Open Access

On the stability of a mixed type quadratic-additive functional equation

Advances in Difference Equations20132013:198

https://doi.org/10.1186/1687-1847-2013-198

Received: 28 December 2012

Accepted: 19 June 2013

Published: 5 July 2013

Abstract

In this paper we establish the general solutions of the following mixed type quadratic-additive functional equation:

9 f ( x + y + z 3 ) + 4 [ f ( x y 2 ) + f ( y z 2 ) + f ( z x 2 ) ] = 3 [ f ( x ) + f ( y ) + f ( z ) ]

in the class of functions between real vector spaces. Moreover, we prove the generalized Hyers-Ulam-Rassias stability of this equation in Banach spaces.

MSC:39B82, 39B52.

Keywords

quadratic functional equationadditive functional equationmixed type functional equationstability

1 Introduction

The stability problems of functional equations go back to 1940 when Ulam [1] proposed the following question: This question was solved affirmatively by Hyers [2] under the assumption that G 2 is a Banach space. He proved that if f is a mapping between Banach spaces satisfying f ( x + y ) f ( x ) f ( y ) ϵ for some fixed ϵ 0 , then there exists a unique additive mapping A such that f ( x ) A ( x ) ϵ . In 1978, Rassias [3] generalized Hyers’ result to the unbounded Cauchy difference. Since then, the stability problems of various functional equations have been extensively studied and generalized by a number of authors (see [49]).

Let f be a mapping from a group G 1 to a metric group G 2 with the metric d ( , ) such that
d ( f ( x y ) , f ( x ) f ( y ) ) ϵ .
Then does there exist a group homomorphism L : G 1 G 2 and δ ϵ > 0 such that
d ( f ( x ) , L ( x ) ) δ ϵ

for all x G 1 ?

In particular, Kannappan [10] introduced the following mixed type quadratic-additive functional equation:
f ( x + y + z ) + f ( x ) + f ( y ) + f ( z ) = f ( x + y ) + f ( y + z ) + f ( z + x )
(1.1)
and proved that a function on a real vector space is a solution of (1.1) if and only if there exist a symmetric biadditive function B and an additive function A such that f ( x ) = B ( x , x ) + A ( x ) . In addition, Jung [11] investigated the Hyers-Ulam stability of (1.1) on restricted domains and applied the result to the study of an interesting asymptotic behavior of the quadratic functions. More generally, Jun and Kim [12] solved the general solutions and proved the stability of the following functional equation, which is a generalization of (1.1):
f ( i = 1 n x i ) + ( n 2 ) i = 1 n f ( x i ) = 1 i < j n f ( x i + x j ) ( n > 2 ) .
Najati and Moghimi [13] introduced another mixed type quadratic-additive functional equation
f ( 2 x + y ) + f ( 2 x y ) = f ( x + y ) + f ( x y ) + 2 f ( 2 x ) 2 f ( x )

and investigated the generalized Hyers-Ulam-Rassias stability of this equation in quasi-Banach spaces.

In this paper, we introduce the following quadratic-additive functional equation:
9 f ( x + y + z 3 ) + 4 [ f ( x y 2 ) + f ( y z 2 ) + f ( z x 2 ) ] = 3 [ f ( x ) + f ( y ) + f ( z ) ]
(1.2)
to establish the general solutions and stability problems of this equation. For real vector spaces X and Y, we prove in Section 2 that a mapping f : X Y satisfies (1.2) if and only if there exist a quadratic mapping Q : X Y satisfying
Q ( x + y ) + Q ( x y ) = 2 Q ( x ) + 2 Q ( y )
(1.3)
and an additive mapping A : X Y satisfying
A ( x + y ) = A ( x ) + A ( y )
(1.4)
such that
f ( x ) = Q ( x ) + A ( x )

for all x X . We refer to [1424] for the stability results of other mixed type functional equations. In Section 3, we prove the generalized Hyers-Ulam-Rassias stability of (1.2) in Banach spaces.

2 General solutions of (1.2)

Throughout this section, X and Y will be real vector spaces. In order to solve the general solutions of (1.2), we need the following two lemmas.

Lemma 2.1 If an even mapping f : X Y satisfies (1.2) for all x , y X , then f is quadratic.

Proof Putting x = y = z = 0 in (1.2), we have f ( 0 ) = 0 . Putting z = x in (1.2) yields
9 f ( y 3 ) + 4 [ f ( x y 2 ) + f ( x + y 2 ) + f ( x ) ] = 6 f ( x ) + 3 f ( y )
(2.1)
for all x , y X . Replacing y with x in (2.1) gives
9 f ( x 3 ) = f ( x )
(2.2)
for all x X . Putting y = 0 in (2.1), we have
4 f ( x 2 ) = f ( x )
(2.3)
for all x X . Using (2.2) and (2.3), we can rewrite (1.2) as
f ( x + y + z ) + f ( x y ) + f ( y z ) + f ( z x ) = 3 [ f ( x ) + f ( y ) + f ( z ) ]
(2.4)
for all x , y , z X . Putting z = 0 in (2.4), we obtain
f ( x + y ) + f ( x y ) = 2 f ( x ) + 2 f ( y )

for all x , y X . □

Lemma 2.2 If an odd mapping f : X Y satisfies (1.2) for all x , y X , then f is additive.

Proof Putting x = y = z = 0 in (1.2), we have f ( 0 ) = 0 . Putting z = x in (1.2) yields
9 f ( y 3 ) + 4 [ f ( x y 2 ) + f ( x + y 2 ) f ( x ) ] = 3 f ( y )
(2.5)
for all x , y X . Replacing y by x in (2.5) gives
9 f ( x 3 ) = 3 f ( x )
(2.6)
for all x X . Putting y = 0 in (2.5), we have
2 f ( x 2 ) = f ( x )
(2.7)
for all x X . It follows from (2.5), (2.6) and (2.7) that
f ( x + y 2 ) + f ( x y 2 ) = 2 f ( x )
(2.8)
for all x , y X . Replacing x with x + y and y with x y in (2.8), we obtain
f ( x + y ) = f ( x ) + f ( y )

for all x , y X . □

Now we are ready to establish the general solutions of (1.2).

Theorem 2.3 A function f : X Y satisfies (1.2) for all x , y , z X if and only if there exist a symmetric biadditive mapping B : X × X Y and an additive mapping A : X Y such that
f ( x ) = B ( x , x ) + A ( x )

for all x X .

Proof (Necessity) We decompose f into the even part and the odd part by putting
f e ( x ) = 1 2 ( f ( x ) + f ( x ) ) , f o ( x ) = 1 2 ( f ( x ) f ( x ) )

for all x X . By Lemmas 2.1 and 2.2 we have the result.

(Sufficiency) This is obvious. □

3 Stability of (1.2)

In what follows, X and Y will be a real normed linear space and a real Banach space, respectively. For convenience, we define
D f ( x , y , z ) : = 9 f ( x + y + z 3 ) + 4 [ f ( x y 2 ) + f ( y z 2 ) + f ( z x 2 ) ] 3 [ f ( x ) + f ( y ) + f ( z ) ]
for all x , y , z X . Let φ : X × X × X [ 0 , ) be a mapping satisfying one of the conditions ( ), () and one of the conditions ( ), ( ):
( A ) Φ 1 ( x , y , z ) : = k = 1 1 9 k φ ( 3 k x , 3 k y , 3 k z ) < , ( B ) Φ 2 ( x , y , z ) : = k = 0 9 k φ ( x 3 k , y 3 k , z 3 k ) < , ( C ) Ψ 1 ( x , y , z ) : = k = 1 1 3 k φ ( 3 k x , 3 k y , 3 k z ) < , ( D ) Ψ 2 ( x , y , z ) : = k = 0 3 k φ ( x 3 k , y 3 k , z 3 k ) < ,

for all x , y , z X . We note that the condition ( ) implies ( ). Similarly, the condition () implies ( ). One of the conditions ( ), () will be needed to derive a quadratic mapping, and one of the conditions ( ), ( ) will be required to derive an additive mapping in the following theorem.

Theorem 3.1 Suppose that a mapping f : X Y satisfies
D f ( x , y , z ) φ ( x , y , z )
(3.1)
for all x , y , z X . Then there exist a unique quadratic mapping Q : X Y satisfying (1.3) and an additive mapping A : X Y satisfying (1.4) such that
f ( x ) Q ( x ) A ( x ) + 1 2 f ( 0 ) 1 2 [ Φ i ( x , x , x ) + Φ i ( x , x , x ) ] + 1 6 [ Ψ j ( x , x , x ) + Ψ j ( x , x , x ) ] , f ( x ) + f ( x ) 2 Q ( x ) + 1 2 f ( 0 ) 1 2 [ Φ i ( x , x , x ) + Φ i ( x , x , x ) ] ,
and
f ( x ) f ( x ) 2 A ( x ) 1 6 [ Ψ j ( x , x , x ) + Ψ j ( x , x , x ) ]
for all x X and for i = 1 or 2, j = 1 or 2. The mappings Q and A are given by
{ Q ( x ) = lim n 1 2 9 n [ f ( 3 n x ) + f ( 3 n x ) ] if condition  ( A holds , Q ( x ) = lim n 9 n 2 [ f ( x 3 n ) + f ( x 3 n ) ] , f ( 0 ) = 0 if condition  ( B holds , A ( x ) = lim n 1 2 3 n [ f ( 3 n x ) f ( 3 n x ) ] if condition  ( C holds , A ( x ) = lim n 3 n 2 [ f ( x 3 n ) f ( x 3 n ) ] , f ( 0 ) = 0 if condition  ( D holds

for all x X .

Proof We first consider the even part of f. Let f e : X Y be a function defined by f e ( x ) : = f ( x ) + f ( x ) 2 for all x X . Then f e ( x ) = f e ( x ) and
D f e ( x , y , z ) 1 2 [ φ ( x , y , z ) + φ ( x , y , z ) ]
(3.2)
for all x , y , z X . Putting y = x , z = x in (3.2), we have
9 g ( x 3 ) g ( x ) 1 2 [ φ ( x , x , x ) + φ ( x , x , x ) ]
(3.3)

for all x X , where g ( x ) : = f e ( x ) + 1 2 f ( 0 ) .

Case 1. Assume that φ satisfies the condition ( ). Replacing x by 3x in (3.3) and dividing by 9 yield
g ( x ) g ( 3 x ) 9 1 2 9 [ φ ( 3 x , 3 x , 3 x ) + φ ( 3 x , 3 x , 3 x ) ]
(3.4)
for all x X . Making use of an induction argument in (3.4) implies
g ( x ) g ( 3 n x ) 9 n 1 2 k = 1 n 1 9 k [ φ ( 3 k x , 3 k x , 3 k x ) + φ ( 3 k x , 3 k x , 3 k x ) ]
(3.5)
for all n N and x X . From (3.5) we figure out
g ( 3 m x ) 9 m g ( 3 n x ) 9 n = 1 9 m g ( 3 m x ) g ( 3 n m 3 m x ) 9 n m 1 2 k = m + 1 n 1 9 k [ φ ( 3 k x , 3 k x , 3 k x ) + φ ( 3 k x , 3 k x , 3 k x ) ]
for all n , m N with n > m and x X . The right-hand side of the inequality above tends to 0 as m , the sequence { g ( 3 n x ) / 9 n } is a Cauchy sequence for all x X and thus converges by the completeness of Y. Therefore, we can define a mapping Q : X Y by
Q ( x ) : = lim n g ( 3 n x ) 9 n = lim n f ( 3 n x ) + f ( 3 n x ) 2 9 n
for all x X . Note that Q ( 0 ) = 0 , Q ( x ) = Q ( x ) for all x X . It follows from the condition ( ) and (3.2) that Q satisfies
9 Q ( x + y + z 3 ) + 4 [ Q ( x y 2 ) + Q ( y z 2 ) + Q ( z x 2 ) ] = 3 [ Q ( x ) + Q ( y ) + Q ( z ) ]
for all x , y , z X . According to Lemma 2.1, the mapping Q satisfies (1.3). Letting n in (3.5), we have
g ( x ) Q ( x ) 1 2 [ Φ 1 ( x , x , x ) + Φ 1 ( x , x , x ) ]
(3.6)
for all x X . Now we are going to prove the uniqueness of Q. Assume that Q is another quadratic function satisfying (1.3) and (3.6). Obviously, we have Q ( 3 n x ) = 9 n Q ( x ) and Q ( 3 n x ) = 9 n Q ( x ) for all x X . Then we figure out
Q ( x ) Q ( x ) = 9 n Q ( 3 n x ) Q ( 3 n x ) 9 n Q ( 3 n x ) g ( 3 n x ) + 9 n g ( 3 n x ) Q ( 3 n x ) k = n + 1 1 9 k [ φ ( 3 k x , 3 k x , 3 k x ) + φ ( 3 k x , 3 k x , 3 k x ) ]

for all n N and x X . Taking the limit as n , we conclude that Q ( x ) = Q ( x ) for all x X .

Case 2. If φ satisfies the condition () (and hence implies ( )), the proof is analogous to that of Case 1. By virtue of the condition () and (3.1), we have φ ( 0 , 0 , 0 ) = 0 and f ( 0 ) = 0 . An induction argument on (3.3) implies
9 n f e ( x 3 n ) f e ( x ) 1 2 k = 0 n 1 9 k [ φ ( x 3 k , x 3 k , x 3 k ) + φ ( x 3 k , x 3 k , x 3 k ) ]
(3.7)
for all n N and x X . Using a similar argument to that of Case 1, we see that the sequence { 9 n f e ( x 3 n ) } is a Cauchy sequence for all x X . Thus we can define a mapping Q : X Y by
Q ( x ) : = lim n 9 n f e ( x 3 n ) = lim n 9 n 2 [ f ( x 3 n ) + f ( x 3 n ) ]
for all x X . Note that Q ( 0 ) = 0 and Q ( x ) = Q ( x ) for all x X . From the condition () and (3.2), we see that Q satisfies
9 Q ( x + y + z 3 ) + 4 [ Q ( x y 2 ) + Q ( y z 2 ) + Q ( z x 2 ) ] = 3 [ Q ( x ) + Q ( y ) + Q ( z ) ]
for all x , y , z X . By Lemma 2.1 the mapping Q satisfies (1.3). Taking the limit as n in (3.7), we obtain
f e ( x ) Q ( x ) 1 2 [ Φ 2 ( x , x , x ) + Φ 2 ( x , x , x ) ]

for all x X . The rest of the proof is similar to that of Case 1.

Next, we consider the odd part of f. Now, let f o : X Y be a function defined by f o ( x ) : = 1 2 [ f ( x ) f ( x ) ] for all x X . Then f o ( 0 ) = 0 , f o ( x ) = f o ( x ) and
D f o ( x , y , z ) 1 2 [ φ ( x , y , z ) + φ ( x , y , z ) ]
(3.8)
for all x , y , z X . Putting y = x , z = x in (3.8) and dividing by 3 yield
3 f o ( x 3 ) f o ( x ) 1 6 [ φ ( x , x , x ) + φ ( x , x , x ) ]
(3.9)

for all x X .

Case 3. Assume that φ satisfies the condition ( ) (and hence implies ( )). Replacing x by 3x in (3.9) and dividing by 3, we have
f o ( x ) f o ( 3 x ) 3 1 6 3 [ φ ( 3 x , 3 x , 3 x ) + φ ( 3 x , 3 x , 3 x ) ]
(3.10)
for all x X . Making use of an induction argument in (3.10) implies
f o ( x ) f o ( 3 n x ) 3 n 1 6 k = 1 n 1 3 k [ φ ( 3 k x , 3 k x , 3 k x ) + φ ( 3 k x , 3 k x , 3 k x ) ]
(3.11)
for all n N and x X . From (3.11) we can show that the sequence { f ( 3 n x ) / 3 n } is a Cauchy sequence for all x X . Define a mapping A : X Y by
A ( x ) : = lim n f o ( 3 n x ) 3 n = lim n f ( 3 n x ) f ( 3 n x ) 2 3 n
for all x X . By the oddness of f, we see that A ( x ) = A ( x ) for all x X . Also, from the condition ( ) and (3.9), we verify that A satisfies
9 A ( x + y + z 3 ) + 4 [ A ( x y 2 ) + A ( y z 2 ) + A ( z x 2 ) ] = 3 [ A ( x ) + A ( y ) + A ( z ) ]
for all x , y , z X . According to Lemma 2.2, the mapping A satisfies (1.4). Letting n in (3.11), we have
f o ( x ) A ( x ) 1 6 [ Ψ 1 ( x , x , x ) + Ψ 1 ( x , x , x ) ]

for all x X . Using a similar argument to that of Case 1, we can easily see the uniqueness of A.

Case 4. If φ satisfies the condition ( ), the proof is analogous to that of Case 3. An induction argument on (3.9) implies
f o ( x ) 3 n f o ( x 3 n ) k = 0 n 1 3 k 6 [ φ ( x 3 k , x 3 k , x 3 k ) + φ ( x 3 k , x 3 k , x 3 k ) ]
(3.12)
for all n N and x X . It follows from (3.12) that { 3 n f ( 3 n x ) } is a Cauchy sequence for all x X . Thus we can define a mapping A : X Y by
A ( x ) : = lim n 3 n f o ( x 3 n ) = lim n 3 n 2 [ f ( x 3 n ) f ( x 3 n ) ]
for all x X . Note that A ( x ) = A ( x ) for all x X . Also, from the condition ( ) and (3.8), we verify that A satisfies
9 A ( x + y + z 3 ) + 4 [ A ( x y 2 ) + A ( y z 2 ) + A ( z x 2 ) ] = 3 [ A ( x ) + A ( y ) + A ( z ) ]
for all x , y , z X . According to Lemma 2.2, the mapping A satisfies (1.4). Taking the limit as n in (3.12), we have
f o ( x ) A ( x ) 1 6 [ Ψ 2 ( x , x , x ) + Ψ 2 ( x , x , x ) ]

for all x X . Similarly, we can show the uniqueness of A. □

From the theorem above, we have the following corollary immediately.

Corollary 3.2 Let p 1 , p 2 and ϵ 0 be real numbers. Suppose that a mapping f : X Y satisfies
D f ( x , y , z ) ϵ ( x p + y p + z p )
for all x , y , z X ( x , y , z X { 0 } if p < 0 ). Then for each three cases p < 1 , 1 < p < 2 and p > 2 , there exist a unique quadratic mapping Q : X Y satisfying (1.3) and an additive mapping A : X Y satisfying (1.4) such that
f ( x ) Q ( x ) A ( x ) + 1 2 f ( 0 ) 3 p ϵ ( 3 | 9 3 p | + 1 | 3 3 p | ) x p , f ( x ) + f ( x ) 2 Q ( x ) + 1 2 f ( 0 ) 3 p + 1 ϵ | 9 3 p | x p ,
and
f ( x ) f ( x ) 2 A ( x ) 3 p ϵ | 3 3 p | x p

for all x X ( x X { 0 } if p < 0 ), where f ( 0 ) = 0 if p > 1 .

Proof Let φ ( x , y , z ) : = ϵ ( x p + y p + z p ) for all x , y , z X . Then φ ( x , x x ) = φ ( x , x , x ) = 3 ϵ x p for all x X ( x X { 0 } if p < 0 ). If p < 2 , the mapping φ satisfies ( ). Thus, we figure out
1 2 [ Φ 1 ( x , x , x ) + Φ 1 ( x , x , x ) ] = k = 1 3 ϵ x p 3 ( p 2 ) k = ϵ x p 3 p + 1 9 3 p
for all x X ( x X { 0 } if p < 0 ). If p > 2 , the mapping φ satisfies (). Thus, we have
1 2 [ Φ 2 ( x , x , x ) + Φ 2 ( x , x , x ) ] = k = 0 3 ϵ x p 3 ( 2 p ) k = ϵ x p 3 p + 1 3 p 9
for all x X . If p < 1 , the mapping φ satisfies ( ). Thus, we get
1 2 [ Ψ 1 ( x , x , x ) + Ψ 1 ( x , x , x ) ] = k = 1 ϵ x p 3 ( p 1 ) k = ϵ x p 3 p 3 3 p
for all x X ( x X { 0 } if p < 0 ). If p > 1 , the mapping φ satisfies ( ). Thus, we obtain
1 2 [ Ψ 2 ( x , x , x ) + Ψ 2 ( x , x , x ) ] = k = 0 ϵ x p 3 ( 1 p ) k = ϵ x p 3 p 3 p 3
for all x X . Therefore, we have
f ( x ) Q ( x ) A ( x ) + 1 2 f ( 0 ) { 3 p ϵ x p ( 3 3 p 9 + 1 3 p 3 ) if  p > 2 , 3 p ϵ x p ( 3 9 3 p + 1 3 p 3 ) if  1 < p < 2 , 3 p ϵ x p ( 3 9 3 p + 1 3 3 p ) if  p < 1

for all x X ( x X { 0 } if p < 0 ). □

Corollary 3.3 Let ϵ 0 be a real number. Suppose that a mapping f : X Y satisfies
D f ( x , y , z ) ϵ
for all x , y , z X . Then there exist a unique quadratic mapping Q : X Y satisfying (1.3) and an additive mapping A : X Y satisfying (1.4) such that
f ( x ) Q ( x ) A ( x ) + 1 2 f ( 0 ) 7 24 ϵ , f ( x ) + f ( x ) 2 Q ( x ) + 1 2 f ( 0 ) 1 8 ϵ ,
and
f ( x ) f ( x ) 2 A ( x ) 1 6 ϵ

for all x X .

Declarations

Authors’ Affiliations

(1)
Hana Academy Seoul, Seoul, Republic of Korea

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