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

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Abstract

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

9f ( 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.

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 ϵ0, 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 ) +(n2) 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(2x+y)+f(2xy)=f(x+y)+f(xy)+2f(2x)2f(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:XY satisfies (1.2) if and only if there exist a quadratic mapping Q:XY satisfying

Q(x+y)+Q(xy)=2Q(x)+2Q(y)
(1.3)

and an additive mapping A:XY satisfying

A(x+y)=A(x)+A(y)
(1.4)

such that

f(x)=Q(x)+A(x)

for all xX. 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:XY satisfies (1.2) for all x,yX, 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

9f ( y 3 ) +4 [ f ( x y 2 ) + f ( x + y 2 ) + f ( x ) ] =6f(x)+3f(y)
(2.1)

for all x,yX. Replacing y with x in (2.1) gives

9f ( x 3 ) =f(x)
(2.2)

for all xX. Putting y=0 in (2.1), we have

4f ( x 2 ) =f(x)
(2.3)

for all xX. Using (2.2) and (2.3), we can rewrite (1.2) as

f(x+y+z)+f(xy)+f(yz)+f(zx)=3 [ f ( x ) + f ( y ) + f ( z ) ]
(2.4)

for all x,y,zX. Putting z=0 in (2.4), we obtain

f(x+y)+f(xy)=2f(x)+2f(y)

for all x,yX. □

Lemma 2.2 If an odd mapping f:XY satisfies (1.2) for all x,yX, 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

9f ( y 3 ) +4 [ f ( x y 2 ) + f ( x + y 2 ) f ( x ) ] =3f(y)
(2.5)

for all x,yX. Replacing y by x in (2.5) gives

9f ( x 3 ) =3f(x)
(2.6)

for all xX. Putting y=0 in (2.5), we have

2f ( x 2 ) =f(x)
(2.7)

for all xX. It follows from (2.5), (2.6) and (2.7) that

f ( x + y 2 ) +f ( x y 2 ) =2f(x)
(2.8)

for all x,yX. Replacing x with x+y and y with xy in (2.8), we obtain

f(x+y)=f(x)+f(y)

for all x,yX. □

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

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

f(x)=B(x,x)+A(x)

for all xX.

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 xX. 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,zX. 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,zX. 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:XY satisfies

D f ( x , y , z ) φ(x,y,z)
(3.1)

for all x,y,zX. Then there exist a unique quadratic mapping Q:XY satisfying (1.3) and an additive mapping A:XY 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 xX 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 xX.

Proof We first consider the even part of f. Let f e :XY be a function defined by f e (x):= f ( x ) + f ( x ) 2 for all xX. 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,zX. 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 xX, 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 xX. 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 nN and xX. 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,mN with n>m and xX. 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 xX and thus converges by the completeness of Y. Therefore, we can define a mapping Q:XY 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 xX. Note that Q(0)=0, Q(x)=Q(x) for all xX. It follows from the condition () and (3.2) that Q satisfies

9Q ( 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,zX. 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 xX. 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 xX. 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 nN and xX. Taking the limit as n, we conclude that Q(x)= Q (x) for all xX.

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 nN and xX. 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 xX. Thus we can define a mapping Q:XY 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 xX. Note that Q(0)=0 and Q(x)=Q(x) for all xX. From the condition () and (3.2), we see that Q satisfies

9Q ( 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,zX. 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 xX. The rest of the proof is similar to that of Case 1.

Next, we consider the odd part of f. Now, let f o :XY be a function defined by f o (x):= 1 2 [f(x)f(x)] for all xX. 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,zX. 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 xX.

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 xX. 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 nN and xX. From (3.11) we can show that the sequence {f( 3 n x)/ 3 n } is a Cauchy sequence for all xX. Define a mapping A:XY 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 xX. By the oddness of f, we see that A(x)=A(x) for all xX. Also, from the condition () and (3.9), we verify that A satisfies

9A ( 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,zX. 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 xX. 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 nN and xX. It follows from (3.12) that { 3 n f( 3 n x)} is a Cauchy sequence for all xX. Thus we can define a mapping A:XY 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 xX. Note that A(x)=A(x) for all xX. Also, from the condition () and (3.8), we verify that A satisfies

9A ( 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,zX. 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 xX. Similarly, we can show the uniqueness of A. □

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

Corollary 3.2 Let p1, p2 and ϵ0 be real numbers. Suppose that a mapping f:XY satisfies

D f ( x , y , z ) ϵ ( x p + y p + z p )

for all x,y,zX (x,y,zX{0} if p<0). Then for each three cases p<1, 1<p<2 and p>2, there exist a unique quadratic mapping Q:XY satisfying (1.3) and an additive mapping A:XY 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 xX (xX{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,zX. Then φ(x,xx)=φ(x,x,x)=3ϵ x p for all xX (xX{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 3ϵ x p 3 ( p 2 ) k =ϵ x p 3 p + 1 9 3 p

for all xX (xX{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 3ϵ x p 3 ( 2 p ) k =ϵ x p 3 p + 1 3 p 9

for all xX. 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 xX (xX{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 xX. 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 xX (xX{0} if p<0). □

Corollary 3.3 Let ϵ0 be a real number. Suppose that a mapping f:XY satisfies

D f ( x , y , z ) ϵ

for all x,y,zX. Then there exist a unique quadratic mapping Q:XY satisfying (1.3) and an additive mapping A:XY 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 xX.

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Correspondence to Young-Su Lee.

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The first author discovered the topic and the main ideas for the proof of the paper and made the actual writing. All authors discussed the paper together. The second and the third authors discovered some helpful ideas for the proof of this paper and checked the proof of the paper. All authors read and approved the final manuscript.

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Keywords

  • quadratic functional equation
  • additive functional equation
  • mixed type functional equation
  • stability