On a functional equation on groups

ON A FUNCTIONAL EQUATION ON GROUPS

Jukang K. Chung, Soon-Mo Jung, and Prasanna K. Sahoo

Abstract. We present the general solution of the functional equation f (x

y

, x

y

) + f (x

y

, x

) + f (x

, x

y

)

= f (x

y

, x

y

) + f (x

y

, x

) + f (x

, x

y

).

Furthermore, we also prove the Hyers-Ulam stability of the above func- tional equation.

1. Introduction

In a paper of Kannappan and Sahoo [8], it was shown that if the qua- dratic polynomial f (x, y) = ax ² + bx + cy ² + dy + exy + α with a ̸= 0 and c ̸= 0, is a solution of the functional dierential equation

f (x + h, y + k) − f(x, y)

= hf _x (x + θh, y + θk) + kf _y (x + θh, y + θk), 0 < θ < 1,

for all x, y, h, k ∈ R (the set of reals) with h ² + k ² ̸= 0, then θ = 1/2.

Conversely, if a function f satises the above equation with θ = 1/2, then the only solution is a quadratic polynomial. Here f _x and f _y represent the rst partial derivatives of f with respect to x and y, respectively.

To establish this result, Kannappan and Sahoo [8] used the following functional equation

f (x ₁ + y ₁ , x ₂ + y ₂ ) + f (x ₁ − y 1 , x ₂ ) + f (x ₁ , x ₂ − y 2 )

= f (x 1 − y 1 , x 2 − y 2 ) + f (x 1 + y 1 , x 2 ) + f (x 1 , x 2 + y 2 )

for all x 1 , x 2 , y 1 , y 2 ∈ R. In the present paper we solve the above functional equation on groups without any regularity assumption about f . We shall not assume the group to be abelian, so we write the above equation as

f (x 1 y 1 , x 2 y 2 ) + f (x 1 y ₁ ⁻¹ , x 2 ) + f (x 1 , x 2 y ₂ ⁻¹ )

= f (x ₁ y ⁻¹ ₁ , x ₂ y ₂ ⁻¹ ) + f (x ₁ y ₁ , x ₂ ) + f (x ₁ , x ₂ y ₂ ) (1)

Received July 4, 1999. Revised March 9, 2000.

2000 Mathematics Subject Classication: 39B32, 39B82.

Key words and phrases: functional equations, Hyers-Ulam stability.

for all x ₁ , x ₂ , y ₁ , y ₂ ∈ G. Throughout this paper, C will denote the set of complex numbers.

This paper is organized as follows. In Section 2, we give some preliminary results that will be used in solving the functional equation (1). In Section 3, we present the general solution of (1). In Section 4, we prove the Hyers- Ulam stability of the functional equations (2) and (3), and then apply these results to prove the stability of the functional equation (1).

2. Some preliminary results

The following lemma was given by Aczel, Chung and Ng [1] and can also be obtained from Chung, Ebanks, Ng and Sahoo [2].

Lemma 1. Let G be a group in which x ² = y has a solution for all x, y ∈ G. The general solution f : G → C of the functional equation (2) f (xy) + f (xy ⁻¹ ) = 2f (x) (x, y ∈ G)

satisfying also

f (xy) = f (yx) (x, y ∈ G) is given by

f (x) = ϕ(x) + a,

where a ∈ C is an arbitrary constant and ϕ an arbitrary homomorphism of G into the additive group ( C, +) of C.

Next, we generalize the above lemma. In fact, this lemma can also be deduced from Chung, Ebanks, Ng and Sahoo [2]. However, we give a short proof for the sake of completeness.

Lemma 2. Let G be a group in which x ² = y has a solution for all x, y ∈ G. The general solution, f 1 , f 2 , f 3 : G → C, of

(3) f ₁ (xy) + f ₂ (xy ⁻¹ ) = f ₃ (x) (x, y ∈ G) satisfying also

(4) f ₁ (xy) = f ₁ (yx), f ₂ (xy) = f ₂ (yx) (x, y ∈ G) are given by

(5) f 1 (x) = ϕ(x) + a, f 2 (x) = ϕ(x) + b, f 3 (x) = 2ϕ(x) + a + b, where a, b ∈ C are arbitrary constants and ϕ : G → C is a solution of

(6) ϕ(xy) = ϕ(x) + ϕ(y) (x, y ∈ G).

Proof. It is obvious that (5) with (6) satises (3) and (4). In order to prove the converse, we rst replace y by y ⁻¹ in (3) to obtain

(7) f 1 (xy ⁻¹ ) + f 2 (xy) = f 3 (x).

Setting y = e (the identity of G) in (7), we get (8) f ₁ (x) + f ₂ (x) = f ₃ (x).

Adding (3) and (7) and then using (8), we have

(9) f 1 (xy) + f 2 (xy) + f 1 (xy ⁻¹ ) + f 2 (xy ⁻¹ ) = 2(f 1 (x) + f 2 (x)).

Using Lemma 1, from (9) and (4) we obtain

(10) f ₁ (x) + f ₂ (x) = 2ϕ(x) + a + b.

Subtracting (7) from (3), we get

f 1 (xy) − f 2 (xy) = f 1 (xy ⁻¹ ) − f 2 (xy ⁻¹ ) for all x, y ∈ G. Hence

(11) f ₁ (x) − f 2 (x) = a − b.

From (10), (11) and (8) we obtain (5). This completes the proof of the lemma.

3. General solutions of Eq. (1)

Theorem 1. Let G be a group in which x ² = y has a solution for all x, y ∈ G. The general solution f : G ² → C of the functional equation (1) satisfying

(12) f (x 1 y 1 , x 2 ) = f (y 1 x 1 , x 2 ) and f (x 1 , x 2 y 2 ) = f (x 1 , y 2 x 2 ) for all x 1 , x 2 , y 1 , y 2 ∈ G, is given by

(13) f (x ₁ , x ₂ ) = A(x ₁ , x ₂ ) + α(x ₁ ) + β(x ₂ ),

where α, β : G → C are arbitrary functions satisfying α(x 1 x 2 ) = α(x 2 x 1 ) and β(x ₁ x ₂ ) = β(x ₂ x ₁ ) for all x ₁ , x ₂ ∈ G, and A : G ² → C is additive in each variable.

Proof. It is easy to check that (13) satises (1) and (12). In order to prove the converse, we rewrite (1) as

(14) f (x ₁ y ₁ , x ₂ y ₂ ) − f(x 1 y ₁ , x ₂ ) − (

f (x ₁ y ⁻¹ ₁ , x ₂ y ⁻¹ ₂ ) − f(x 1 y ₁ ⁻¹ , x ₂ ) )

= f (x 1 , x 2 y 2 ) − f(x 1 , x 2 y ₂ ⁻¹ )

and dene F _i : G ³ → C (i = 1, 2, 3) by

(15) F ₁ (x ₁ , x ₂ , y ₂ ) := f (x ₁ , x ₂ y ₂ ) − f(x 1 , x ₂ ), (16) F ₂ (x ₁ , x ₂ , y ₂ ) := −f(x 1 , x ₂ y ₂ ⁻¹ ) + f (x ₁ , x ₂ ), (17) F 3 (x 1 , x 2 , y 2 ) := f (x 1 , x 2 y 2 ) − f(x 1 , x 2 y ⁻¹ ₂ ).

The equation (14) can be written as

F ₁ (x ₁ y ₁ , x ₂ , y ₂ ) + F ₂ (x ₁ y ₁ ⁻¹ , x ₂ , y ₂ ) = F ₃ (x ₁ , x ₂ , y ₂ ).

From (12), (15) and (16), we get

F 1 (x 1 y 1 , x 2 , y 2 ) = F 1 (y 1 x 1 , x 2 , y 2 ), F ₂ (x ₁ y ₁ , x ₂ , y ₂ ) = F ₂ (y ₁ x ₁ , x ₂ , y ₂ ) for x ₁ , x ₂ , y ₁ , y ₂ ∈ G. So by Lemma 2, (15) and (16) we get (18) f (x 1 , x 2 y 2 ) − f(x 1 , x 2 ) = ϕ 1 (x 1 , x 2 , y 2 ) + b(x 2 , y 2 ) and

f (x ₁ , x ₂ y ₂ ⁻¹ ) − f(x 1 , x ₂ ) (19)

= −ϕ 1 (x 1 , x 2 , y 2 ) + a(x 2 , y 2 ) − b(x 2 , y 2 ), where ϕ 1 (x 1 , x 2 , y 2 ) satises

ϕ ₁ (x ₁ y ₁ , x ₂ , y ₂ ) = ϕ ₁ (x ₁ , x ₂ , y ₂ ) + ϕ ₁ (y ₁ , x ₂ , y ₂ ).

Adding (18) and (20) we get

(20) f (x 1 , x 2 y 2 ) + f (x 1 , x 2 y ⁻¹ ₂ ) − 2f(x 1 , x 2 ) = a(x 2 , y 2 ).

Setting x 1 = e in (20) we have

(21) a(x ₂ , y ₂ ) = β ₁ (x ₂ y ₂ ) + β ₁ (x ₂ y ₂ ⁻¹ ) − 2β 1 (x ₂ ),

where β ₁ (x ₂ ) := f (e, x ₂ ). Further, by (12), we also have β ₁ (x ₂ y ₂ ) = β 1 (y 2 x 2 ). Substituting (21) into (20), we get

F (x 1 , x 2 y 2 ) + F (x 1 , x 2 y ₂ ⁻¹ ) = 2F (x 1 , x 2 ),

where F (x ₁ , x ₂ ) := f (x ₁ , x ₂ ) − β 1 (x ₂ ) satises F (x ₁ , x ₂ y ₂ ) = F (x ₁ , y ₂ x ₂ ).

By Lemma 1, we obtain

F (x 1 , x 2 ) = ϕ 2 (x 1 , x 2 ) + α(x 1 ), that is

(22) f (x ₁ , x ₂ ) = ϕ ₂ (x ₁ , x ₂ ) + α(x ₁ ) + β ₁ (x ₂ ), where ϕ 2 (x 1 , x 2 ) satises

(23) ϕ ₂ (x ₁ , x ₂ y ₂ ) = ϕ ₂ (x ₁ , x ₂ ) + ϕ ₂ (x ₁ , y ₂ ).

Since f satises (12), using (22) we obtain

(24) ϕ ₂ (x ₁ y ₁ , x ₂ ) + α(x ₁ y ₁ ) = ϕ ₂ (y ₁ x ₁ , x ₂ ) + α(y ₁ x ₁ ) for all x ₁ , y ₁ , x ₂ ∈ G. Replacing x 2 by x ₂ y ₂ in (24), we get (25) ϕ ₂ (x ₁ y ₁ , x ₂ y ₂ ) + α(x ₁ y ₁ ) = ϕ ₂ (y ₁ x ₁ , x ₂ y ₂ ) + α(y ₁ x ₁ ).

Hence from (23) and (25), we see that

ϕ ₂ (x ₁ y ₁ , x ₂ ) + ϕ ₂ (x ₁ y ₁ , y ₂ ) + α(x ₁ y ₁ )

= ϕ 2 (y 1 x 1 , x 2 ) + ϕ 2 (y 1 x 1 , y 2 ) + α(y 1 x 1 ).

Thus the last equation together with (24) yields (26) ϕ 2 (x 1 y 1 , y 2 ) = ϕ 2 (y 1 x 1 , y 2 )

for all x 1 , y 1 , y 2 ∈ G. From (24) and (26), we see that α(x 1 x 2 ) = α(x 2 x 1 ) for all x ₁ , x ₂ ∈ G. Substituting (22) into (1) and using (23), we get (27) ϕ ₂ (x ₁ y ₁ , y ₂ ) + ϕ ₂ (x ₁ y ₁ ⁻¹ , y ₂ ) = 2ϕ ₂ (x ₁ , y ₂ ).

From Lemma 1, (27) and (26), we get

(28) ϕ ₂ (x ₁ , x ₂ ) = A(x ₁ , x ₂ ) + β ₂ (x ₂ ), where A : G ² → C satises

(29) A(x ₁ y ₁ , x ₂ ) = A(x ₁ , x ₂ ) + A(y ₁ , x ₂ ).

Rewriting (28), we have

(30) A(x ₁ , x ₂ ) = ϕ ₂ (x ₁ , x ₂ ) − β 2 (x ₂ ).

Using (30) in (29), we obtain

(31) ϕ 2 (x 1 y 1 , x 2 ) = ϕ 2 (x 1 , x 2 ) + ϕ 2 (y 1 , x 2 ) − β 2 (x 2 ).

Replacing x 1 by x 1 y 1 in (23) and using (31), we see that (32) β 2 (x 2 y 2 ) = β 2 (x 2 ) + β 2 (y 2 ).

From (23), (30) and (32), we obtain

A(x 1 , x 2 y 2 ) = A(x 1 , x 2 ) + A(x 1 , y 2 ).

Putting (28) into (22), we get

f (x 1 , x 2 ) = A(x 1 , x 2 ) + α(x 1 ) + β(x 2 ),

where β(x 2 ) = β 1 (x 2 ) + β 2 (x 2 ) which proves (13). It is easy to check that β satises β(x ₁ x ₂ ) = β(x ₂ x ₁ ) for all x ₁ , x ₂ ∈ G. This concludes the proof of the theorem.

Remark 1. Theorem 1 also remains true if C is replaced by a 2-

cancellative abelian group H.

4. Stability results

In this section, we rst prove the Hyers-Ulam stability of the functional equations (2) and (3) and apply these results to the proof of the Hyers- Ulam stability of the equation (1). Throughout this section, we assume G to be an abelian group.

Lemma 3. Let G be an abelian group. If a function f : G → C satises the functional inequality

(33) |f(xy) + f(xy ⁻¹ ) − 2f(x)| ≤ δ (x, y ∈ G)

for some given δ ≥ 0, then there exist an arbitrary constant a ∈ C and a group homomorphism ϕ of G into the additive group ( C, +) of C such that

|f(x) − ϕ(x) − a| ≤ 2δ for all x in G.

Proof. By e we will denote the unit element of G. If we dene φ(x) = f (x) − f(e), then φ(e) = 0. From (33) we have

(34) |φ(xy) + φ(xy ⁻¹ ) − 2φ(x)| ≤ δ (x, y ∈ G).

We put y = x in (34) to get

(35) |φ(x ² ) − 2φ(x)| ≤ δ (x ∈ G).

Substituting x ⁻¹ y ⁻¹ for y in (34), we have

(36) |φ(y ⁻¹ ) + φ(x ² y) − 2φ(x)| ≤ δ (x, y ∈ G).

By (35) and (36), we obtain

|φ(y ⁻¹ ) + φ(x ² y) − φ(x ² ) | (37)

≤ |φ(y ⁻¹ ) + φ(x ² y) − 2φ(x)| + |2φ(x) − φ(x ² ) |

≤ 2δ (x, y ∈ G).

If we put u = y ⁻¹ and v = x ² y in (37), then we have

|φ(u) + φ(v) − φ(uv)| ≤ 2δ (u, v ∈ G).

By a theorem of R¨ atz [9] (or see also Hyers and Rassias [6], Forti [4], Jung [7] or Hyers, Isac and Rassias [5]), there exists a group homomorphism ϕ of G into the additive group (C, +) of C such that

|φ(x) − ϕ(x)| ≤ 2δ (x ∈ G), and hence

|f(x) − ϕ(x) − a| ≤ 2δ (x ∈ G),

where a = f (e).

Lemma 4. Let G be an abelian group in which x ² = y has a solution for all x, y ∈ G. If a function f : G → C satises the functional inequality (38) |f 1 (xy) + f ₂ (xy ⁻¹ ) − f 3 (x) | ≤ δ (x, y ∈ G)

for some δ ≥ 0, then there exist constants a, b ∈ C and a group homomor- phism ϕ : G → (C, +) such that

|f 1 (x) − ϕ(x) − a| ≤ 5δ,

|f 2 (x) − ϕ(x) − b| ≤ 5δ,

|f 3 (x) − 2ϕ(x) − a − b| ≤ 11δ for all x ∈ G.

Proof. If we replace y by y ⁻¹ in (38), then we have (39) |f 1 (xy ⁻¹ ) + f ₂ (xy) − f 3 (x) | ≤ δ.

Put y = e in (38) to get

(40) |f 1 (x) + f 2 (x) − f 3 (x)| ≤ δ.

This inequality, together with (38) and (39), yields

|f 1 (xy) + f ₂ (xy) + f ₁ (xy ⁻¹ ) + f ₂ (xy ⁻¹ ) − 2f 1 (x) − 2f 2 (x) |

≤ |f 1 (xy) + f ₂ (xy ⁻¹ ) − f 3 (x) | + |f 1 (xy ⁻¹ ) + f ₂ (xy) − f 3 (x) | + 2 |f 3 (x) − f 1 (x) − f 2 (x) |

≤ 4δ.

By Lemma 3, we can conclude that there exists a group homomorphism ϕ : G → (C, +) such that

(41) |f 1 (x) + f ₂ (x) − 2ϕ(x) − f 1 (e) − f 2 (e) | ≤ 8δ for any x in G. By (38) and (39), we get

|f 1 (xy) − f 2 (xy) − f 1 (xy ⁻¹ ) + f ₂ (xy ⁻¹ ) |

≤ |f 1 (xy) + f ₂ (xy ⁻¹ ) − f 3 (x) | + |f 3 (x) − f 1 (xy ⁻¹ ) − f 2 (xy) |

≤ 2δ.

Putting y = x in the above inequality, we obtain

|f 1 (x ² ) − f 2 (x ² ) − f 1 (e) + f 2 (e)| ≤ 2δ.

Since the equation x ² = y is solvable for all x and y in G, the last inequality can be rewritten as

(42) |f 1 (x) − f 2 (x) − f 1 (e) + f ₂ (e) | ≤ 2δ

for each x ∈ G. It then follows from (41) and (42) that 2 |f 1 (x) − ϕ(x) − f 1 (e) |

≤ |f 1 (x) + f ₂ (x) − 2ϕ(x) − f 1 (e) − f 2 (e) | + |f 1 (x) − f 2 (x) − f 1 (e) + f ₂ (e) |

≤ 10δ, that is,

(43) |f 1 (x) − ϕ(x) − f 1 (e) | ≤ 5δ for all x ∈ G. Analogously, we have

(44) |f 2 (x) − ϕ(x) − f 2 (e) | ≤ 5δ.

By (40), (43) and (44), we see

|f 3 (x) − 2ϕ(x) − f 1 (e) − f 2 (e) |

≤ |f 3 (x) − f 1 (x) − f 2 (x) | + |f 1 (x) − ϕ(x) − f 1 (e) | + |f 2 (x) − ϕ(x) − f 2 (e)|

≤ 11δ which completes the proof.

We are now ready to prove the main theorem of this section concerning the Hyers-Ulam stability of the functional equation (1).

Theorem 2. Let G be an abelian group in which x ² = y has a solution for all x, y ∈ G. If a function f : G ² → C satises the functional inequality

|f(x 1 y ₁ , x ₂ y ₂ ) + f (x ₁ y ⁻¹ ₁ , x ₂ ) + f (x ₁ , x ₂ y ⁻¹ ₂ ) (45)

− f(x 1 y ₁ ⁻¹ , x 2 y ₂ ⁻¹ ) − f(x 1 y 1 , x 2 ) − f(x 1 , x 2 y 2 ) | ≤ δ

for some δ ≥ 0 and for all x 1 , y 1 , x 2 , y 2 ∈ G, then there exist arbitrary functions α, β : G → C and A : G ² → C additive in each variable such that (46) |f(x 1 , x 2 ) − A(x 1 , x 2 ) − α(x 1 ) − β(x 2 )| ≤ 14982δ

for all x ₁ , x ₂ ∈ G.

Proof. Dene F _i : G ³ → C (i = 1, 2, 3) by (15), (16) and (17). Then, the inequality (45) can be written as

|F 1 (x ₁ y ₁ , x ₂ , y ₂ ) + F ₂ (x ₁ y ⁻¹ ₁ , x ₂ , y ₂ ) − F 3 (x ₁ , x ₂ , y ₂ ) | ≤ δ

for any x 1 , y 1 , x 2 , y 2 ∈ G. According to Lemma 4, by considering (15), (16)

and (17), the last inequality implies that there exists a function ϕ ₁ : G ³ →

C such that

ϕ ₁ (x ₁ y ₁ , x ₂ , y ₂ ) = ϕ ₁ (x ₁ , x ₂ , y ₂ ) + ϕ ₁ (y ₁ , x ₂ , y ₂ ), (47) |f(x 1 , x 2 y 2 ) − f(x 1 , x 2 ) − ϕ 1 (x 1 , x 2 , y 2 ) − b(x 2 , y 2 ) | ≤ 5δ and

(48) |f(x 1 , x ₂ y ⁻¹ ₂ ) −f(x 1 , x ₂ ) + ϕ ₁ (x ₁ , x ₂ , y ₂ ) −a(x 2 , y ₂ ) + b(x ₂ , y ₂ ) | ≤ 5δ for all x 1 , y 1 , x 2 , y 2 ∈ G. Combining (47) and (48), we get

(49) |f(x 1 , x 2 y 2 ) + f (x 1 , x 2 y ₂ ⁻¹ ) − 2f(x 1 , x 2 ) − a(x 2 , y 2 ) | ≤ 10δ for any x ₁ , x ₂ , y ₂ ∈ G. Putting x 1 = e and setting β ₁ (x ₂ ) = f (e, x ₂ ) in (49) yield

(50) |a(x 2 , y 2 ) − β 1 (x 2 y 2 ) − β 1 (x 2 y ₂ ⁻¹ ) + 2β 1 (x 2 ) | ≤ 10δ.

By dening F (x ₁ , x ₂ ) = f (x ₁ , x ₂ ) − β 1 (x ₂ ), it follows from (49) and (50) that

|F (x 1 , x ₂ y ₂ ) + F (x ₁ , x ₂ y ₂ ⁻¹ ) − 2F (x 1 , x ₂ ) | ≤ 20δ

for any x ₁ , x ₂ , y ₂ ∈ G. Hence, by Lemma 3, there exist an arbitrary function α : G → C and a function ϕ 2 : G ² → C such that

(51) ϕ 2 (x 1 , x 2 y 2 ) = ϕ 2 (x 1 , x 2 ) + ϕ 2 (x 1 , y 2 ) and

(52) |f(x 1 , x 2 ) − ϕ 2 (x 1 , x 2 ) − α(x 1 ) − β 1 (x 2 )| ≤ 40δ for all x 1 , x 2 , y 2 ∈ G. Using (45), (51) and (52), we obtain

| − ϕ 2 (x 1 y 1 , y 2 ) − ϕ 2 (x 1 y ₁ ⁻¹ , y 2 ) + 2ϕ 2 (x 1 , y 2 ) |

≤ |f(x 1 y ₁ , x ₂ y ₂ ) − ϕ 2 (x ₁ y ₁ , x ₂ y ₂ ) − α(x 1 y ₁ ) − β 1 (x ₂ y ₂ ) | + |f(x 1 y ₁ ⁻¹ , x ₂ ) − ϕ 2 (x ₁ y ⁻¹ ₁ , x ₂ ) − α(x 1 y ⁻¹ ₁ ) − β 1 (x ₂ ) | + |f(x 1 , x 2 y ₂ ⁻¹ ) − ϕ 2 (x 1 , x 2 y ₂ ⁻¹ ) − α(x 1 ) − β 1 (x 2 y ₂ ⁻¹ ) |

+ | − f(x 1 y ₁ ⁻¹ , x 2 y ⁻¹ ₂ ) + ϕ 2 (x 1 y ₁ ⁻¹ , x 2 y ⁻¹ ₂ ) + α(x 1 y ₁ ⁻¹ ) + β 1 (x 2 y ₂ ⁻¹ )|

+ | − f(x 1 y 1 , x 2 ) + ϕ 2 (x 1 y 1 , x 2 ) + α(x 1 y 1 ) + β 1 (x 2 ) | + | − f(x 1 , x ₂ y ₂ ) + ϕ ₂ (x ₁ , x ₂ y ₂ ) + α(x ₁ ) + β ₁ (x ₂ y ₂ ) | + | − f(x 1 y ₁ , x ₂ y ₂ ) − f(x 1 y ₁ ⁻¹ , x ₂ ) − f(x 1 , x ₂ y ₂ ⁻¹ )

+ f (x 1 y ⁻¹ ₁ , x 2 y ⁻¹ ₂ ) + f (x 1 y 1 , x 2 ) + f (x 1 , x 2 y 2 ) |

≤ 241δ.

By Lemma 3 again, there exist an arbitrary function β ₂ : G → C and a function A 1 : G ² → C such that

(53) A ₁ (x ₁ y ₁ , x ₂ ) = A ₁ (x ₁ , x ₂ ) + A ₁ (y ₁ , x ₂ )

and

(54) |ϕ 2 (x 1 , x 2 ) − A 1 (x 1 , x 2 ) − β 2 (x 2 )| ≤ 482δ

for all x ₁ , y ₁ , x ₂ ∈ G. By considering (53), it follows from (54) that

|ϕ 2 (x 1 y 1 , x 2 ) − ϕ 2 (x 1 , x 2 ) − ϕ 2 (y 1 , x 2 ) + β 2 (x 2 )|

(55)

≤ |ϕ 2 (x 1 y 1 , x 2 ) − A 1 (x 1 y 1 , x 2 ) − β 2 (x 2 ) | + | − ϕ 2 (x ₁ , x ₂ ) + A ₁ (x ₁ , x ₂ ) + β ₂ (x ₂ ) | + | − ϕ 2 (y ₁ , x ₂ ) + A ₁ (y ₁ , x ₂ ) + β ₂ (x ₂ ) |

≤ 1446δ.

By replacing x ₂ by y ₂ respectively by x ₂ y ₂ in (55) and combining (55) and the resulting inequalities and then considering (51), we have

|β 2 (x ₂ ) + β ₂ (y ₂ ) − β 2 (x ₂ y ₂ ) | ≤ 4338δ

for any x 2 , y 2 ∈ G. By a theorem of Forti [3] or R¨atz [9] (see also Forti [4], Hyers and Isac and Rassias [5], Hyers and Rassias [6] and Jung [7]), there exists an additive function A ₂ : G → C such that

(56) |β 2 (x 2 ) − A 2 (x 2 ) | ≤ 4338δ for all x ₂ ∈ G. By (54) and (56), we have

(57) |ϕ 2 (x 1 , x 2 ) − A 1 (x 1 , x 2 ) − A 2 (x 2 ) | ≤ 4820δ

for any x ₁ , x ₂ ∈ G. By replacing x 2 by y ₂ and by x ₂ y ₂ separately in (57), and combining (57) and the resulting inequalities, and then considering (51) and the fact that A ₂ is additive, we get

|A 1 (x 1 , x 2 y 2 ) − A 1 (x 1 , x 2 ) − A 1 (x 1 , y 2 )| ≤ 14460δ

for any x ₁ , x ₂ , y ₂ ∈ G. According to a theorem of Forti [3] or R¨atz [9] again, if we dene a function A : G ² → C by

A(x 1 , x 2 ) = lim

n →∞ 2 ⁻ⁿ A 1

( x 1 , x ² ₂

)

for any x ₁ , x ₂ ∈ G, then A is additive in the second variable and also in the rst variable, since A 1 has this property (see (53)). Furthermore, (58) |A 1 (x ₁ , x ₂ ) − A(x 1 , x ₂ ) | ≤ 14460δ

for all x 1 , x 2 ∈ G. By (52), (54), (58) and putting β(x 2 ) = β 1 (x 2 ) + β 2 (x 2 ),

we see that the inequality (46) holds true.

References

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[3] G. L. Forti, The stability of homomorphisms and amenability, with applications to functional equations, Abh. Math. Sem. Univ. Hamburg 57 (1987), 215–226.

[4] , Hyers-Ulam stability of functional equations in several variables, Aequationes Math. 50 (1995), 143–190.

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[9] J. R¨ atz, On approximately additive mappings, In “General Inequalities 2,” E. F.

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Jukang K. Chung

Department of Applied Mathematics South China University of Technology Guanzhou, P. R. China

Soon-Mo Jung Mathematics Section

College of Science and Technology Hong-Ik University

Chochiwon 339–701, Korea Prasanna K. Sahoo

Department of Mathematics University of Louisville

Louisville, Kentucky 40292, USA