1 Introduction

In 1940, Ulam [31] raised a question concerning the stability of group homomorphisms as follows: The case of approximately additive mappings was solved by Hyers [16] under the assumption that G 2 is a Banach space. In 1978, Rassias [25] generalized Hyers’ result to the unbounded Cauchy difference.

Let G 1 be a group and let G 2 be a metric group with the metric d(,). Given ϵ>0, does there exist a δ>0 such that if a function h: G 1 G 2 satisfies the inequality d(h(xy),h(x)h(y))<δ for all x,y G 1 , then there exists a homomorphism H: G 1 G 2 with d(h(x),H(x))<ϵ for all x G 1 ?

During the last decades, stability problems of various functional equations have been extensively studied and generalized by a number of authors (see [13, 14, 17, 19, 24, 27, 30]). In particular, the stability problem of the following quadratic functional equation

f(x+y)+f(xy)=2f(x)+2f(y)
(1.1)

was proved by Skof [29]. Thereafter, many authors studied the stability problems of (1.1) in various settings (see [3, 4, 12, 18]). Usually, quadratic functional equations are used to characterize the inner product spaces. Note that a square norm on an inner product space satisfies the parallelogram equality

x + y 2 + x y 2 =2 x 2 +2 y 2

for all vectors x, y. By virtue of this equality, the quadratic functional equation (1.1) is induced. It is well known that a function f between real vector spaces satisfies (1.1) if and only if there exists a unique symmetric biadditive function B such that f(x)=B(x,x) (see [1, 13, 17, 19, 27]). The biadditive function B is given by

B(x,y)= 1 4 ( f ( x + y ) f ( x y ) ) .

Recently, Lee et al. [21] introduced the following quadratic functional equation which is equivalent to (1.1):

f(kx+y)+f(kxy)=2 k 2 f(x)+2f(y),
(1.2)

where k is a fixed positive integer. They proved the Hyers-Ulam-Rassias stability of this equation in Banach spaces. Wang [32] considered the intuitionistic fuzzy stability of (1.2) by using the fixed-point alternative. Saadati and Park [26] proved the Hyers-Ulam-Rassias stability of (1.2) in non-Archimedean L-fuzzy normed spaces.

In this paper, we solve the general solution and the stability problem of (1.2) in the spaces of generalized functions such as S of tempered distributions and F of Fourier hyperfunctions. Using pullbacks, Chung and Lee [8] reformulated (1.1) as the equation for generalized functions and proved that every solution of (1.1) in S (or F , resp.) is a quadratic form. Also, Chung [7, 11] proved the stability problem of (1.1) in the spaces S and F . Making use of the similar methods as in [711, 22], we reformulate (1.2) and the related inequality in the spaces of generalized functions as follows:

(1.3)
(1.4)

where A, B, P, and Q are the functions defined by

A(x,y)=kx+y,B(x,y)=kxy,P(x,y)=x,Q(x,y)=y.

Here, ∘ denotes the pullback of generalized functions and the inequality vϵ in (1.4) means that |v,φ|ϵ φ L 1 for all test functions φ. We refer to [15] for pullbacks and to [2, 711] for more details of the spaces of generalized functions.

As results, we shall prove that every solution u in S (or F , resp.) of Eq. (1.3) is a quadratic form

u= 1 i j n a i j x i x j ,

where a i j C. Also, we shall prove that every solution u in S (or F , resp.) of the inequality (1.4) can be written uniquely in the form

u= 1 i j n a i j x i x j +μ(x),

where μ is a bounded measurable function such that

μ L { ϵ 2 , k = 1 , ( k 2 + 1 ) ϵ 2 k 2 ( k 2 1 ) , k 2 .

2 Preliminaries

In this section, we introduce the spaces of tempered distributions and Fourier hyperfunctions. Here, we use the n-dimensional notations. If α=( α 1 ,, α n ) N 0 n , where N 0 n is the set of nonnegative integers, then |α|= α 1 ++ α n , α!= α 1 ! α n !. For x=( x 1 ,, x n ) R n , we denote x α = x 1 α 1 x n α n and α = ( / x 1 ) α 1 ( / x n ) α n .

2.1 Tempered distributions

We present a very useful space of test functions for the tempered distributions as follows.

Definition 2.1 ([15, 28])

An infinitely differentiable function φ in R n is called rapidly decreasing if

φ α , β = sup x R n | x α β φ ( x ) | <
(2.1)

for all α,β N 0 n . The vector space of such functions is denoted by S( R n ). A linear functional u on S( R n ) is said to be a tempered distribution if there exists the constant C0 and the nonnegative integer N such that

| u , φ | C | α | , | β | N sup x R n | x α β φ |

for all φS( R n ). The set of all tempered distributions is denoted by S ( R n ).

We note that, if φS( R n ), then each derivative of φ decreases faster than | x | N for all N>0 as |x|. It is easy to see that the function φ(x)=exp(a | x | 2 ), where a>0 belongs to S( R n ), but ψ(x)= ( 1 + | x | 2 ) 1 is not a member of S( R n ). It is known from [5] that (2.1) is equivalent to

sup x R n | x α φ ( x ) | <, sup ξ R n | ξ β φ ˆ ( ξ ) | <

for all α,β N 0 n , where φ ˆ is the Fourier transform of φ.

For example, every polynomial p(x)= | α | m a α x α , where a α C, defines a tempered distribution by

p ( x ) , φ = R n p(x)φ(x)dx,φS ( R n ) .

Note that tempered distributions are generalizations of L p -functions. These are very useful for the study of Fourier transforms in generality, since all tempered distributions have a Fourier transform.

2.2 Fourier hyperfunctions

Imposing the growth condition on α , β in (2.1) Sato and Kawai introduced the new space of test functions for the Fourier hyperfunctions as follows.

Definition 2.2 ([6])

We denote by F( R n ) the set of all infinitely differentiable functions φ in R n such that

φ A , B = sup x , α , β | x α β φ ( x ) | A | α | B | β | α ! β ! <
(2.2)

for some positive constants A, B depending only on φ. The strong dual of F( R n ), denoted by F ( R n ), is called the Fourier hyperfunction.

It can be verified that the seminorm (2.2) is equivalent to

φ h , k = sup x , α | α φ ( x ) | exp k | x | h | α | α ! <

for some constants h,k>0. Furthermore, it is shown in [6] that (2.2) is equivalent to

sup x R n | φ ( x ) | expk|x|<, sup ξ R n | φ ˆ ( ξ ) | exph|ξ|<

for some h,k>0.

Fourier hyperfunctions were introduced by Sato in 1958. The space F ( R n ) is a natural generalization of the space S ( R n ) and can be thought informally as distributions of a infinite order. Observing the above growth conditions, we can easily see the following topological inclusions:

F ( R n ) S ( R n ) , S ( R n ) F ( R n ) .

3 General solution in generalized functions

In order to solve the general solution of (1.3), we employ the n-dimensional heat kernel, fundamental solution of the heat equation,

E t (x)={ ( 4 π t ) n / 2 exp ( | x | 2 / 4 t ) , x R n , t > 0 , 0 , x R n , t 0 .

Since for each t>0, E t () belongs to the space F( R n ), the convolution

u ˜ (x,t)=(u E t )(x)= u y , E t ( x y )

is well defined for all u in F ( R n ), which is called the Gauss transform of u. Subsequently, the semigroup property

( E t E s )(x)= E t + s (x)

of the heat kernel is very useful to convert Eq. (1.3) into the classical functional equation defined on upper-half plane. We also use the following famous result, the so-called heat kernel method, which is stated as follows.

Theorem 3.1 ([23])

Let u S ( R n ). Then its Gauss transform u ˜ is a C -solution of the heat equation

(/tΔ) u ˜ (x,t)=0

satisfying

  1. (i)

    There exist positive constants C, M, and N such that

    | u ˜ ( x , t ) | C t M ( 1 + | x | ) N in R n ×(0,δ).
    (3.1)
  2. (ii)

    u ˜ (x,t)u as t 0 + in the sense that for every φS( R n ),

    u,φ= lim t 0 + u ˜ (x,t)φ(x)dx.

Conversely, every C -solution U(x,t) of the heat equation satisfying the growth condition (3.1) can be uniquely expressed as U(x,t)= u ˜ (x,t) for some u S ( R n ).

Similarly, we can represent Fourier hyperfunctions as a special case of the results as in [20]. In this case, the estimate (3.1) is replaced by the following:

For every ϵ>0, there exists a positive constant C ϵ such that

| u ˜ ( x , t ) | C ϵ exp ( ϵ ( | x | + 1 / t ) ) in  R n ×(0,δ).

Here, we need the following lemma to solve the general solution of (1.3).

Lemma 3.2 Suppose that f: R n ×(0,)C is a continuous function satisfying the equation

f ( k x + y , k 2 t + s ) +f ( k x y , k 2 t + s ) =2 k 2 f(x,t)+2f(y,s)
(3.2)

for all x,y R n , t,s>0. Then the solution f is the quadratic-additive function

f(x,t)= 1 i j n a i j x i x j +bt

for some a i j ,bC.

Proof Define a function h: R n ×(0,)C as h(x,t):=f(x,t)f(0,t). We immediately have h(0,t)=0 and

h ( k x + y , k 2 t + s ) +h ( k x y , k 2 t + s ) =2 k 2 h(x,t)+2h(y,s)
(3.3)

for all x,y R n , t,s>0. Putting y=0 in (3.3) yields

h ( k x , k 2 t + s ) = k 2 h(x,t)
(3.4)

for all x R n , t,s>0. Letting s 0 + in (3.4) gives

h ( k x , k 2 t ) = k 2 h(x,t)
(3.5)

for all x R n , t>0. Replacing s by k 2 s in (3.4) and then using (3.5), we obtain

h(x,t+s)=h(x,t)

for all x R n , t,s>0. This shows that h(x,t) is independent with respect to the second variable. Thus, we see that H(x):=h(x,t) satisfies (1.2). Using the induction argument on the dimension n, we verify that every continuous solution of (1.2) in R n is a quadratic form

H(x)=h(x,t)= 1 i j n a i j x i x j ,

where a i j C.

On the other hand, putting x=y=0 in (3.2) yields

f ( 0 , k 2 t + s ) = k 2 f(0,t)+f(0,s)
(3.6)

for all t,s>0. In view of (3.6), we verify that lim s 0 + f(0,s)=0 and

f ( 0 , k 2 t ) = k 2 f(0,t)
(3.7)

for all t>0. It follows from (3.6) and (3.7) that we see that f(0,t) satisfies the Cauchy functional equation

f(0,t+s)=f(0,t)+f(0,s)

for all t,s>0. Given the continuity, we have

f(0,t)=bt

for some bC. Therefore, we finally obtain

f(x,t)=h(x,t)+f(0,t)= 1 i j n a i j x i x j +bt

for all x R n , t>0. □

As a direct consequence of the above lemma, we present the general solution of the quadratic functional equation (1.3) in the spaces of generalized functions.

Theorem 3.3 Every solution u in S ( R n ) (or F ( R n ), resp.) of Eq. (1.3) is the quadratic form

u= 1 i j n a i j x i x j

for some a i j C.

Proof Convolving the tensor product E t (ξ) E s (η) of n-dimensional heat kernels in both sides of (1.3), we have

[ ( u A ) ( E t ( ξ ) E s ( η ) ) ] ( x , y ) = u A , E t ( x ξ ) E s ( y η ) = u ξ , k n E t ( x ξ η k ) E s ( y η ) d η = u ξ , k n E t ( k x + y ξ η k ) E s ( η ) d η = u ξ , E k 2 t ( k x + y ξ η ) E s ( η ) d η = u ξ , ( E k 2 t E s ) ( k x + y ξ ) = u ξ , E k 2 t + s ( k x + y ξ ) = u ˜ ( k x + y , k 2 t + s )

and similarly we get

Thus, (1.3) is converted into the classical functional equation

u ˜ ( k x + y , k 2 t + s ) + u ˜ ( k x y , k 2 t + s ) =2 k 2 u ˜ (x,t)+2 u ˜ (y,s)

for all x,y R n , t,s>0. We note that the Gauss transform u ˜ is a C function and so, by Lemma 3.2, the solution u ˜ is of the form

u ˜ (x,t)= 1 i j n a i j x i x j +bt
(3.8)

for some a i j ,bC. By the heat kernel method, we obtain

u= 1 i j n a i j x i x j

as t 0 + in (3.8). □

4 Stability in generalized functions

In this section, we are going to solve the stability problem of (1.4). For the case of k=1 in (1.4), the result is known as follows.

Theorem 4.1 ([7, 10])

Suppose that u in S ( R n ) (or F ( R n ), resp.) satisfies the inequality

uA+uB2uP2uQϵ.

Then there exists a unique quadratic form

T(x)= 1 i j n a i j x i x j

such that

u T ( x ) ϵ 2 .

We here need the following lemma to solve the stability problem of (1.4).

Lemma 4.2 Let k be a fixed positive integer with k2. Suppose that f: R n ×(0,)C is a continuous function satisfying the inequality

f ( k x + y , k 2 t + s ) + f ( k x y , k 2 t + s ) 2 k 2 f ( x , t ) 2 f ( y , s ) L ϵ.
(4.1)

Then there exist a unique function g(x,t) satisfying the quadratic-additive functional equation

g ( k x + y , k 2 t + s ) +g ( k x y , k 2 t + s ) =2 k 2 g(x,t)+2g(y,s)

such that

f ( x , t ) g ( x , t ) L k 2 + 1 2 k 2 ( k 2 1 ) ϵ.

Proof Putting x=y=0 in (4.1) yields

| f ( 0 , k 2 t + s ) k 2 f ( 0 , t ) f ( 0 , s ) | ϵ 2
(4.2)

for all t,s>0. In view of (4.2), we see that

c:= lim sup t 0 + f(0,t)

exists. Letting t= t n 0 + so that f(0, t n )c in (4.2) gives

|c| ϵ 2 k 2 .
(4.3)

Putting y=0 and letting s= s n 0 + so that f(0, s n )c in (4.1) we have

| f ( k x , k 2 t ) k 2 f ( x , t ) c | ϵ 2
(4.4)

for all x R n , t>0. Using (4.3), we can rewrite (4.4) as

| f ( k x , k 2 t ) k 2 f ( x , t ) | k 2 + 1 2 k 4 ϵ

for all x R n , t>0. By the induction argument yields

| f ( k n x , k 2 n t ) k 2 n f ( x , t ) | k 2 + 1 2 k 2 ( k 2 1 ) ϵ
(4.5)

for all nN, x R n , t>0. We claim that the sequence { k 2 n f( k n x, k 2 n t)} converges. Replacing x by k m x and t by k 2 m t in (4.5), respectively, where mn, we get

| f ( k m + n x , k 2 ( m + n ) t ) k 2 ( m + n ) f ( k m x , k 2 m t ) k 2 m | k 2 + 1 2 k 2 ( m + 1 ) ( k 2 1 ) ϵ.

Letting n, by Cauchy convergence criterion, we see that the sequence { k 2 n f( k n x, k 2 n t)} is a Cauchy sequence. We can now define a function h: R n ×(0,)C by

g(x,t):= lim n f ( k n x , k 2 n t ) k 2 n .

Letting n in (4.5) we obtain

f ( x , t ) g ( x , t ) L k 2 + 1 2 k 2 ( k 2 1 ) ϵ.
(4.6)

Replacing x, y, t, s by k n x, k n y, k 2 n t, k 2 n s in (4.1), dividing both sides by k 2 n and letting n we have

g ( k x + y , k 2 t + s ) +g ( k x y , k 2 t + s ) =2 k 2 g(x,t)+2g(y,s)
(4.7)

for all x,y R n , t,s>0. Next, we shall prove that g is unique. Suppose that there exists another function h: R n ×(0,)C such that h satisfies (4.6) and (4.7). Since g and h satisfy (4.7), we see from Lemma 3.2 that

g ( k n x , k 2 n t ) = k 2 n g(x,t),h ( k n x , k 2 n t ) = k 2 n h(x,t)

for all nN, x R n , t>0. One gets from (4.6) that

for all nN, x R n , t>0. Taking the limit as n, we conclude that g(x,t)=h(x,t) for all x R n , t>0. □

We now state and prove the main theorem of this paper.

Theorem 4.3 Suppose that u in S ( R n ) (or F ( R n ), resp.) satisfies the inequality (1.4). Then there exists a unique quadratic form

T(x)= 1 i j n a i j x i x j

such that

u T ( x ) { ϵ 2 , k = 1 , ( k 2 + 1 ) ϵ 2 k 2 ( k 2 1 ) , k 2 .

Proof As discussed above, it is done for the case of k=1. We assume that k is a fixed-positive integer with k2. Convolving the tensor product E t (ξ) E s (η) of n-dimensional heat kernels in both sides of (1.4), we have

u ˜ ( k x + y , k 2 t + s ) + u ˜ ( k x y , k 2 t + s ) 2 k 2 u ˜ ( x , t ) 2 u ˜ ( y , s ) L ϵ.

By Lemma 4.2, there exists a unique function g(x,t) satisfying the quadratic-additive functional equation

g ( k x + y , k 2 t + s ) +g ( k x y , k 2 t + s ) =2 k 2 g(x,t)+2g(y,s)

such that

u ˜ ( x , t ) g ( x , t ) L k 2 + 1 2 k 2 ( k 2 1 ) ϵ.
(4.8)

It follows from Lemma 3.2 that g(x,t) is of the form

g(x,t)= 1 i j n a i j x i x j +bt

for some a i j ,bC. Letting t 0 + in (4.8), we have

u 1 i j n a i j x i x j k 2 + 1 2 k 2 ( k 2 1 ) ϵ.

This completes the proof. □

Remark 4.4 The resulting inequality in Theorem 4.3 implies that uT(x) is a measurable function. Thus, all of the solution u in S ( R n ) (or F ( R n ), resp.) can be written uniquely in the form

u=T(x)+μ(x),

where

μ L { ϵ 2 , k = 1 , ( k 2 + 1 ) ϵ 2 k 2 ( k 2 1 ) , k 2 .