On the quotient stability of a family of functional equations

Nonlinear Analysis 71 (2009) 4396–4404

Contents lists available at ScienceDirect

Nonlinear Analysis journal homepage: www.elsevier.com/locate/na

On the quotient stability of a family of functional equations Janusz Brzdek ¸ ∗ Department of Mathematics, Pedagogical University, Podchorażych ¸ 2, 30-084 Kraków, Poland

article

info

Article history: Received 20 January 2009 Accepted 26 February 2009

abstract We show some stability results for a family of functional equations that contains the exponential and the Gołab–Schinzel ¸ functional equations. © 2009 Elsevier Ltd. All rights reserved.

MSC: 39B82 Keywords: Stability Superstability Exponential equation Gołab–Schinzel ¸ functional equation

1. Introduction Let R, Z and N stand, as usual, for the sets of reals, integers and positive integers, respectively; moreover, we write R0 := R \ {0}, D + E := {d + e : d ∈ D, e ∈ E } and cD := {cd : d ∈ D} for c ∈ R, D, E ⊂ R. In what follows λ ∈ R0 and M : R → R denotes a continuous function that is multiplicative (i.e., M (xy) = M (x)M (y) for x, y ∈ R) and M (R) 6= {0}. Several particular forms of the functional equation f (x + M (f (x))y) = λf (x)f (y)

(1)

(with the unknown function f mapping, e.g., a real linear space into R) have already found nontrivial applications in meteorology and fluid mechanics, theories of near rings and quasialgebras, finding algebraic substructures, theory of algebraic objects, and description of associative operations; for suitable references see the monograph [1] and the survey paper [2]. Note that, in the particular case: λ = 1 and M (x) = 1 for x ∈ R, Eq. (1) becomes the well-known exponential equation f (x + y) = f (x)f (y)

(2)

and, if λ = 1 and M (x) = x for x ∈ R, (1) has the form f (x + f (x)y) = f (x)f (y),

(3)

which is now called the Gołab–Schinzel ¸ functional equation. Some recent results concerning solutions of Eq. (3) and its generalizations can be found in [3–7]. The investigation of stability of functional equations started with a problem posed by S.M. Ulam and a solution of it given by Hyers [8]. Among other results, the superstability of the exponential equation (2) has been discovered and in connection with that phenomenon, a new approach has been suggested (see [9]) that is now called: stability in the sense of Ger (for more information on stability and superstability of functional equations we refer the reader to, e.g., [10–13]; for some recent

∗

Tel.: +48 12 358 39 23; fax: +48 12 637 22 43. E-mail address: [email protected].

0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.02.123

J. Brzdek ¸ / Nonlinear Analysis 71 (2009) 4396–4404

4397

results on these subjects see, e.g., [14–27,37]). Motivated by this approach, Chudziak [28] (see also [29]) has proved that, in the class of functions f , mapping a linear space Y over a field K ∈ {R, C} into K, such that the set Af := {x ∈ Y ; f (x) 6= 0} has an algebraically interior point, the functional equation f (x + f (x)n y) = λf (x)f (y)

(4)

(with fixed n ∈ N) is superstable in the sense of Ger. Namely he has shown that every function f : Y → K such that the set Af possesses an algebraically interior point and

f (x + f (x)n y) ≤ ε whenever f (x)f (y) 6= 0, − 1 λf (x)f (y) λf (x)f (y) n f (x + f (x)n y) − 1 ≤ ε whenever f (x + f (x) y) 6= 0,

(5)

(6)

with some ε ∈ (0, 1), is a solution of Eq. (4) or it is bounded. Some further results related to or concerning the stability of (3) or (4) can be found in [30–33]. The result in [28] (and in [29]) does not remain valid if we replace inequalities (5) and (6) by the following weaker condition

f (x + f (x)n y) ≤ ε and λf (x)f (y) − 1 ≤ ε − 1 λf (x)f (y) f (x + f (x)n y) n whenever f (x + f (x) y)f (x)f (y) 6= 0.

(7)

√ n

For instance, for λ = 1, the function f : R → R, given by the formulas f (x) = x + 1 for x > 0 and f (x) = 0 for x ≤ 0, satisfies (7) (with ε = 0 and λ = 1), but it is not a solution of (4) (take x > 0 and y = 0). However, it is easy to check that f satisfies (with λ = 1) the functional equation f (x + f (x)n y)f (x)f (y)[f (x + f (x)n y) − λf (x)f (y)] = 0.

(8)

In this paper we complete (and generalize to some extent) the results in [29,28] by showing that every continuous solution f : R → R of (7) satisfies Eq. (8) and, moreover, in the case where f (x0 + f (x0 )n y0 )f (x0 )f (y0 ) 6= 0 for some x0 , y0 ∈ R, it must be a solution of (4). We also extend these results to the case of Eq. (1); in particular, we prove analogous results for the exponential equation (2). In view of Remark 3, for M (R) 6= {1}, we should not expect to obtain such results under assumptions weaker than continuity of f (at each point of the domain); in the case M (R) = {1} the situation is somewhat different, as it is shown in Proposition 1 and Corollaries 1 and 2. The present paper gives a starting point for investigations of analogous problems in more general classes of functions, e.g., mapping a linear topological space Y over a field K ∈ {R, C} into K. However, since some additional longer arguments are necessary in such a case, they will be studied in next publications. At the end of the paper (see Proposition 3) we show that a description of the continuous solutions f : R → R of the equation f (x + M (f (x))y)f (x)f (y)[f (x + M (f (x))y) − λf (x)f (y)] = 0

(9)

can be easily derived from our main theorem (Theorem 2). In this way we generalize the main results in [34], where the continuous solutions f : R → R of the functional equation f (x + f (x)y)f (x)f (y)[f (x + f (x)y) − f (x)f (y)] = 0

(10)

have been determined. 2. The case M (x) ≡ x and λ = 1 In what follows µ ≥ 1 is a fixed real number, f : R → R denotes a continuous solution of the inequality

f (x + f (x)y) ≤ µ for x, y ∈ R, f (x + f (x)y)f (x)f (y) 6= 0, ≤ µ f (x)f (y) 1

A := {x ∈ R : f (x) 6= 0}, B := R \ A, τx (y) := x + f (x)y for x, y ∈ R, and F (x, y) :=

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

for x, y ∈ R, f (x)f (y) 6= 0.

Moreover, we assume that f (x0 + f (x0 )y0 )f (x0 )f (y0 ) 6= 0 for some x0 , y0 ∈ R. Lemma 1. The following three statements are valid. (i) A ∈ {(−∞, b), (a, ∞), (−∞, b) ∪ (a, ∞), R} for some a, b ∈ R0 , b ≤ a.

(11)

4398

J. Brzdek ¸ / Nonlinear Analysis 71 (2009) 4396–4404

(ii) The function f : R → R, given by: f (x) = f (−x) for x ∈ R, is a solution of (11). (iii) If A = (a, ∞) for some a ∈ R, then τx (y) ∈ A for x, y ∈ A. Proof. (i) Let B 6= ∅. First we show that B is connected. Clearly B is a closed set, so for the proof by contradiction suppose that there exist α, β ∈ B, α < β , such that (α, β) ⊂ A. Since limx→α τx (α) = α and limx→β τx (α) = β , there is t0 ∈ (α, β) with τt0 (α) ∈ (α, β). Note that limy→α− |F (t0 , y)| = ∞, which contradicts to (11). Thus we have proved that there exist a, b ∈ R, a ≤ b, such that A ∈ {(−∞, b), (a, ∞), (−∞, b) ∪ (a, ∞)}. Suppose that 0 ∈ B and (0, ∞) ⊂ A ((−∞, 0) ⊂ A, respectively). Then, for every x > 0, limy→0+ |F (x, y)| = ∞ (limy→0− |F (x, y)| = ∞, resp.). This contradiction shows that a 6= 0 6= b. (ii) It is easy to check that the statement is valid. (iii) Note that x0 , y0 , τx0 (y0 ) ∈ A. First we show

τx0 (y) ∈ A for y ∈ A.

(12)

For the proof by contradiction suppose that τx0 (z ) ≤ a for some z > a. Then there is u > a with τx0 (u) = a (because τx0 (y0 ) > a). Assume that u < y0 (u > y0 , respectively). Let z0 := sup {z ∈ (a, y0 ) : τx0 (z ) = a} (z0 := inf {z ∈ (y0 , u] : τx0 (z ) = a}, resp.). Clearly, τx0 (z ) > a for each z ∈ (z0 , y0 ) (z ∈ (y0 , z0 ), resp.), z0 > a, τx0 (z0 ) = a, and consequently limz →z0 + |F (x0 , z )| = 0 (limz →z0 − |F (x0 , z )| = 0, resp.), which is a contradiction. Thus we have proved (12). Suppose that τv (y) ≤ a for some v, y ∈ (a, ∞). Since, by (12), τx0 (y) > a, there is s ∈ (a, ∞) with τs (y) = a. Assume that s < x0 (s > x0 , respectively) and write w0 := sup{w ∈ [s, x0 ) : τw (y) = a} (w0 := inf{w ∈ (x0 , s] : τw (y) = a}, resp.). Then w0 > a, τw0 (y) = a, and consequently limz →w0 + |F (z , y)| = 0 (limz →w0 − |F (z , y)| = 0, resp.). This is a contradiction.  Lemma 2. The following three statements are valid.

(α) τx (A) = A and τx (B) = B for every x ∈ A. (β) F (x, y)A = A and F (x, y)B = B for x, y ∈ A. (γ ) If B 6= ∅, then |F (x, y)| = 1 for x, y ∈ A with F (x, y) 6= 0. Proof. (α) Take x ∈ A. The case A = R is trivial, so assume that B 6= ∅. First we show that τx (y) ∈ B for y ∈ B. For the proof by contradiction suppose that τx (y) ∈ A for some y ∈ B. In view of Lemma 1(i), (ii), it is enough to consider the case where there is a ∈ B with (a, ∞) ⊂ A and τx (y) > a. If f (x) > 0, then τx (a) ≥ τx (y) > a. Hence limz →a+ |F (x, z )| = ∞, which contradicts (11). If f (x) < 0, then limz →∞ τx (z ) = −∞ and consequently, according to Lemma 1(i), (iii), there is b ∈ B with (−∞, b) ⊂ A. Clearly y ≥ b, whence τx (b) ≥ τx (y) > a. Hence limz →b− |F (x, z )| = ∞, which is a contradiction. Thus we have proved that τx (B) ⊂ B. So now it remains to show that τx (A) ⊂ A. On account of Lemma 1, we can only consider the case: A = (−∞, b) ∪ (a, ∞) with some a, b ∈ B, b ≤ a, and x > a. For the proof by contradiction suppose that there is y ∈ A with τx (y) ∈ B. Then, if f (x) < 0 and y > a (y < b, respectively), τx (t0 ) = b for some t0 ≥ y (τx (t0 ) = a for some t0 ≤ y, resp.); if f (x) > 0 and y > a (y < b, resp.), there is t0 ≥ y (t0 ≤ y, resp.) with τx (t0 ) = a (τx (t0 ) = b, resp.). In each of those cases limz →t0 |F (x, z )| = 0. This yields a contradiction to (11), which ends the proof of (α).

(β) Take x, y ∈ A. Then F (x, y)u = τy−1 τx−1 ττx (y) (u) for u ∈ B. Since, by (α), τx (y) ∈ A and τz (B) = B for z ∈ A, we get F (x, y)B = B. Clearly this implies F (x, y)A = A. (γ ) For the proof by contradiction suppose 1 ) 6∈ {−1, 0, 1} for some x1 , y1 ∈ A. Lemma 1(i) yields S that ρ := F (x1 , yS 0 ∈ int B or 0 ∈ int A, whence, by (β), R = n∈Z ρ n B = B or R = n∈Z ρ n A = A. It is a contradiction.  Finally we are in a position to prove the following. Theorem 1. Let f : R → R h be a icontinuous solution of (11), with some µ ∈ [1, ∞). Then either B 6= ∅ and f satisfies functional equation (10), or |f (x)| ∈

1

µ

, µ for x ∈ R.

Proof. First assume that B 6= ∅. If A ∈ {(−∞, a), (a, ∞)} with some a ∈ R, then by Lemma 2(α ) we have f (x) > 0 for x ∈ A. Hence Lemma 2(γ ) implies that f is a solution of (10). According to Lemma 1(i) it remains to consider the case: A = (−∞, b)∪(a, ∞) with some a, b ∈ R0 , b ≤ a. If b < a, from Lemma 2(α ) we derive that, for every x ∈ A, τx (a), τx (b) ∈ {a, b}. Clearly the functions x → τx (a), x → τx (b) are continuous and τa (a) = a, τb (b) = b. Consequently, for each x ∈ A, τx (a) = a and τx (b) = b, whence τx ((−∞, b)) = (−∞, b) and τx ((a, ∞)) = (a, ∞). This implies that f (x) > 0 for x ∈ A. Hence, in view of Lemma 2(γ ), (10) holds for every x, y ∈ R. If a = b, then x + f (x)a = a for x ∈ A, because otherwise limy→a |F (x, y)| = ∞. Hence f (x) = ax + 1 for x ∈ R, which means that f satisfies functional equation (3) and consequently it is a solution to (10). To complete the proof suppose A = R. Let x ∈ R, f (x) 6= 1, and write y := 1−xf (x) . Then x + f (x)y = y, whence, by (11),

|f (x)| ∈

h

1

µ

i ,µ .



J. Brzdek ¸ / Nonlinear Analysis 71 (2009) 4396–4404

4399

3. The case M (x) ≡ 1 and λ = 1 In this part X denotes a real linear space. We start with a result which follows from [35, Theorem 1] and will be useful in what follows. Lemma 3. Let δ > 0, D ⊂ X , D 6= ∅, h : D → R, and

|h(x + y) − h(x) − h(y)| ≤ δ for x, y ∈ D, x + y ∈ D.

(13)

Suppose that 2D ⊂ D or D ⊂ 2D. Then there is a unique h0 : D → R with h0 (x + y) = h0 (x) + h0 (y) for x, y ∈ D, x + y ∈ D, |h(x) − h0 (x)| ≤ δ for x ∈ D.

(14) (15)

Now we are in a position to prove the following. Proposition 1. Let c1 , c2 ∈ [1, ∞), f : X → R, S := {x ∈ X : f (x) 6= 0} 6= ∅,

f (x + y) ≤ c2 for x, y ∈ X , f (x)f (y)f (x + y) 6= 0, ≤ c1 f (x)f (y) 1

S ⊂ 2S

or

2S ⊂ S .

(16) (17)

Then there exists a unique function F : S → (0, ∞) such that F (x + y) = F (x)F (y)

for x, y ∈ S , x + y ∈ S , f (x) 1 ≤ c1 for x ∈ S . ≤ c F (x)

(18) (19)

2

Proof. Observe that (16) yields

√ f (x + y) c1 √ ≤ √ ≤ c1 c2 for x, y ∈ X , f (x)f (y)f (x + y) 6= 0. c1 c2 f (x)f (y) c2 √ √ Let C := ln( c2 ) − ln( √c1 ) and define h : S → R by: h(x) := C + ln |f (x)| for x ∈ S. Then, by (16), condition (13) is valid with D := S and δ := ln c1 c2 . Hence, according to Lemma 3, there exists a unique function h0 : S → R such that (14) and (15) hold. Now it is enough to write F := exp ◦h0 . The uniqueness of F follows from the uniqueness of h0 . √

1

Corollary 1. Let c1 , c2 ∈ [1, ∞), f : X → R, S := {x ∈ X : f (x) 6= 0} 6= ∅ and

f (x + y) ≤ c2 for x, y ∈ X , f (x)f (y) 6= 0. ≤ c1 f (x)f (y) 1

(20)

Then there is a unique F : S → (0, ∞) such that (18) and (19) are valid. Proof. It is easily seen that (16) follows from (20). Moreover, (20) implies that 2S ⊂ S. Hence Proposition 1 completes the proof.  Corollary 2. Let c1 , c2 ∈ [1, ∞), f : X → R, f (X ) 6= {0} and

f (x)f (y) ≤ c2 for x, y ∈ X , f (x + y) 6= 0. ≤ c1 f (x + y) 1

(21)

Then there is a unique F : X → (0, ∞) such that (18) and (19) are valid with S = X . Proof. Take z ∈ X with f (z ) 6= 0. Then, by (21), 0 6= f (z − y)f (y) for y ∈ X , which means that 0 6∈ f (X ). Hence Proposition 1 completes the proof.  Lemma 4. Let c > 0, f : R → R be continuous, S := {x ∈ R : f (x) 6= 0} 6= ∅, S 6= R, and

f (x + y) f (x)f (y) ≤ c for x ∈ R, f (x)f (y) 6= 0.

(22)

Then every connected component of the set S is a finite interval and f (x)f (y)f (x + y) = 0 for x, y ∈ R.

(23)

4400

J. Brzdek ¸ / Nonlinear Analysis 71 (2009) 4396–4404

Proof. Assume that there is z ∈ R with f (z ) = 0. Take x, y ∈ R such that f (x)f (y) 6= 0 and let Ix , Iy be connected components of S with x ∈ Ix , y ∈ Iy . First we show that xi := inf Ix > −∞, xs := sup Ix < ∞, yi := inf Iy > −∞, ys := sup Ix < ∞. For the proof by contradiction suppose that this is not the case. We consider only the situation xs = ∞ (because the remaining possibilities are analogous). Then xi ≥ z and Ix = (xi , ∞). Let w ∈ Ix and w > 0. Clearly f (w) 6= 0 and f (xi + w) 6= 0. Hence

f (t + w) = ∞, t →xi + f (t )f (w) lim

which contradicts to (22). Thus we have shown that the intervals Ix = (xi , xs ) and Iy = (yi , ys ) are finite. Take u ∈ Iy , w ∈ Ix and suppose that f (xs + u) 6= 0 or f (yi + w) 6= 0. Then

f (t + u) = ∞ or lim t →xs − f (t )f (u)

f (t + w) = ∞, lim t →yi + f (t )f (w)

which is a contradiction. So we have proved that f (xs +u) = 0 = f (yi +w) for u ∈ Iy , w ∈ Ix and consequently f (Ix +Iy ) = {0}, whence f (x + y) = 0.  4. The general case Let us begin with the following remark. Remark 1. Note that there is a ≥ 0 such that M (x) = xa for x ≥ 0 and either M (z ) = M (−z ) for z < 0 or M (z ) = −M (−z ) for z < 0, because M is continuous and multiplicative (cf. [1]). Now we are in a position to prove the main result. Theorem 2. Let ε1 , ε2 ∈ [0, ∞), g : R → R be continuous, and

g (x + M (g (x))y) ≤ ε2 + 1 for x, y ∈ R, ≤ ε1 + 1 λg (x)g (y) g (x + M (g (x))y)g (x)g (y) 6= 0. 1

(24)

Then g is a solution of the functional equation g (x + M (g (x))y)g (x)g (y) = 0

(25)

or the following three conditions are valid. (a) If M is odd, then

|λg (x)| ∈



1

ε2 + 1

 , ε1 + 1

for x ∈ R

(26)

or g satisfies the equation g (x + M (g (x))y) = g (x)g (y).

(27)

(b) If M is even and M (R) 6= {1}, then (26) holds or g fulfils the equation g (x + M (g (x))y) = σ g (x)g (y)

(28)

with some σ ∈ {−1, 1}. (c) If M (R) = {1}, then there exists a unique α ∈ R with 1

ε2 + 1

λg (x) ≤ αx ≤ ε1 + 1 for x ∈ R. e

(29)

Proof. First consider the case M (R) 6= {1}. Write f := M ◦ g. Then (24) and Remark 1 imply that (11) holds with

n ε + 1 o 1 µ := max M , M (|λ|(ε2 + 1)) . |λ| Consequently, by Theorem 1, f satisfies functional equation (10) or

|f (x)| ∈



1

µ

 ,µ

for x ∈ R.

(30)

J. Brzdek ¸ / Nonlinear Analysis 71 (2009) 4396–4404

4401

If f is a solution of (10), then, according to [34, Theorem 2 and Corollary 2], f fulfils Eq. (3) or f (x + f (x)y)f (x)f (y) = 0 for x, y ∈ R. In the latter case g is a solution of (25). Next, if M is odd (and therefore bijective), then (3) implies that (27) holds for every x, y ∈ R. So, assume now that M is even (and f satisfies Eq. (3)). Then (see [2]) there is c ∈ R such that M ◦ g (x) = f (x) = max {cx + 1, 0} for x ∈ R. Clearly this and the continuity of g and M imply that g (x) = σ M0−1 (max {cx + 1, 0}) for x ∈ R, with some σ ∈ {−1, 1}, where M0 := M |[0,∞) . Now it is easy to check that g is a solution to (28). If (30) is valid, then 0 6∈ f (R), whence 0 6∈ g (R). Hence, for h every x ∈ Risuch that g (x) 6= 1, we have M (g (x)) 6= 1 and consequently from (24), with y = 1−Mx(g (x)) , we get |λg (x)| ∈

|λg (x)| ∈



1

ε2 +1

, ε1 + 1 . So



1

ε2 + 1

, ε1 + 1 ∪ {|λ|} for x ∈ R.

Since g is continuous and (24) does not hold when |λg (x)| ≡ |λ| and

 |λ| 6∈

1

ε2 + 1



, ε1 + 1 ,

this completes the proof in the case M (R) 6= {1}. Now assume that M (R) = {1}. Then, in view of Lemma 4, 0 6∈ g (R) or g is a solution to (25). Further, in the case 0 6∈ g (R), Proposition 1 (with f (x) := λg (x ) for x ∈ R) implies that there exists a unique function F : R → (0, ∞) such

λg (x) that F (x + y) = F (x)F (y) and ε 1+1 ≤ F (x) ≤ ε2 + 1 for x, y ∈ R. Since g is continuous, there is α ∈ R with F (x) = eα x for 1 x ∈ R (cf. [1]). 

Remark 2. Actually, the statement of Theorem 2 can be easily generalized a bit. Namely, let δ1 , δ2 ∈ (0, ∞), δ1 ≤ δ2 , ρ ∈ R0 , g : R → R be continuous and

g (x + M (g (x))y) ≤ δ2 for x, y ∈ R, g (x + M (g (x))y)g (x)g (y) 6= 0. δ1 ≤ ρ g (x)g (y) δ

Then, for each δ ∈ (δ1 , δ2 ), (24) holds with λ = ρδ , ε1 = δδ − 1 and ε2 = δ2 − 1 and consequently the statement of 1 Theorem 2 is valid. Clearly (26) yields |ρ g (x)| ∈ [ δ1 , δ1 ] for x ∈ R and (29) takes the form: 2

ρ g ( x) 1 ≤ αx ≤ δ2 e δ1 1

1

for x ∈ R.

Theorem 2 implies the following corollary, which corresponds to the results in [29,28]. Corollary 3. Let ε1 , ε2 ∈ [0, ∞), g : R → R be continuous, and

λg (x)g (y) g (x + M (g (x))y) − 1 ≤ ε1

g (x + M (g (x))y) and − 1 ≤ ε2 λg (x)g (y) for x, y ∈ R, g (x + M (g (x))y)g (x)g (y) 6= 0.

(31)

Then g is a solution of Eq. (25) or statements (a)–(c) of Theorem 2 are valid. Moreover, in the case M (R) = {1} and εi ≤ 1 for some i ∈ {1, 2},

−ε1 ≤

λg (x) eα x

− 1 ≤ ε2 and

− ε2 ≤

eα x

λg (x)

− 1 ≤ ε1 for x ∈ R.

Proof. First we show that (31) implies (24). To this end take x, y ∈ R with g (x + M (g (x))y)g (x)g (y) 6= 0. Clearly, from (31), we get

λg (x)g (y) ≤ ε1 + 1 g (x + M (g (x))y)

and

g (x + M (g (x))y)

λg (x)g (y)

≤ ε2 + 1 ,

which yields (24) in the case where g (x + M (g (x))y)

λg (x)g (y)

> 0.

(32)

Since it is easily seen that (31) implies (32) when ε1 ≤ 1 or ε2 ≤ 1, it remains to consider the case where εi > 1 for i = 1, 2 and g (x + M (g (x))y)

λg (x)g (y)

< 0.

4402

J. Brzdek ¸ / Nonlinear Analysis 71 (2009) 4396–4404

Then, by (31),

λg (x)g (y) g (x + M (g (x))y)

g (x + M (g (x))y)

≥ −ε1 + 1 and

λg (x)g (y)

≥ −ε2 + 1,

whence

λg (x)g (y) g (x + M (g (x))y) ≤ ε1 − 1 and

g (x + M (g (x))y) λg (x)g (y) ≤ ε2 − 1

and consequently 1

ε1 + 1

g (x + M (g (x))y) ≤ ε2 − 1 < ε2 + 1. ≤ ε1 − 1 λg (x)g (y) 1

<

In this way we have proved that (24) is valid. Hence, by Theorem 2, g is a solution of (25) or statements (a)–(c)are valid. To complete the proof suppose that M (R) = {1} and ε1 ≤ 1 or ε2 ≤ 1. Then (29) implies that 0 6∈ g (R) and, by the continuity of g, λg (R) ⊂ (0, ∞) or λg (R) ⊂ (−∞, 0). The latter case is not possible in view of (31), because ε1 ≤ 1 or ε2 ≤ 1. Hence, for each x ∈ R, λg (x) > 0 and consequently, on account of (29), 1 − ε1 ≤

1

ε1 + 1

≤

λg (x) eα x

≤ ε2 + 1 and 1 − ε2 ≤

Remark 3. Let M (R) 6= {1}, ε > 0 and λ ∈ s :=

1 1+ε

1

ε2 + 1

≤

eα x

λg (x)

≤ ε1 + 1. 

 , 1 + ε . Since M0 := M |[0,∞) is bijective (onto [0, ∞)), we may write

1 M (λ)M (1 + ε) − 1

and define g : R → R by: g (x) = 0 for x < s and g (x) = M0−1 (x) for x ≥ s. It is easy to check that g is not a solution of (25) or of (28) with any real σ , but for every x, y ∈ [s, ∞) 1 1+ε

≤

g (x + M (g (x))y)

λg (x)g (y)

=

1

λ

M0−1

1 y

 + 1 ≤ 1 + ε,

whence

−ε ≤ 1 −

λg (x)g (y) g (x + M (g (x))y)

≤1−

1 1+ε

< ε.

Consequently (24) and (31) are valid (with ε1 = ε2 := ε ). In this way we have proved that the assumption of continuity of g is essential in Theorem 2 and Corollary 3 when M (R) 6= {1}. If M (R) = {1}, then the situation is different, which shows Proposition 1 (see also Corollaries 1 and 2). 5. Solutions of (1), (9) and (25) We end the paper with three propositions that describe continuous solutions g : R → R of functional equations (1), (9) and (25). Proposition 2. Let M (R) 6= {1}, λ ∈ R \ {0} and g : R → R be continuous. Then g is a solution of functional equation (1) if and only if one of the following five conditions holds.

(α) g (x) = λ1 for x ∈ R. (β) g (x) = 0 for x ∈ R. (γ ) λ = 1, M is odd and there is c ∈ R0 such that either g (x) = M −1 (cx + 1) for x ∈ R or g (x) = M −1 (max {cx + 1, 0}) for x ∈ R. (δ) |λ| = 1, M is even, M (R) 6= {1}, and there is c ∈ R0 with g (x) = λM0−1 (max {cx + 1, 0}) for x ∈ R, where M0 := M |[0,∞) . () M (R) = {1} and there is c ∈ R0 with g (x) = λ1 ecx for x ∈ R. Proof. It is easily seen that the only constant solutions of (1) are described by (α) and (β). Next, the case M (R) = {1} is trivial. So, let g be a non-constant solution of (1) and M (R) 6= {1}. Then f := M ◦ g is a continuous and non-constant solution of the equation f (x + f (x)y) = M (λ)f (x)f (y). According to [36, Theorem 3] this means that M (λ) = 1 and there is c ∈ R0 such that either f (x) = cx + 1 for x ∈ R or f (x) = max {cx + 1, 0} for x ∈ R. If M is odd (and consequently bijective), then clearly this yields (γ ).

J. Brzdek ¸ / Nonlinear Analysis 71 (2009) 4396–4404

4403

If M is even (and consequently M (R) ⊂ [0, ∞)), then we have only f (x) = max {cx + 1, 0} for x ∈ R and the continuity of M and g implies that g (x) = M0−1 (max {cx + 1}) for x ∈ R or g (x) = −M0−1 (max {cx + 1, 0}) for x ∈ R. It is easy to check that the first form of g satisfies (1) only when λ = 1 and the second one is a solution of (1) if and only if λ = −1. Hence we have (δ). The converse is easy to check.  Proposition 3. Let λ ∈ R \ {0} and g : R → R be continuous. Then g is a solution of functional equation (9) if and only if it satisfies one of Eq. (1), (25). Proof. Let g be a non-constant solution of (9) and there exist x0 , y0 ∈ R such that g (x0 )g (y0 )g (x0 + M (g (x0 ))y0 ) 6= 0. Then (24) holds with ε1 = ε2 = 0. If M is odd, then by Theorem 2, g satisfies Eq. (27). Clearly, it is possible only when λ = 1, which yields the statement. If M is even and M (R) 6= {1}, then by Theorem 2, g satisfies functional equation (28) with some σ ∈ {−1, 1}. Consequently, λ = σ . Finally, if M (R) = {1}, then in view of Theorem 2 and (9), there is c ∈ R0 with g (x) = λ1 ecx for x ∈ R, which means that g satisfies Eq. (1). The converse is easy to check.  Proposition 4. A continuous g : R → R is a solution of Eq. (25) if and only if either g (x) = 0 for x ∈ R or, (a) in the case M (R) 6= {1}, M is bijective and there exists a > 0 such that  x g (x) ≤ M −1 1 − for x > a and g (x) = 0 for x ≤ a a or  x g (x) ≤ M −1 1 + for x < −a and g (x) = 0 for x ≥ −a; a (b) in the case M (R) = {1}, there exists a nonempty open set A ⊂ R such that A + A ⊂ R \ A and f (x) = 0 for x ∈ R \ A. Proof. The cases M (R) = {1} and g (R) = {0} are trivial. So suppose that M (R) 6= {1} and g : R → R, g (R) 6= {0}, is a continuous solution of (25). Write f := M ◦ g. Then f satisfies the equation f (x + f (x)y)f (x)f (y) = 0 and, according to [34, Theorem 2 and Corollary 2], there is d > 0 such that f ( x) ≤ 1 −

x d

for x > d and

f (x) = 0

for x ≤ d

or f ( x) ≤ 1 +

x d

for x < −d and

f (x) = 0

for x ≥ −d.

Since f (R) = (−∞, 0], M must be bijective (cf. Remark 1). The converse is easy to check.  Example 1. It is easily seen that
A + A ⊂ R \ A. A less trivial example of Sif, for instance, A =: (a, 2a) with some a > 0, then such a set A is following: A := n∈Z (2n + 1 − rn , 2n + 1 + rn ), where rn ∈ 0, 14 for n ∈ Z. References [1] J. Aczél, J. Dhombres, Functional Equations in Several Variables, in: Encyclopedia of Mathematics and its Applications, vol. 31, Cambridge University Press, Cambridge, New York, New Rochelle, Melbourne, Sidney, 1989. [2] J. Brzdek, ¸ The Gołab–Schinzel ¸ equation and its generalizations, Aequationes Math. 70 (2005) 14–24. [3] E. Jabłońska, On solutions of a generalization of the Gołab–Schinzel ¸ equation, Aequationes Math. 71 (2006) 269–279. [4] E. Jabłońska, Continuity of Lebesgue measurable solutions of a generalized Gołab–Schinzel ¸ equation, Demonstratio Math. 39 (2006) 91–96. [5] E. Jabłońska, A short note concerning solutions of a generalization of the Gołab–Schinzel ¸ equation, Aequationes Math. 74 (2007) 318–320. [6] A. Mureńko, On solutions of a conditional generalization of the Gołab–Schinzel ¸ equation, Publ. Math. Debrecen 63 (2003) 693–702. [7] A. Mureńko, On solutions of a common generalization of the Gołab–Schinzel ¸ equation and of the addition formulae, J. Math. Anal. Appl. 341 (2008) 1236–1240. [8] D.H. Hyers, On the stability of the linear functional equation, Proc. Natl. Acad. Sci. USA 27 (1941) 222–224. [9] R. Ger, Superstability is not natural, Rocznik Nauk.-Dydakt. Prace Mat. 13 (1993) 109–123. [10] R.P. Agarwal, B. Xu, W. Zhang, Stability of functional equations in single variable, J. Math. Anal. Appl. 288 (2003) 852–869. [11] S. Czerwik, Functional Equations and Inequalities in Several Variables, World Scientific, London, 2002. [12] D.H. Hyers, G. Isac, Th.M. Rassias, Stability of Functional Equations in Several Variables, Birkhäuser, 1998. [13] S.-M. Jung, Hyers–Ulam–Rassias Stability of Functional Equations in Mathematical Analysis, Hadronic Press, Palm Harbor, FL, 2001. [14] M. Amyari, M.S. Moslehian, Approximate homomorphisms of ternary semigroups, Lett. Math. Phys. 77 (2006) 1–9. [15] R. Badora, J. Chmieliński, Decomposition of mappings approximately inner product preserving, Nonlinear Anal. 62 (2005) 1015–1023. [16] V.A. Fa˘ıziev, Th.M. Rassias, P.K. Sahoo, The space of (ψ, γ )-additive mappings on semigroups, Trans. Amer. Math. Soc. 354 (2002) 4455–4472.

4404

J. Brzdek ¸ / Nonlinear Analysis 71 (2009) 4396–4404

[17] G.L. Forti, Comments on the core of the direct method for proving Hyers–Ulam stability of functional equations, J. Math. Anal. Appl. 295 (2004) 127–133. [18] A. Gilányi, Hyers–Ulam stability of monomial functional equations on a general domain, Proc. Natl. Acad. Sci. USA 96 (1999) 10588–10590. [19] Y.-S. Jung, I.-S. Chang, The stability of a cubic functional equation and fixed point alternative, J. Math. Anal. Appl 306 (2005) 752–760. [20] S.-M. Jung, T.-S. Kim, A fixed point approach to the stability of the cubic functional equation, Bol. Soc. Mat. Mexicana (3) 12 (2006) 51–57. [21] G.H. Kim, On the stability of functional equations with a square-symmetric operation, Math. Inequal. Appl. 4 (2001) 257–266. [22] G.H. Kim, Addendum to ‘On the stability of functional equations on square-symmetric groupoid’, Nonlinear Anal. 62 (2005) 365–381. [23] M. Mirzavaziri, M.S. Moslehian, A fixed point approach to stability of a quadratic equation, Bull. Braz. Math. Soc., New Series 37 (3) (2006) 361–376. [24] Zs. Páles, Hyers–Ulam stability of the Cauchy functional equation on square-symmetric groupoids, Publ. Math. Debrecen 58 (2001) 651–666. [25] Zs. Páles, P. Volkmann, R.D. Luce, Hyers–Ulam stability of functional equations with a square-symmetric operation, Proc. Natl. Acad. USA 95 (1998) 12772–12775. [26] J. Tabor, J. Tabor, General stability of functional equations of linear type, J. Math. Anal. Appl. 328 (2007) 192–200. [27] S.-E. Takahasi, T. Miura, H. Takagi, Exponential type functional equation and its Hyers–Ulam stability, J. Math. Anal. Appl. 329 (2007) 1191–1203. [28] J. Chudziak, Stability problem for the Gołab–Schinzel ¸ type functional equations, J. Math. Anal. Appl. 339 (2008) 454–460. [29] J. Chudziak, Approximate solutions of the Gołab–Schinzel ¸ functional equation, J. Approx. Theory 136 (2005) 21–25. [30] J. Chudziak, On a functional inequality related to the stability problem for the Gołab–Schinzel ¸ equation, Publ. Math. Debrecen 67 (2005) 199–208. [31] J. Chudziak, Approximate solutions of the generalized Gołab-Schinzel ¸ equation, J. Inequal. Appl. 2006 (2006). Art. ID 89402, 8 p. doi:10.1155/JIA/2006/89402. [32] J. Chudziak, Stability of the generalized Gołab-Schinzel ¸ equation, Acta Math. Hungar. 113 (2006) 133–144. [33] J. Chudziak, J. Tabor, On the stability of the Gołab–Schinzel ¸ functional equation, J. Math. Anal. Appl. 302 (2005) 196–200. [34] N. Brillouët-Belluot, J. Brzdek, ¸ On continuous solutions and stability of a conditional Gołab–Schinzel ¸ equation, Publ. Math. Debrecen 72 (2008) 441–450. [35] J. Brzdek, ¸ A. Pietrzyk, A note on stability of the general linear equation, Aequationes Math. 75 (2008) 267–270. [36] J. Brzdek, ¸ Some remarks on solutions of the functional equation f (x + f (x)n y) = tf (x)f (y), Publ. Math. Debrecen 43 (1993) 147–160. [37] T. Kochanek, M. Lewicki, Stability problem for number-theoretically multiplicative functions, Proc. Amer. Math. Soc. 135 (2007) 2591–2597.