1. Introduction and preliminaries

In 1940, Ulam [1] proposed the general stability problem: Let G1 be a group, G2 be a metric group with the metric d. Given ε > 0, does there exists δ > 0 such that if a function h: G1 G2 satisfies the inequality

d h ( x y ) - h ( x ) h ( y ) < δ , ( x , y G 1 ) ,

then there is a homomorphism H: G 1 G 2 with

d ( h ( x ) , H ( x ) ) < ε , ( x G 1 ) ?

Hyers [2] gave a partial affirmative answer to the question of Ulam in the context of Banach spaces. In 1950, Aoki [3] extended the theorem of Hyers by considering the unbounded cauchy difference inequality

f ( x + y ) - f ( x ) - f ( y ) ε ( x p + y p ) ( ε > 0 , p ε [ 0 , 1 ) ) .

In 1978, Rassias [4] also generalized the Hyers' theorem for linear mappings under the assumption tf (tx) is continuous in t for each fixed x.

Recently, Adam and Czerwik [5] investigated the problem of the Hyers-Ulam stability of a generalized quadratic functional equation in linear topological spaces. Najati and Moghimi [6] investigated the Hyers-Ulam stability of the functional equation

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

in quasi-Banach spaces. In this article, we prove that the Pexiderized functional equation

f ( 2 x + y ) + f ( 2 x - y ) = g ( x + y ) + g ( x - y ) + 2 g ( 2 x ) - 2 g ( x )

is stable for functions f, g defined on an abelian group and taking values in a topological vector space.

Throughout this article, let G be an abelian group and X be a sequentially complete Hausdorff topological vector space over the field ℚ of rational numbers.

A mapping f: GX is said to be quadratic if and only if it satisfies the following functional equation

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

for all x yG. A mapping f GX is said to be additive if and only if it satisfies f (x + y) = f (x) + f (y) for all x y ε G. For a given f: GX, we will use the following notation

D f ( x , y ) : = f ( 2 x + y ) + f ( 2 x - y ) - f ( x + y ) - f ( x - y ) - 2 f ( 2 x ) + 2 f ( x ) .

For given sets A BX and a number k ∈ ℝ, we define the well known operations

A + B : = { a + b : a A , b B } , k A : = { k a : a A } .

We denote the convex hull of a set UX by conv(U) and by U ¯ the sequential closure of U. Moreover it is well know that:

  1. (1)

    If AX are bounded sets, then conv(A) and A ¯ are bounded subsets of X.

  2. (2)

    If A, BX and α β ∈ ℝ, then α conv(A) + β conv(B) = conv(αA + βB).

  3. (3)

    Let X1 and X2 be linear spaces over ℝ. If f: X1 X2 is a additive (quadratic) function, then f (rx) = rf (x) (f (rx) = r2f (x)), for all xX1 and all r ∈ ℚ.

2. Main results

We start with the following lemma.

Lemma 2.1. Let G be a 2-divisible abelian group and BX be a nonempty set. If the functions f, g: G X satisfy

f ( 2 x + y ) + f ( 2 x - y ) - g ( x + y ) - g ( x - y ) - 2 g ( 2 x ) + 2 g ( x ) B
(2.1)

for all x, yG, then

D f ( x , y ) 2  conv ( B - B ) ,
(2.2)
D g ( x , y ) conv ( B - B )
(2.3)

for all x, yG.

Proof. Putting y = 0 in (2.1), we get

2 f ( 2 x ) - 2 g ( 2 x ) B
(2.4)

for all xG. If we replace x by 1 2 x in (2.4), then we have

f ( x ) - g ( x ) 1 2 B
(2.5)

for all xG. It follows from (2.5) and (2.1) that

D f ( x , y ) = f ( 2 x + y ) + f ( 2 x - y ) - g ( x + y ) - g ( x - y ) - 2 g ( 2 x ) + 2 g ( x ) - [ f ( x + y ) - g ( x + y ) ] - [ f ( x - y ) - g ( x - y ) ] - [ 2 f ( 2 x ) - 2 g ( 2 x ) ] + [ 2 f ( x ) - 2 g ( x ) ] 2  conv ( B - B ) .

Moreover, we have

D g ( x , y ) = f ( 2 x + y ) + f ( 2 x - y ) - g ( x + y ) - g ( x - y ) - 2 g ( 2 x ) + 2 g ( x ) - [ f ( 2 x + y ) - g ( 2 x + y ) ] - [ f ( 2 x - y ) - g ( 2 x - y ) ] conv ( B - B ) .

Theorem 2.2. Let G be a 2-divisible abelian group and BX be a bounded set. Suppose that the odd functions f, g: G X satisfy (2 1) for all x, yG. Then there exists exactly one additive function A : G X such that

A ( x ) - f ( x ) 4 conv ( B - B ) ¯ , A ( x ) - g ( x ) 2 conv ( B - B ) ¯
(2.6)

for all xG. Moreover the function A is given by

A ( x ) = lim n 1 2 n f ( 2 n x ) = lim n 1 2 n g ( 2 n x )

for all xG. Moreover, the convergence of the sequences are uniform on G.

Proof. By Lemma 2.1, we get (2.2). Setting y = x, y = 3x and y = 4x in (2.2), we get

f ( 3 x ) - 3 f ( 2 x ) + 3 f ( x ) 2  conv ( B - B ) ,
(2.7)
f ( 5 x ) - f ( 4 x ) - f ( 2 x ) + f ( x ) 2 conv ( B - B ) ,
(2.8)
f ( 6 x ) - f ( 5 x ) + f ( 3 x ) - 3 f ( 2 x ) + 2 f ( x ) 2 conv ( B - B )
(2.9)

for all xG. It follows from (2.7), (2.8), and (2.9) that

f ( 6 x ) - f ( 4 x ) - f ( 2 x ) 6 conv ( B - B )

for all xG. So

f ( 3 x ) - f ( 2 x ) - f ( x ) 6 conv ( B - B )
(2.10)

for all xG. Using (2.7) and (2.10), we obtain

1 2 f ( 2 x ) - f ( x ) 2 conv ( B - B )

for all xG. Therefore

1 2 n f ( 2 n x ) - 1 2 m f ( 2 m x ) = k = m n - 1 1 2 k + 1 f ( 2 k + 1 x ) - 1 2 k f ( 2 k x ) k = m n - 1 2 2 k conv ( B - B ) 4 2 m conv ( B - B )
(2.11)

for all xG and all integers n > m ≥ 0. Since B is bounded, we conclude that conv(B - B) is bounded. It follows from (2.11) and boundedness of the set conv(B - B) that the sequence { 1 2 n f ( 2 n x ) } is (uniformly) Cauchy in X for all xG. Since X is a sequential complete topological vector space, the sequence { 1 2 n f ( 2 n x ) } is convergent for all xG, and the convergence is uniform on G. Define

A 1 : G X , A 1 ( x ) : = lim n 1 2 n f ( 2 n x ) .

Since conv(B - B) is bounded, it follows from (2.2) that

D A 1 ( x , y ) = lim n 1 2 n D f ( 2 n x , 2 n y ) = 0

for all x yG. So A 1 is additive (see [6]). Letting m = 0 and n ∞ in (2.11), we get

A 1 ( x ) - f ( x ) 4 conv ( B - B ) ¯
(2.12)

for all xG. Similarly as before applying (2.3) we have an additive mapping A 2 : G X defined by A 2 ( x ) := lim n 1 2 n g ( 2 n x ) which is satisfying

A 2 ( x ) - g ( x ) 2 conv ( B - B ) ¯
(2.13)

for all xG. Since B is bounded, it follows from (2.5) that A 1 = A 2 . Letting A : = A 1 , we obtain (2.6) from (2.12) and (2.13).

To prove the uniqueness of A , suppose that there exists another additive function A : G X satisfying (2.6). So

A ( x ) - A ( x ) = [ A ( x ) - f ( x ) ] + [ f ( x ) - A ( x ) ] 8 conv ( B - B ) ¯

for all xG. Since A and A are additive, replacing x by 2nx implies that

A ( x ) - A ( x ) 8 2 n conv ( B - B ) ¯

for all xG and all integers n. Since conv ( B - B ) ¯ is bounded, we infer A = A . This completes the proof of theorem.

Theorem 2.3 Let G be a 2, 3-divisible abelian group and BX be a bounded set. Suppose that the even functions f, g: G X satisfy (2 1) for all x, yG. Then there exists exactly one quadratic function Q : G X such that

Q ( x ) - f ( x ) + f ( 0 ) 4 conv ( B - B ) ¯ , Q ( x ) - g ( x ) + g ( 0 ) 2 conv ( B - B ) ¯

for all xG. Moreover, the function Q is given by

Q ( x ) = lim n 1 4 n f ( 2 n x ) = lim n 1 4 n g ( 2 n x )

for all xG. Moreover, the convergence of the sequences are uniform on G.

Proof. By replacing y by x + y in (2.2), we get

f ( 3 x + y ) + f ( x - y ) - f ( 2 x + y ) - f ( y ) - 2 f ( 2 x ) + 2 f ( x ) 2 c o n v ( B - B )
(2.14)

for all x, yG. Replacing y by - y in (2.14), we get

f ( 3 x - y ) + f ( x + y ) - f ( 2 x - y ) - f ( y ) - 2 f ( 2 x ) + 2 f ( x ) 2 conv ( B - B )
(2.15)

for all x, yG. It follows from (2.2), (2.14), and (2.15) that

f ( 3 x + y ) + f ( 3 x - y ) - 2 f ( y ) - 6 f ( 2 x ) + 6 f ( x ) 6 conv ( B - B )
(2.16)

for all x, yG. By letting y = 0 and y = 3x in (2.16), we get

2 f ( 3 x ) - 6 f ( 2 x ) + 6 f ( x ) - 2 f ( 0 ) 6 conv ( B - B ) ,
(2.17)
f ( 6 x ) - 2 f ( 3 x ) - 6 f ( 2 x ) + 6 f ( x ) + f ( 0 ) 6 conv ( B - B )
(2.18)

for all xG. Using (2.17) and (2.18), we obtain

f ( 6 x ) - 4 f ( 3 x ) + 3 f ( 0 ) 12 conv ( B - B )
(2.19)

for all xG. If we replace x by 1 3 x in (2.19), then

f ( 2 x ) - 4 f ( x ) + 3 f ( 0 ) 12 conv ( B - B )

for all xG. Therefore

1 4 n + 1 f ( 2 n + 1 x ) - 1 4 n f ( 2 n x ) + 3 4 n + 1 f ( 0 ) 3 4 n conv ( B - B )
(2.20)

for all xG and all integers n. So

1 4 n f ( 2 n x ) - 1 4 m f ( 2 m x ) = k = m n - 1 1 4 k + 1 f ( 2 k + 1 x ) - 1 4 k f ( 2 k x ) - k = m n - 1 3 4 k + 1 f ( 0 ) + k = m n - 1 3 4 k conv ( B - B ) - k = m n - 1 3 4 k + 1 f ( 0 ) + 1 4 m - 1 conv ( B - B )
(2.21)

for all xG and all integers n >m ≥ 0. It follows from (2.21) and boundedness of the set conv(B - B) that the sequence { 1 4 n f ( 2 n x ) } is (uniformly) Cauchy in X for all xG. The rest of the proof is similar to proof of of Theorem 2.2.

Remark 2.4. If the functions f, g: G X satisfy (2.1), where f is even (odd) and g is odd (even), then it is easy to show that f and g are bounded.