Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 7 (2014), 18–27
Research Article
Solution and stability of a reciprocal type functional
equation in several variables
K. Ravia,∗, E. Thandapanib , B.V. Senthil Kumarc
a
b
c
PG & Research Department of Mathematics, Sacred Heart College, Tirupattur - 635 601, TamilNadu, India
Ramanujan Institute of Advance Study in Mathematics, University of Madras, Chepauk, Chennai- 600 005. Tamil Nadu, India
Department of Mathematics, C. Abdul Hakeem College of Engg. and Tech., Melvisharam - 632 509, TamilNadu, India
Communicated by P. Gavruta
Abstract
In this paper, we obtain the general solution and investigate the generalized Hyers-Ulam stability of a
reciprocal type functional equation in several variables of the form
Qm
Qm
r(x1 + xi )
)
i=2
i=1 r(x
ii
i=P
hQ
Qm
Pm hQm
m
m
i=2
j=2,j6=i r(x1 + xj )
i=2 r(x1 )
j=2,j6=i r(xj ) + (m − 1)
i=2 r(xi )
c
where m is a positive integer with m ≥ 3. 2014
All rights reserved.
Keywords: Rassias reciprocal functional equation, General reciprocal functional equations, Adjoint and
difference functional equations
2010 MSC: Primary 39B22, 39B52, 39B72
1. Introduction
About seventy years ago, Ulam [13] raised the well known stability problem of functional equations. In
the next year, Ulam’s problem was partially answered by Hyers [4] in Banach spaces. T. Aoki [1] generalized
Hyers’ theorem for additive mappings in the year 1950. In the year 1978, a generalized version of the theorem
of Hyers for approximately linear mappings was given by Th.M. Rassias [12]. During 1982-1989, J.M. Rassias
([5], [6], [7]) treated the Ulam-Gavruta-Rassias stability on linear and non-linear mappings and generalized
∗
Corresponding author
Email addresses: shckravi@yahoo.co.in (K. Ravi), ethandapani@yahoo.co.in (E. Thandapani), bvssree@yahoo.co.in
(B.V. Senthil Kumar)
Received 2012-12-10
K. Ravi, E. Thandapani, B.V. Senthil Kumar, J. Nonlinear Sci. Appl. 7 (2014), 18–27
19
Hyers’ result. In 1994, a further generalization of the Th.M. Rassias’ theorem was obtained by P. Gavruta
[3], who replaced the bound θ (kxkp + kykp ) by a general control function φ(x, y).
In the year 2010, K. Ravi and B.V. Senthil Kumar [8] investigated the generalized Hyers-Ulam stability
for the reciprocal functional equation
r(x)r(y)
(1.1)
r(x + y) =
r(x) + r(y)
where r : X → Y is a mapping on the spaces of non-zero real numbers. The reciprocal function r(x) = xc is
a solution of the functional equation (1.1).
Later, K. Ravi, J.M. Rassias and B.V. Senthil Kumar [10] introduced the reciprocal difference functional
equation
x+y
r(x)r(y)
r
− r(x + y) =
(1.2)
2
r(x) + r(y)
and the reciprocal adjoint functional equation
3r(x)r(y)
x+y
+ r(x + y) =
r
2
r(x) + r(y)
(1.3)
and investigated the generalized Hyers-Ulam stability for the above two functional equations (1.2) and (1.3).
Recently, K. Ravi, J.M. Rassias and B.V. Senthil Kumar [9] discussed the generalized Hyers-Ulam stability
for the generalized reciprocal functional equation
!
Qm
m
X
r(xi )
i
r
(1.4)
αi xi = P h i=1
Q
m
m
α
r(x
)
i
j
i=1
i=1
j=1,j6=i
for
Pmarbitrary but fixed real numbers αi
i=1 αi 6= 1 and r : X → Y with X and
Very recently, K. Ravi, E. Thandapani
investigated the generalized Hyers-Ulam
6= 0 for i = 1, 2, . . . , m, so that 0 < α = α1 + α2 + · · · + αm =
Y are the sets of non-zero real numbers.
and B.V. Senthil Kumar [11] obtained the general solution and
stability for the reciprocal type functional equations
r((k1 − k2 )x + (k1 − k2 )y) =
r(k1 x − k2 y)r(k1 y − k2 x)
r(k1 x − k2 y) + r(k1 y − k2 x)
(1.5)
r(k1 x + k2 y)r(k1 y + k2 x)
r(k1 x + k2 y) + r(k1 y + k2 x)
(1.6)
where k1 and k2 are any integers with k1 6= k2 and
r((k1 + k2 )x + (k1 + k2 )y) =
where k1 and k2 are any integers with k1 6= −k2 .
In this paper, we obtain the solution and investigate the generalized Hyers-Ulam stability for a reciprocal
type functional equation in several variables of the form
Qm
Qm
r(x1 + xi )
)
i=2
i=1 r(x
i=P
hQ
ii
Pm hQm
Qm
m
m
i=2
j=2,j6=i r(x1 + xj )
i=2 r(x1 )
j=2,j6=i r(xj ) + (m − 1)
i=2 r(xi )
(1.7)
where m is a positive integer with m ≥ 3.
Throughout this paper, we assume that X is the set of non-zero real numbers. For convenience, we define
the difference operator Dm r : X m → R such that
Qm
i=2 r(x1 + xi )
h
i
Dm r(x1 , x2 , . . . , xm ) = P
Qm
m
r(x
+
x
)
1
j
i=2
j=2,j6=i
Qm
)
i=1 r(x
hQ
ii
−P
Qm
m
m
i=2 r(x1 )
j=2,j6=i r(xj ) + (m − 1)
i=2 r(xi )
K. Ravi, E. Thandapani, B.V. Senthil Kumar, J. Nonlinear Sci. Appl. 7 (2014), 18–27
20
for x1 , x2 , . . . , xm ∈ X.
m−1
In the following results, we will set 00m−2 = 0 for m ≥ 3 and assume




m
m
m
m
m
X
Y
X
Y
Y

r(x1 + xj ) 6= 0,
r(x1 ) 
r(xj ) + (m − 1)
r(xi ) 6= 0
i=2
i=2
j=2,j6=i
i=2
j=2,j6=i
for all xi ∈ X; i = 1, 2, . . . , m; m ≥ 3 and x1 6= −xi , for all i; 2 ≤ i ≤ m; m ≥ 3.
2. General solution of functional equation (1.7)
Theorem 2.1. A mapping r : X → R satisfies the functional equation (1.7) for all x1 , x2 , . . . , xm ∈ X if
and only if there exists a reciprocal mapping r : X → R satisfying the reciprocal functional equation (1.1)
for all x, y ∈ X.
Proof. Let the mapping r : X → R satisfy the functional equation (1.7). Replacing x1 by x and xi by y for
i = 2, 3 . . . , m in (1.7), we arrive (1.1).
Conversely, let the mapping r : X → R satisfy the functional equation (1.1). Replacing (x, y) by
(x1 , x2 + x3 ) in (1.1), we obtain
r(x1 )r(x2 + x3 )
r(x1 ) + r(x2 + x3 )
r(x1 )r(x2 )r(x3 )
=
r(x1 )r(x2 ) + r(x1 )r(x3 ) + r(x2 )r(x3 )
Q3
r(xi )
hQ i=1
i Q
=P
3
3
r(x
)
r(x
)
+ 3i=2 r(xi )
1
j
i=2
j=2,j6=i
r(x1 + x2 + x3 ) =
for all x1 , x2 , x3 ∈ X. Using induction on a positive integer m − 1, we have
Qm−1
r(xi )
hQ i=1
i Q
r(x1 + x2 + · · · + xm−1 ) = P
m−1
m−1
m−1
i=2 r(xi )
i=2 r(x1 )
j=2,j6=i r(xj ) +
(2.1)
for all x1 , x2 , . . . , xm−1 ∈ X. Now, replacing xi by x for i = 1, 2, . . . , m − 1 in (2.1), we get r((m − 1)x) =
1
m−1 r(x), for all x ∈ X. Replacing xi by xi+1 for i = 1, 2, . . . , m − 1 in (2.1), we obtain
Qm
r(xi )
hQ i=2
i Q
(2.2)
r(x2 + x3 + · · · + xm ) = P
m
m
m
i=3 r(x2 )
j=3,j6=i r(xj ) +
i=3 r(xi )
for all x2 , x3 , . . . , xm ∈ X. Now, replacing xi by x + xi for i = 2, 3 . . . , m in (2.2), we get
Qm
r(x1 + xi )
hQ i=2
i Q
Pm
m
r(x
+
x
)
r(x
+
x
)
+ m
1
2
1
j
i=3
j=3,j6=i
i=3 r(x1 + xj )
= r((m − 1)x1 + x2 + · · · + xm )
=
=
Qm
r(xi )
1
Pm
Qm
Qm i=2
r(x
)
1
m−1
i=3 r(x2 )[ j=3,j6=i r(xj )]+ i=3 r(xi )
Qm
r(xi )
1
Qm
Pm
Qm i=2
m−1 r(x1 ) +
i=3 r(x2 )[ j=3,j6=i r(xj )]+ i=3 r(xi )
Qm
i=1 r(xi )
=P
m
i=2 r(x1 )
hQ
i
m
j=2,j6=i r(xj )
+ (m − 1)
Qm
(2.3)
i=2 r(xi )
for all x1 , x2 , . . . , xm ∈ X. On further simplification of the above equation (2.3) yields the equation (1.7).
This completes the proof of Theorem 2.1.
K. Ravi, E. Thandapani, B.V. Senthil Kumar, J. Nonlinear Sci. Appl. 7 (2014), 18–27
21
3. Generalized Hyers-Ulam stability of equation (1.7)
Theorem 3.1. Let ϕ : X m → R be a function satisfying
∞
x
X
x2
xm 1
1
ϕ
,
,
.
.
.
,
< +∞
2i+1
2i+1 2i+1
2i+1
(3.1)
i=0
for all x1 , x2 , . . . , xm ∈ X. If a function f : X → R satisfies the functional inequality
|Dm f (x1 , x2 , . . . , xm )| ≤ ϕ(x1 , x2 , . . . , xm )
(3.2)
for all x1 , x2 , . . . , xm ∈ X, then there exists a unique reciprocal mapping r : X → R which satisfies (1.7)
and the inequality
∞
x
X
1
x
x |r(x) − f (x)| ≤ 2(m − 1)
ϕ
,
,
.
.
.
,
(3.3)
2i+1
2i+1 2i+1
2i+1
i=0
for all x ∈ X.
Proof. Replacing xi by
x
2
for i = 1, 2, . . . , m in (3.2) and multiplying by (m − 1), we get
f (x) − 1 f x ≤ (m − 1)ϕ x , x , . . . , x
2
2 2 2
2
(3.4)
for all x ∈ X. Now, replacing x by x2 in (3.4), dividing by 2 and summing the resulting inequality with
(3.4), we obtain
1
x
X
1
x
x f (x) − 1 f x ≤ 2(m − 1)
ϕ
,
,
.
.
.
,
22
22 2i+1
2i+1 2i+1
2i+1
i=0
for all x ∈ X. Proceeding further and using induction arguments on a positive integer n, we arrive
n−1
x
X 1
x
x f (x) − 1 f x ≤ 2(m − 1)
ϕ
,
,
.
.
.
,
2n
2n 2i+1
2i+1 2i+1
2i+1
(3.5)
i=0
for all x ∈ X. For any positive integer i and x ∈ X, we have
x 1 x 1
1 x 1 x 2i+1 f 2i+1 − 2i f 2i = 2i f 2i − 2 f 2i+1 x
1
x
x ≤ 2(m − 1) i+1 ϕ i+1 , i+1 , . . . , i+1 .
2
2
2
2
Hence for any integers l, k with l > k > 0, we obtain by using the triangular inequality
1 x
1 x f
− kf k 2l
2l
2
2
x x x 1
x
1
1
1
= l f l − l−1 f l−1 + l−1 f l−1 − · · · + k+1 f k+1 2
2
2
2
2
2
2
2
1
x x
x
1
x
x
x ≤ 2(m − 1) l ϕ l , l , . . . , l + · · · + 2(m − 1) k+1 ϕ k+1 , k+1 , . . . , k+1
2
2 2
2
2
2
2
2
l
X 1
x x
x
≤ 2(m − 1)
ϕ i, i,..., i
i
2
2 2
2
i=k+1
≤ 2(m − 1)
l−1
x
X
1
x
x ϕ
,
,
.
.
.
,
2i+1
2i+1 2i+1
2i+1
i=k
(3.6)
K. Ravi, E. Thandapani, B.V. Senthil Kumar, J. Nonlinear Sci. Appl. 7 (2014), 18–27
22
for all x ∈ X. Taking the limit as k → +∞ in (3.6) and considering (3.1), it follows that the sequence
{ 21n f 2xn } is a Cauchy
sequence for each x ∈ X. Since R is complete, we can define r : X → R by
r(x) = lim 21n f 2xn . To show that r satisfies (1.7), replacing (x1 , x2 , . . . , xm ) by (2−n x1 , 2−n x2 , . . . , 2−n xm )
n→∞
in (3.2) and dividing by 2n , we obtain
|2−n Dm f (2−n x1 , 2−n x2 , . . . , 2−n xm )| ≤ 2−n ϕ(2−n x1 , 2−n x2 , . . . , 2−n xm )
(3.7)
for all x1 , x2 , . . . , xm ∈ X and for all positive integer n. Using (3.1) and (3.5) in (3.7), we see that r satisfies
(1.7), for all x1 , x2 , . . . , xm ∈ X. Taking limit n → ∞ in (3.5), we arrive (3.3). Now, it remains to show that
r is uniquely defined. Let r1 : X → R be another reciprocal mapping which satisfies (1.7) and the inequality
(3.3). Clearly, r1 (2−n x) = 2n r1 (x), r(2−n x) = 2n r(x) and using (3.3), we arrive
|r1 (x) − r(x)| = 2−n r1 (2−n x) − r(2−n x)
∞
x
X
1
x
x ≤ 4(m − 1)
ϕ n+i+1 , n+i+1 , . . . , n+i+1
2n+i+1
2
2
2
i=0
∞
x
X
x
x 1
ϕ
,
,
.
.
.
,
≤ 4(m − 1)
2i+1
2i+1 2i+1
2i+1
(3.8)
i=n
for all x ∈ X. Allowing n → ∞ in (3.8), we find that r is unique. This completes the proof of Theorem
3.1.
Theorem 3.2. Let ϕ : X m → R be a function satisfying
∞
X
2i ϕ(2i x1 , 2i x2 , . . . , 2i xm ) < +∞
(3.9)
i=0
for all x1 , x2 , . . . , xm ∈ X. If a function f : X → R satisfies the functional inequality
|Dm f (x1 , x2 , . . . , xm )| ≤ ϕ(x1 , x2 , . . . , xm )
(3.10)
for all x1 , x2 , . . . , xm ∈ X, then there exists a unique reciprocal mapping r : X → R which satisfies (1.7)
and the inequality
∞
X
|r(x) − f (x)| ≤ 2(m − 1)
2i ϕ(2i x, 2i x, . . . , 2i x)
(3.11)
i=0
for all x ∈ X.
Proof. The proof is obtained by replacing xi by x for i = 1, 2, . . . , m in (3.10) and proceeding further by
similar arguments as in Theorem 3.1.
Corollary 3.3. For any fixed c1 ≥ 0 and p > −1 or p < −1, if f : X → R satisfies
!
m
X
|Dm f (x1 , x2 , . . . , xm )| ≤ c1
|xi |p
i=1
for all x1 , x2 , . . . , xm ∈ X, then there exists a unique reciprocal mapping r : X → R such that
(
2m(m−1)c1
|x|p for p > −1
2p+1 −1
|r(x) − f (x)| ≤ 2m(m−1)c
1
|x|p for p < −1
1−2p+1
for all x ∈ X.
K. Ravi, E. Thandapani, B.V. Senthil Kumar, J. Nonlinear Sci. Appl. 7 (2014), 18–27
23
P
p
Proof. If we choose ϕ(x1 , x2 , . . . , xm ) = c1 ( m
i=1 |xi | ), for all x1 , x2 , . . . , xm ∈ X, then by Theorem 3.1, we
arrive
2m(m − 1)c1 p
|r(x) − f (x)| ≤
|x| , for all x ∈ X and p > −1
2p+1 − 1
and using Theorem 3.2, we arrive
|r(x) − f (x)| ≤
2m(m − 1)c1 p
|x| , for all x ∈ X and p < −1.
1 − 2p+1
Corollary 3.4. Let f : X → R be a mapping and there exists p such that p > −1 or p < −1. If there exists
c2 ≥ 0 such that
!
m
Y
p
|Dm f (x1 , x2 , . . . , xm )| ≤ c2
|xi | m
i=1
for all x1 , x2 , . . . , xm ∈ X, then there exists a unique reciprocal mapping r : X → R satisfying the functional
equation (1.7) and
(
2(m−1)c2
|x|p for p > −1
2p+1 −1
|r(x) − f (x)| ≤ 2(m−1)c
2
|x|p for p < −1
1−2p+1
for all x ∈ X.
Q
p
m
m
Proof. Considering ϕ(x1 , x2 , . . . , xm ) = c2
|x
|
, for all x1 , x2 , . . . , xm ∈ X, then by Theorem 3.1,
i=1 i
we arrive
2(m − 1)c2 p
|r(x) − f (x)| ≤ p+1
|x| , for all x ∈ X and p > −1
2
−1
and using Theorem 3.2, we arrive
|r(x) − f (x)| ≤
2(m − 1)c2 p
|x| , for all x ∈ X and p < −1.
1 − 2p+1
1
1
or α < − m
be real numbers, and f : X → R be a mapping
Corollary 3.5. Let c3 > 0 and α > − m
satisfying the functional inequality
(m
!)
m
X
Y
|Dm f (x1 , x2 , . . . , xm )| ≤ c3
|xi |mα +
|xi |α
i=1
i=1
for all x1 , x2 , . . . , xm ∈ X. Then there exists a unique reciprocal mapping r : X → R satisfying the functional
equation (1.7) and
(
2(m−1)(m+1)c3
1
|x|mα for α > − m
2mα+1 −1
|r(x) − f (x)| ≤ 2(m−1)(m+1)c
1
3
|x|mα for α < − m
1−2mα+1
for all x ∈ X.
P
Q
mα + ( m |x |α )}, for all x , x , . . . , x
Proof. Choosing ϕ(x1 , x2 , . . . , xm ) = c3 { m
1 2
m ∈ X, then by
i=1 |xi |
i=1 i
Theorem 3.1, we arrive
|r(x) − f (x)| ≤
2(m − 1)(m + 1)c3 mα
1
|x| , for all x ∈ X and α > −
mα+1
2
−1
m
and using Theorem 3.2, we arrive
|r(x) − f (x)| ≤
1
2(m − 1)(m + 1)c3 mα
|x| , for all x ∈ X and α < − .
mα+1
1−2
m
K. Ravi, E. Thandapani, B.V. Senthil Kumar, J. Nonlinear Sci. Appl. 7 (2014), 18–27
24
4. Counter-examples
The following example illustrates the fact that the functional equation (1.7) is not stable for p = −1 in
Corollary 3.3. We present the following counter-example modified by the well-known counter-example of Z.
Gajda [2].
Example 4.1. Let ϕ : X → R be a mapping defined by
(
µ
for x ∈ (1, ∞)
ϕ(x) = x
µ otherwise
where µ > 0 is a constant, and define a mapping f : X → R by
f (x) =
∞
X
ϕ(2−n x)
2n
n=0
, for all x ∈ X.
Then the mapping f satisfies the inequality
6µ
|Dm f (x1 , x2 , . . . , xm )| ≤
m−1
m
X
!
|xi |
−1
(4.1)
i=1
for all x1 , x2 , . . . , xm ∈ X. Therefore there do not exist a reciprocal mapping r : X → R and a constant
β > 0 such that
|f (x) − r(x)| ≤ β|x|−1
(4.2)
for all x ∈ X.
Proof. |f (x)| ≤
P∞
n=0
|ϕ(2−n x)|
|2n |
≤
P∞
µ
n=0 2n
=µ 1−
m
X
1 −1
2
= 2µ. Hence f is bounded by 2µ. If
!
|xi |
−1
≥ 1,
i=1
6µ
then the left hand side of (4.1) is less than m−1
. Now, suppose that 0 <
exists a positive integer k such that
m
X
1
1
≤
|xi |−1 < k .
2k+1
2
i=1
Pm
1
−1
Hence i=1 |xi | < 2k implies
2k
m
X
Pm
i=1 |xi |
|xi |−1 < 1
i=1
xi
> 1 for i = 1, 2, . . . , m
2k
xi
1
> 1 > for i = 1, 2, . . . , m
2
2k
xi
> 2 > 1 for i = 1, 2, . . . , m
2k−1
or
or
or
and consequently
1
2k−1
(x1 ),
1
2k−1
(xi ),
1
2k−1
(x1 + xi ) > 1 for i = 2, 3 . . . , m.
Therefore, for each value of n = 0, 1, 2, . . . , k − 1, we obtain
1
1
1
(x1 ), n (xi ), n (x1 + xi ) > 1 for i = 2, 3, . . . , m
n
2
2
2
−1
< 1. Then there
(4.3)
K. Ravi, E. Thandapani, B.V. Senthil Kumar, J. Nonlinear Sci. Appl. 7 (2014), 18–27
25
and Dm ϕ(2−n x1 , 2−n x2 , . . . , 2−n xm ) = 0 for n = 0, 1, 2, . . . , k − 1. Using (4.3) and the definition of f , we
obtain
Qm
f (x1 + xi )
i
|Dm f (x1 , x2 , . . . , xm )| = P hQi=2
m
m
i=2
j=2,j6=i f (x1 + xj )
Qm
)
i=1 f (x
hQ
ii
−P
Qm
m
m
i=2 f (x1 )
j=2,j6=i f (xj ) + (m − 1)
i=2 f (xi )
Qm P∞ µ
Qm P∞ µ
n
i=1
i=2
n=k 2n
Qm n=k
P2∞ µ ≤
Qm−1 P∞ µ +
2(m − 1) i=2
(m − 1) i=2
n=k 2n
n=k 2n
−1
3
µ
1
≤
1−
k
2(m − 1) 2
2
1
6µ
≤
m − 1 2k+1
!
m
X
6µ
≤
|xi |−1
m−1
i=1
for all x1 , x2 , . . . , xm ∈ X. Therefore the inequality (4.1) holds true.
We claim that the reciprocal functional equation (1.7) is not stable for p = −1 in Corollary 3.3.
Assume that there exists a reciprocal mapping r : X → R satisfying (4.2). Therefore, we have
|f (x)| ≤ (β + 1)|x|−1 .
(4.4)
However, we can choose a positive integer m with mµ > β + 1. If x ∈ 1, 2m−1 then 2−n x ∈ (1, ∞)
for all n = 0, 1, 2, . . . , m − 1 and therefore
|f (x)| =
∞
X
ϕ(2−n x)
n=0
2n
≥
m−1
X 2n µ
x
2n
n=0
=
mµ
> (β + 1)x−1
x
which contradicts (4.4). Therefore, the reciprocal type functional equation (1.7) is not stable for p = −1 in
Corollary 3.3.
1
The following example illustrates the fact that the functional equation (1.7) is not stable for α = − m
in Corollary 3.5.
Example 4.2. Let φ : X → R be a mapping defined by
(
δ
for x ∈ (1, ∞)
φ(x) = x
δ otherwise
where δ > 0 is a constant, and define a mapping f : X → R by
f (x) =
∞
X
φ(2−n x)
n=0
2n
, for all x ∈ X.
Then the mapping f satisfies the inequality
6δ
|Dm f (x1 , x2 , . . . , xm )| ≤
m−1
(m
X
i=1
|xi |−1 +
m
Y
!)
|xi |
1
−m
(4.5)
i=1
for all x1 , x2 , . . . , xm ∈ X. Therefore there do not exist a reciprocal mapping r : X → R and a constant
β > 0 such that
|f (x) − r(x)| ≤ β|x|−1
(4.6)
for all x ∈ X.
K. Ravi, E. Thandapani, B.V. Senthil Kumar, J. Nonlinear Sci. Appl. 7 (2014), 18–27
26
Proof. It is easy to show that f is bounded by 2δ, by similar arguments as in Example 4.1. If
(m
!)
m
X
Y
1
−1
−m
|xi | +
≥ 1,
|xi |
i=1
i=1
6δ
then the left hand side of (4.5) is less than m−1
. Now, suppose that
!)
(m
m
X
Y
1
< 1.
0<
|xi |−1 +
|xi |− m
i=1
i=1
Then there exists a positive integer k such that
(m
X
1
≤
|xi |−1 +
2k+1
i=1
Hence
nP
m
−1 +
i=1 |xi |
Q
1
m
−m
i=1 |xi |
(
2
k
o
m
X
1
2k
<
−1
|xi |
m
Y
!)
|xi |
1
−m
i=1
<
1
.
2k
(4.7)
implies
or
or
m
Y
!)
1
−m
<1
|xi |
i=1
2k x−1
i < 1 for i = 1, 2, . . . , m
xi
> 1 for i = 1, 2, . . . , m
2k
+2
i=1
k
xi
1
> 1 > for i = 1, 2, . . . , m
k
2
2
xi
> 2 > 1 for i = 1, 2, . . . , m
2k−1
or
or
and consequently
1
2k−1
(x1 ),
1
2k−1
(xi ),
1
2k−1
(x1 + xi ) > 1 for i = 2, 3, . . . , m.
Therefore, for each value of n = 0, 1, 2, . . . , k − 1, we obtain
1
1
1
(x1 ), n (xi ), n (x1 + xi ) > 1 for i = 2, 3, . . . , m
n
2
2
2
and Dm φ(2−n x1 , 2−n x2 , . . . , 2−n xm ) = 0 for n = 0, 1, 2, . . . , k − 1. Using (4.7), the definition of f and similar
arguments as in Example 4.1, we obtain the inequality (4.5). The remaining part of the proof is obtained
by similar arguments as in Example 4.1. Hence, the reciprocal type functional equation (1.7) is not stable
1
for α = − m
in Corollary 3.5.
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