
n
)
g
(
y
)
g
(
y
n
)
=
lim
⁡
n
→
∞
f
(
x
+
σ
y
n
+
σ
y
)
g
(
y
n
)
=
f
(
x
+
σ
y
)
for all x
,
y
∈
S. Putting x
=
0 and replacing y by σ
x in (66) we have f
(
x
)
=
f
(
0
)
m
(
σ
x
) for all x
∈
S. Replacing m
(
x
) by m
(
σ
x
), we get (62). This completes the proof.Using Theorem12 we have the following.Corollary 13.
LetP
(
x
)
,
x
∈
R
n, be a polynomial. Suppose that f
,
g
:
R
n
→
C are unbounded function satisfying
(67)
|
f
(
x
+
σ
y
)
-
f
(
x
)
g
(
y
)
|
≤
|
P
(
y
)
|
for all x
,
y
∈
R
n. Then there exists an exponential function m
:
R
n
→
C such that
(68)
f
(
x
)
=
f
(
0
)
m
(
x
)
,
g
(
x
)
=
m
(
σ
x
)
for all x
∈
R
n.Corollary 14.
Letγ
∈
R
n be fixed. Suppose that f
,
g
:
R
n
→
C are unbounded continuous function satisfying
(69)
|
f
(
x
+
σ
y
)
-
f
(
x
)
g
(
y
)
|
≤
e
γ
·
y
for all x
,
y
∈
R
n. Then there exists c
∈
C
n
,
R
c
≠
0 such that g
(
x
)
=
e
c
σ
·
x for all x
∈
S. If R
c
≠
t
γ for all t
≤
1, then we have
(70)
f
(
x
)
=
f
(
0
)
e
c
·
x
,
g
(
x
)
=
e
c
σ
·
x
for all x
∈
S.


## 3. Applications

In this section we consider the stability of (6)~(8). A function M
:
(
0
,
∞
)
→
R is calledmultiplicative function provided that M
(
x
y
)
=
M
(
x
)
M
(
y
) for all x
,
y
>
0. Let F
(
x
+
y
i
)
=
f
(
x
,
y
)
,
G
(
x
+
i
y
)
=
g
(
x
,
y
), and Φ
(
x
+
y
i
)
=
ϕ
(
x
,
y
) for all (
x
,
y
)
∈
R
0
2. Then the functional inequalities (6) and (7) are converted to
(71)
|
F
(
z
w
¯
)
-
G
(
z
)
F
(
w
)
|
≤
Φ
(
z
)
[
resp
.
,
Φ
(
w
)
]
,
(72)
|
F
(
z
w
¯
)
-
F
(
z
)
G
(
w
)
|
≤
Φ
(
z
)
[
resp
.
,
Φ
(
w
)
]
for all z
,
w
∈
C
0.ViewingC
0 as a multiplicative group, letting σ
(
z
)
=
z
¯, and applying Theorems 2 and 6 to the inequalities (71) we have the following.Theorem 15.
Letf
,
g
:
R
0
2
→
R be unbounded functions satisfying (6). Then f
,
g are of the form(73)
f
(
x
,
y
)
=
f
(
1,0
)
M
(
x
2
+
y
2
)
,
g
(
x
,
y
)
=
M
(
x
2
+
y
2
)
for all x
,
y
∈
R, where M
:
(
0
,
∞
)
→
R is a multiplicative function.Applying Theorems9 and 12 to the inequalities (72) we have the following.Theorem 16.
Letf
,
g
:
R
0
2
→
R be unbounded functions satisfying (7). Then f
,
g are of the form
(74)
f
(
x
,
y
)
=
f
(
1,0
)
M
(
x
2
+
y
2
)
E
(
tan
-
1
(
y
x
)
)
,
g
(
x
,
y
)
=
M
(
x
2
+
y
2
)
E
(
-
tan
-
1
(
y
x
)
)
for all (
x
,
y
)
∈
R
0
2, where M
:
(
0
,
∞
)
→
R is a multiplicative function and E
:
R
→
R is an exponential function satisfying E
(
2
π
)
=
1.LetH
=
{
a
+
b
i
+
c
j
+
d
k
∣
a
,
b
,
c
,
d
∈
R
} be the set of quaternions. Recall that i
2
=
j
2
=
k
2
=
-
1, i
j
=
k, j
k
=
i, k
i
=
j, j
i
=
-
k, k
j
=
-
i, and i
k
=
-
j and the conjugate of q
=
a
+
b
i
+
c
j
+
d
k
∈
H is given by q
*
=
a
-
b
i
-
c
j
-
d
k. We denote ∥
q
∥
=
q
q
*
=
a
2
+
b
2
+
c
2
+
d
2. Let F
(
x
+
y
i
+
u
j
+
v
k
)
=
f
(
x
,
y
,
u
,
v
), G
(
x
+
i
y
+
u
j
+
v
k
)
=
g
(
x
,
y
,
u
,
v
), and Φ
(
x
+
y
i
+
u
j
+
v
k
)
=
ϕ
(
x
,
y
,
u
,
v
) for all (
x
,
y
,
u
,
v
)
∈
R
0
4. Then the functional inequalities (8) are converted to
(75)
|
F
(
q
p
*
)
-
G
(
q
)
F
(
p
)
|
≤
Φ
(
q
)
[
resp
.
,
Φ
(
p
)
]
for all p
,
q
∈
H
∖
{
0
}.Applying Theorems2 and 6 to the inequalities (75) we have the following.Theorem 17.
Letf
,
g
:
R
0
4
→
R be unbounded functions satisfying (8). Then f
,
g are of the form
(76)
f
(
x
,
y
,
u
,
v
)
=
f
(
1,0
,
0,0
)
M
(
x
2
+
y
2
+
u
2
+
v
2
)
,
g
(
x
,
y
,
u
,
v
)
=
M
(
x
2
+
y
2
+
u
2
+
v
2
)
for all (
x
,
y
,
u
,
v
)
∈
R
0
4, where M
:
(
0
,
∞
)
→
R is a multiplicative function.


## 4. Stability inL
∞-Version

Letf
:
R
n
→
C be a locally integrable function and σ
:
R
n
→
R
n an involution. In this section, we consider an L
∞-version of the stability of functional equation
(77)
f
(
x
+
σ
y
)
=
f
(
x
)
f
(
y
)
for almost every (
x
,
y
)
∈
R
2
n. More precisely, we study the functional inequality
(78)
∥
f
(
x
+
σ
y
)
-
f
(
x
)
f
(
y
)
∥
L
∞
(
R
2
n
)
≤
ϵ
.
As is well known, inequality (78) implies
(79)
|
∫
‍
(
f
(
x
+
σ
y
)
-
f
(
x
)
f
(
y
)
)
φ
(
x
,
y
)
d
x
d
y
|
≤
ϵ
∥
φ
∥
L
1
(
R
2
n
)
for all φ
∈
L
1
(
R
4
).We first employδ
:
R
n
→
R defined by
(80)
δ
(
x
)
=
{
q
e
-
(
1
-
|
x
|
2
)
-
1
,
if
|
x
|
<
1
0
,
if
|
x
|
≥
1
,
where q
=
(
∫
|
x
|
<
1
‍
e
-
(
1
-
|
x
|
2
)
-
1
d
x
)
-
1. It is easy to see that δ is an infinitely differentiable function with support {
x
:
|
x
|
≤
1
}.Letf be a locally integrable function on R
n and δ
t
(
x
)
:
=
t
-
n
δ
(
x
/
t
)
,
t
>
0. Then for each t
>
0,
(81)
f
*
δ
t
(
x
)
=
∫
R
n
‍
f
(
ξ
)
δ
t
(
x
-
ξ
)
d
ξ
is a smooth function and f
*
δ
t
(
x
)
→
f
(
x
) for almost every x
∈
R
n as t
→
0
+.In the following, we exclude the case whenf
(
x
)
=
0 for almost every x
∈
R
n.Theorem 18.
Letf
:
R
n
→
C satisfy (78). Then either there exists an unbounded exponential function m
:
R
n
→
C such that
(82)
f
(
x
)
=
m
(
x
+
σ
x
2
)
for almost every x
∈
R
n, or else
(83)
∥
f
∥
L
∞
(
R
n
)
≤
1
2
(
1
+
1
+
4
ϵ
)
.
If  ϵ
<
1
/
4, then either
(84)
1
2
(
1
+
1
-
4
ϵ
)
≤
∥
f
∥
L
∞
(
R
n
)
≤
1
2
(
1
+
1
+
4
ϵ
)
or
(85)
∥
f
∥
L
∞
(
R
n
)
≤
1
2
(
1
-
1
-
4
ϵ
)
.Proof.
Applyingφ
(
x
′
,
y
′
)
=
δ
t
(
x
-
x
′
)
δ
s
(
y
-
y
′
) in (79) we have
(86)
∬
f
(
x
′
+
σ
y
′
)
δ
t
(
x
-
x
′
)
δ
s
(
y
-
y
′
)
d
x
′
d
y
′
=
∫
‍
f
(
z
′
)
(
∫
‍

δ
t
(
x
-
z
′
+
σ
y
′
)
δ
s
(
y
-
y
′
)
d
y
′
)
d
z
′
=
∫
‍
f
(
z
′
)
(
∫
‍
δ
t
(
x
-
z
′
+
σ
y
′
)
(
δ
s
∘
σ
)
×
(
σ
y
-
σ
y
′
)
d
y
′
∫
‍
δ
t
(
x
-
z
′
+
σ
y
′
)
(
δ
s
∘
σ
)
)
d
z
′
=
∫
‍
f
(
z
′
)
(
∫
‍

δ
t
(
x
-
z
′
+
y
′
)
(
δ
s
∘
σ
)
(
σ
y
-
y
′
)
d
y
′
)
d
z
′
=
∫
‍
f
(
z
′
)
(
∫
‍
δ
t
(
y
′
)
(
δ
s
∘
σ
)
(
x
+
σ
y
-
z
′
-
y
′
)
d
y
′
)
d
z
′
=
∫
‍
f
(
z
′
)
δ
t
*
(
δ
∘
σ
)
s
(
x
+
σ
y
-
z
′
)
d
z
′
=
(
f
*
δ
t
*
(
δ
∘
σ
)
s
)
(
x
+
σ
y
)
.
We also have
(87)
∬
f
(
x
′
)
f
(
y
′
)
δ
t
(
x
-
x
′
)
δ
s
(
y
-
y
′
)
d
x
′
d
y
′
=
∫
‍
f
(
y
′
)
(
∫
‍
f
(
x
′
)
δ
t
(
x
-
x
′
)
d
x
′
)
δ
s
(
y
-
y
′
)
d
y
′
=
(
f
*
δ
t
)
(
x
)
∫
‍
f
(
y
′
)
(
f
*
δ
s
)
(
y
-
y
′
)
d
y
=
(
f
*
δ
t
)
(
x
)
(
f
*
δ
s
)
(
y
)
.
Thus, the inequality (78) is converted to the classical functional inequality
(88)
|
(
f
*
δ
t
*
δ
s
σ
)
(
x
+
σ
y
)
-
(
f
*
δ
t
)
(
x
)
(
f
*
δ
s
)
(
y
)
|
≤
ϵ
for all x
,
y
∈
R
n, where δ
σ
=
δ
∘
σ.
Choosingy
0
∈
R
,
s
0
>
0 such that (
f
*
δ
s
0
)
(
y
0
)
≠
0, putting y
=
y
0
,
s
=
s
0 in (88), using the triangle inequality, and dividing the result by |
(
f
*
δ
s
0
)
(
y
0
)
| we have
(89)
|
(
f
*
δ
t
)
(
x
)
|
≤
|
(
f
*
δ
t
*
δ
s
0
σ
)
(
x
+
σ
y
0
)
|
+
ϵ
|
(
f
*
δ
s
0
)
(
y
0
)
|
for all x
∈
R
n. Since (
f
*
δ
t
*
δ
s
0
σ
)
(
x
+
σ
y
0
)
→
(
f
*
δ
s
0
σ
)
(
x
+
σ
y
0
) as t
→
0
+, it follows that
(90)
F
(
x
)
∶
=
limsup
⁡
t
→
0
+
(
f
*
δ
t
)
(
x
)
exists for all x
∈
R
n. Since (
f
*
δ
t
)
(
x
)
→
f
(
x
) for almost every x
∈
R
n, it follows from (90) that
(91)
F
(
x
)
=
f
(
x
)
for almost every x
∈
R
n.
Fixingy
∈
R
n and letting s
→
0
+ so that (
f
*
δ
s
)
(
y
)
→
F
(
y
) in (88), we have
(92)
|
(
f
*
δ
t
)
(
x
+
σ
y
)
-
(
f
*
δ
t
)
(
x
)
F
(
y
)
|
≤
ϵ
for all x
,
y
∈
R
n. We first consider the case when F is unbounded. Let y
n
∈
R
n
,
n
=
1,2
,
3
,
…, be a sequence such that |
F
(
y
n
)
|
→
∞. Putting y
=
y
n in (92), dividing the result by |
F
(
y
n
)
|, and letting n
→
∞ we have
(93)
(
f
*
δ
t
)
(
x
)
=
lim
⁡
n
→
∞
(
f
*
δ
t
)
(
x
+
σ
y
n
)
F
(
y
n
)
for all (
x
,
y
)
∈
R
n. Multiplying F
(
y
) in (93) and using (92) and (93) we have
(94)
(
f
*
δ
t
)
(
x
)
F
(
y
)
=
lim
⁡
n
→
∞
(
f
*
δ
t
)
(
x
+
σ
y
n
)
F
(
y
)
F
(
y
n
)
=
lim
⁡
n
→
∞
(
f
*
δ
t
)
(
x
+
σ
y
+
σ
y
n
)
F
(
y
n
)
=
(
f
*
δ
t
)
(
x
+
σ
y
)
for all x
,
y
∈
R
n
,
t
>
0. Putting x
=
0 in (94) we have
(95)
(
f
*
δ
t
)
(
0
)
F
(
y
)
=
(
f
*
δ
t
)
(
σ
y
)
for all y
∈
R
n
,
t
>
0. From (95) we have (
f
*
δ
t
)
(
0
)
≠
0 for some t
>
0. Putting (95) in (94) we have(96)
F
(
σ
x
)
F
(
y
)
=
F
(
y
+
σ
x
)
for all x
,
y
∈
R
n. From (96) F is an exponential function. Now, we prove that
(97)
F
(
x
)
=
F
(
σ
x
)
for all x
∈
R
n. In view of (94), replacing (
f
*
δ
t
)
(
x
) by (
f
*
δ
t
)
(
0
)
F
(
σ
x
) and (
f
*
δ
s
)
(
y
) by (
f
*
δ
s
)
(
0
)
F
(
σ
y
) in (88) and letting s
→
0
+ so that (
f
*
δ
s
)
(
0
)
→
F
(
0
) we have
(98)
|
(
f
*
δ
t
)
(
x
+
σ
y
)
-
(
f
*
δ
t
)
(
0
)
F
(
σ
x
)
F
(
σ
y
)
|
≤
ϵ
for all x
,
y
∈
R
n
,
t
>
0. Using (95) and (98) we have
(99)
|
(
f
*
δ
t
)
(
0
)
F
(
y
+
σ
x
)
-
(
f
*
δ
t
)
(
0
)
F
(
σ
x
)
F
(
σ
y
)
|
≤
ϵ
.
Letting t
→
0
+ in (99) so that (
f
*
δ
t
)
(
0
)
→
F
(
0
) we have
(100)
|
F
(
y
+
σ
x
)
-
F
(
σ
x
)
F
(
σ
y
)
|
≤
ϵ
for all x
,
y
∈
R. Since F is an exponential function, it follows from (100) that
(101)
|
F
(
σ
x
)
|
|
F
(
y
)
-
F
(
σ
y
)
|
≤
ϵ
for all x
,
y
∈
R
n. Since F is unbounded, from (101) we have F
(
y
)
=
F
(
σ
y
) for all y
∈
R
n. Now, F is written in the form
(102)
F
(
x
)
=
F
(
x
2
+
x
2
)
=
F
(
x
2
)
F
(
x
2
)
=
F
(
x
2
)
F
(
σ
x
2
)
=
F
(
x
+
σ
x
2
)
for all x
,
y
∈
R
n. Conversely, let F
(
x
)
=
m
(
(
x
+
σ
x
)
/
2
), where m
:
R
n
→
C is an arbitrary exponential function. Then F is an exponential function satisfying F
(
x
)
=
F
(
σ
x
) for all x
∈
R
n. Thus, we get (82). From now on, we assume that F is bounded, say |
F
(
x
)
|
≤
M for all x
∈
R
n. Then, it follows from (91) that ∥
f
∥
L
∞
(
R
n
)
≤
M. Thus, we have
(103)
|
(
f
*
δ
t
)
(
x
)
|
=
|
∫
‍
f
(
x
′
)
δ
(
x
-
x
′
)
d
x
′
|
≤
M
∫
‍
|
δ
(
x
-
x
′
)
|
d
x
′
=
M
for all x
∈
R
n. From the inequality (92), using the method in Theorem 10 of [1] we have
(104)
|
(
f
*
δ
t
)
(
x
)
(
|
F
(
y
)
|
-
1
)
|
≤
ϵ
for all x
,
y
∈
R
n
,
t
>
0. Fixing x
∈
R
n and letting t
→
0
+ in (104) we have
(105)
|
F
(
x
)
(
|
F
(
y
)
|
-
1
)
|
≤
ϵ
for all x
,
y
∈
R. From (105), using the method in Theorem 10 of [1] we have
(106)
|
F
(
x
)
|
≤
1
2
(
1
+
1
-
4
ϵ
)
for all x
∈
R, and if ϵ
<
1
/
4, then we have either
(107)
1
2
(
1
-
1
+
4
ϵ
)
≤
|
F
(
x
)
|
≤
1
2
(
1
+
1
+
4
ϵ
)
for all x
∈
R or
(108)
|
F
(
x
)
|
≤
1
2
(
1
-
1
-
4
ϵ
)
for all x
∈
R. Since f
(
x
)
=
F
(
x
) almost every x
∈
R
n, we get (83), (84), and (85). This completes the proof.



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*Source: 763914-2014-10-14.xml*