On Stability of a General Bilinear Functional Equation

We prove the Hyers–Ulam stability of the functional equation *f(a1x1+a2x2,b1y1+b2y2)=C1f(x1,y1)+C2f(x1,y2)+C3f(x2,y1)+C4f(x2,y2)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned}&f(a_1x_1+a_2x_2,b_1y_1+b_2y_2)=C_{1}f(x_1,y_1)\nonumber \\ \nonumber \\&\quad +C_{2}f(x_1,y_2)+C_{3}f(x_2,y_1)+C_{4}f(x_2,y_2) \end{aligned}$$\end{document}in the class of functions from a real or complex linear space into a Banach space over the same field. We also study, using the fixed point method, the generalized stability of (∗)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(*)$$\end{document} in the same class of functions. Our results generalize some known outcomes.


Introduction
Problem of studying the stability of functional equations has begun with a question posed by S. Ulam (see, e.g., [17]) and an answer given by D.H. Hyers [13]. Since then a number of papers investigating the so called now Hyers-Ulam stability have appeared. The results concern also various generalizations of the problem and these kind of research have their origins in the papers by T. Aoki [1], D.G. Bourgin [7], Th. Rassias [16], P. Gavruta [11].
Let X and Y be linear spaces over the same field F ∈ {R, C}, a 1 , a 2 , b 1 , b 2 ∈ F \ {0}, C 1 , C 2 , C 3 , C 4 ∈ F and f : X 2 → Y . In [10], K. Ciepliński starting with a bilinear mapping, i.e., linear in each of its arguments, considered the following functional equation f (a 1 x 1 + a 2 x 2 , b 1 y 1 + b 2 y 2 ) = C 1 f (x 1 , y 1 ) for all x 1 , x 2 , y 1 , y 2 ∈ X, and investigated, among others, its Hyers-Ulam stability in Banach spaces. In fact, he proved the stability without knowing the general solution of (1) and under some additional assumptions. In [6], the authors described the form of solutions of (1). They were also studying relations between (1) and bilinear mappings.
In the present paper, firstly knowing already the form of solutions of (1) we prove its Hyers-Ulam stability, also in the cases excluded in [10]. Secondly, applying the fixed point method, we study the generalized stability of (1) for the same classes of control functions.
For the convenience of the reader we recall here a result describing the solutions of (1) (see [6], and also [12], where Y is an arbitrary field of characteristic different from two). (1), then there exist a biadditive function g : X 2 → Y, additive functions ϕ, ψ : X → Y and a constant δ ∈ Y such that and for all x, y ∈ X, and Conversely, each function f of the form (2) with g biadditive, ϕ, ψ additive, and such that conditions (3), (4), (5), (6) are satisfied, is a solution of (1).
Throughout this paper we keep the standard notation: N, R and C stand for the sets of all positive integers, all real numbers and all complex numbers, respectively. Moreover, we denote R + := [0, ∞), N 0 := N ∪ {0} and we adopt the convention 0 0 = 1.
Then there exists an additive function F : Moreover, F is a unique function satisfying the above condition and it is of Lemma 2. Let (H, +) be an abelian group and (Y, · ) be a Banach space. Given ε > 0 assume that g : H 2 → Y satisfies Then there exists an additive function G : Moreover, G is a unique function satisfying the above condition and it is of the form G(x, y) = lim Proof. Immediately from (7) we obtain the following inequalities for all x 1 , x 2 , y 1 , y 2 ∈ X, and Therefore, the functions ϕ( (12) and respectively. By (7) we also have and, moreover, From (7), (11), (14) and (15) it follows that and, since a 1 a 2 b 1 b 2 = 0, From (9), (11) and (15) we obtain and by (10), (11) and (15) we have so, for all x 1 , x 2 , y 1 , y 2 ∈ X, On account of Lemma 1, there exist a unique additive function Φ and a unique additive function Ψ such that with Therefore using (12) and (13), we derive that the functions F 1 (x, y) := Φ(x) and F 2 (x, y) := Ψ(y), for x, y ∈ X, satisfy (1). Let us define g : Then and by (7), (9), (10) and (11), we get On account of (16), (17), (18) and (20), we obtain By Lemma 2, there exists a unique biadditive function G such that and, moreover, G(x, y) = lim n→∞ 1 4 n g(2 n x, 2 n y). Using (21), we obtain that G satisfies (1).
Let us define Function F satisfies (1) and from (19) and (22) we get For the proof of the uniqueness, assume that C = 1 and let F be another function satisfying (1) and inequality (8). Therefore, F is of the form (cf., Theorem 1) with biadditive G , additive Φ and Ψ , satisfying (3), (4) and (5), respectively, and with δ = 0 in the case C = 1.
We have for all x, y ∈ X, n ∈ N, Dividing the above inequality by n 2 side by side and letting n tend to infinity we obtain G = G , and consequently, It is now enough to set y = 0 and then x = 0 in order to obtain Φ = Φ and Ψ = Ψ , respectively.

Remark 1.
A thorough inspection of the proof of Theorem 2 shows that in the case C = 1 we are able to obtain a better approximation. Namely, if f : X 2 → Y is a mapping satisfying (7) for x 1 , x 2 , y 1 , y 2 ∈ X and C = 1, then there exists a solution F : Remark 2. It is also easy to observe that in the case C = 1 we do not have the uniqueness of function F in (8). Indeed, each function F : X 2 → Y , with G, Φ, Ψ defined as in the proof of Theorem 2, and with δ ∈ Y such that δ ≤ 3ε satisfies, on account of Remark 1, conditions (1) and (8).

Generalized Stability of (1)
In this section we provide some results concerning generalized stability with various approximation functions. In what follows we will use a notation for x 1 , x 2 , y 1 , y 2 ∈ X. Let us also denote a := a 1 + a 2 and b : Our first result reads as follows.
Assume, further, that for an s ∈ {−1, 1} (depending on a, b, C) we have and Then there exists a unique solution F : Proof. Putting x 1 = x 2 = x and y 1 = y 2 = y in (25) we get Similarly, putting Define and for all x, y ∈ X. Then, for any ξ, μ : (29) and (30), x,y∈ X.
Next, put (Λη)(x, y) := 1 |C| s η a s x, b s y , η∈ R X 2 + , x, y ∈ X. As one can check, θ(a n x, b n y, a n x, b n y) |C| n+1 , for s = 1, satisfy the assumptions of Theorem 1 in [8], therefore, there exists a unique fixed point F : X 2 → Y of T such that (28) holds. Moreover, Now, we prove that for any x 1 , x 2 , y 1 , y 2 ∈ X and n ∈ N 0 we have Since the case n = 0 is just (25), fix an n ∈ N 0 and assume that (33) holds for any x 1 , x 2 , y 1 , y 2 ∈ X. Then for any x 1 , x 2 , y 1 , y 2 ∈ X we get and thus, (33) holds for any x 1 , x 2 , y 1 , y 2 ∈ X and n ∈ N 0 . Letting n → ∞ in (33) and using (27) we finally obtain which means that function F satisfies (1). For the proof of uniqueness, assume that F is another function satisfying (1) and (28). We have for all x, y ∈ X, l ∈ N 0 whence letting l → ∞ and using (26) we obtain F (x, y) = F (x, y) for all x, y ∈ X, which finishes the proof.
then there exists a unique solution F : • If a = 0 = b (and |C| > 1, for (26) to be satisfied), we take s = 1, and we have in Theorem 3, and with θ(x, y, x, y) = ε, in Corollary 1. Then (34), it follows that in Theorem 3, f is majorized by the function and in Corollary 1, it is simply bounded.
Then there exists a unique solution F : X 2 → Y of (1) such that condition (28) holds.
Proof. Putting x 1 = x 2a1 , x 2 = x 2a2 , y 1 = y 2b1 and y 2 = y 2b2 in (25) (with x, y ∈ X) we get Then, by (37), we obtain Put also Now, using induction, we show that for any n ∈ N 0 and x, y ∈ X we have Fix x, y ∈ X. Clearly, (38) is true for n = 0. Next, fix an n ∈ N 0 and assume that (38) holds. Then and thus (38) is true for any n ∈ N 0 and x, y ∈ X.
One can now show that the operators T : Y X 2 → Y X 2 and Λ : R + satisfy the assumptions of Theorem 1 in [8] and therefore there exists a unique fixed point F : X 2 → Y of T such that (28) holds. Moreover, F is given by (32).
We prove that for any x 1 , x 2 , y 1 , y 2 ∈ X and n ∈ N 0 we have Since the case n = 0 is just (25), fix an n ∈ N 0 and assume that (39) holds for any x 1 , x 2 , y 1 , y 2 ∈ X. Then for any x 1 , x 2 , y 1 , y 2 ∈ X we get We have thus shown that (39) holds for x 1 , x 2 , y 1 , y 2 ∈ X and n ∈ N 0 . Letting n → ∞ in (39) and using (36) we see that which means that function F satisfies (1). For the proof of uniqueness, assume that F is another function satisfying (1) and (28). Then, for any m ∈ N we have Tending now with m to infinity, on the account of the assumption, it follows that F = F , which completes the proof.
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