Non-periodic solutions of the Goła̧b–Schinzel type functional equation

We determine the solutions of the Goła̧b-Schinzel type functional equation in the class of non-periodic functions. Applying this result we give a positive answer to the problem raised by E. Jabłońska (Aequationes Math 87:125–133, 2014).


Introduction
The paper is inspired by an open problem, posed by Jab lońska [16], concerning the solutions of the Go lab-Schinzel type functional equation where f, g : R → R are unknown functions. Equation (1) is a generalization of the Go lab-Schinzel type equations and where n is a fixed positive integer. Equations (2)-(3) play an important role in determination of substructures of algebraic structures (see e.g. [1,2,4,5,12,13] and [18]). Furthermore, the conditional versions of (2) are strictly related to some problems arising in meteorology and fluid mechanics (cf. [17]). In a more general setting, where an unknown function f maps a real linear space into R, solutions of (2) and (3) have been investigated by several authors under various regularity assumptions. More details concerning properties of the solutions of these equations as well as their applications can be found in a survey paper [6].
In order to formulate the aforementioned problem, recall the main result proved in [16]. and It has been noted in [16] The same form of solutions of equation (2) was established in [19] under the assumption that an unknown function is not microperiodic. This fact inspired E. Jab lońska to raise the following question: does Theorem 1.1 hold, if we replace the assumption that f is locally bounded above at each point of the line by the assumption that f is not periodic? In this paper we give a positive answer to this question. In fact, we determine the form of non-periodic solutions of a significantly more general functional equation, namely where X is a real linear space, (S, •) is a commutative semigroup, f : R → S and g : R → R are unknown functions. In the case X = R equation (6) has been introduced and studied in [7]. The solutions of (6) under some regularity assumptions have been determined in [8]. The following particular case of (6) and some its conditional versions have been investigated under various regularity assumptions in [9][10][11] and [14,15]. It is remarkable that solutions of (6) and their applications to invariance under binomial thinning have been recently considered in [3].

Results
In what follows, X is a non-zero real linear space and (S, •) is a commutative semigroup. Furthermore, E(S) denotes the set of all idempotents of the semigroup (S, •), that is Moreover, for a function g : X → R, we set Remark 1. Let f : X → S and g : X → R. If the pair (f, g) satisfies equation (6) then The following three auxiliary results will play a crucial role in our considerations.
Lemma 2.1. Assume that f : X → S, g : X → R and the pair (f, g) satisfies equation (6). and Proof. Assume that A g = X and fix an x 0 ∈ X \ A g . Since the semigroup (S, •) is commutative, setting in (6) x = x 0 , we obtain (10) with z := f (x 0 ). In view of (8), we have z ∈ E(S). Moreover, applying (6) and (10), for every x ∈ X \ A g , we get that is, (9) holds.

Lemma 2.2.
Assume that a pair (f, g), where f : X → S and g : X → R, satisfies equation (6). If f is non-constant then: (ii) for every x, y ∈ X, it holds: Proof. Assume that f is non-constant. Suppose that 0 ∈ A g . Then A g = X and so, according to Lemma 2.1, there exists a z ∈ E(S) such that (9)-(10) hold. Hence, in view of (6), for every x ∈ X, we get Since f is non-constant, this yields a contradiction and proves (i). In order to prove (ii) note that, applying (6), for every x, y, z ∈ X, we obtain Hence, as f is non-constant, for every x, y ∈ X, we get which implies (11).

Lemma 2.3.
Assume that a pair (f, g), where f : X → S and g : X → R, satisfies equation (6). If f is non-periodic then Proof. In view of (6), for every x, y, z ∈ X, we get Replacing in this equality z by z−g(y)x−y g(x)g(y) , for every x, y ∈ A g and z ∈ X, we obtain Therefore, if f is non-periodic, then (12) holds.
In what follows we call a mapping φ defined on a subset G of R periodic provided there exists a t 0 ∈ R \ {0} such that and The next theorem is the main result of the paper.
Theorem 2.4. Assume that f : X → S and g : X → R. If the pair (f, g) satisfies equation (6) and f is non-periodic then one of the following possibilities holds: Vol. 78 (2023) Non-periodic solutions of the Go lab-Schinzel type Page 5 of 15 28 and Conversely, in any case (a)-(d) the pair (f, g) satisfies equation (6) and f is non-periodic.
Proof. Assume that the pair (f, g) satisfies equation (6) and f is non-periodic. Then f is non-constant and so, according to Lemma 2.2 (i), we have 0 ∈ A g . We shall divide our considerations into the following three cases: A g \ {0} = ∅ and g(y 0 ) = 1 for some y 0 ∈ A g . Case 1. Taking a := g(0), s := f (0) and applying Remark 1 and Lemma 2.1, we obtain that a = 0, s ∈ E(S) and there exists a z ∈ E(S) such that (15)- (16) hold. Moreover, setting in (6) x = 0 and y ∈ X \ {0}, in view of (15)-(16), we get s • z = z. As f is non-constant, we have also s = z. Thus, (a) holds. Case 2. If A g = X then g = 1 and so, in view of (6), f is a homomorphism of the group (X, +) into (S, •). Furthermore, if f (y 1 ) = f (y 2 ) for some y 1 , y 2 ∈ X, then for every x ∈ X, we obtain Since f is non-periodic, this implies that y 1 = y 2 and proves the injectivity of f . Thus, (b) holds.
Assume that A := A g = X. Then 0 ∈ A, g is of the form (18) and, in view of (11), we get Furthermore, since applying (11) again, we conclude that −x ∈ A for x ∈ A. Hence, A is a subgroup of the group (X, +). Moreover, as A \ {0} = ∅ and A = X, the subgroup is non-trivial and proper.
Let ψ : A → S be given by ψ(x) = f (x) for x ∈ A. Then, making use of Lemma 2.1, we obtain that there exists a z ∈ E(S) such that (17) and (19) hold. Furthermore, taking into account (6) and (18), we get Thus, ψ is a homomorphism. In order to show that ψ is injective, suppose that ψ(y 1 ) = ψ(y 2 ) for some y 1 , y 2 ∈ A. Since A is a subgroup of the group (X, +), we have y 1 − y 2 ∈ A. Hence, according to (11) and (18), we obtain Therefore, in view of (19), we get As f is non-periodic, this implies that y 1 = y 2 . Hence, ψ is injective and so, (c) is valid. Case 3. According to Lemma 2.3, we get Vol. 78 (2023) Non-periodic solutions of the Go lab-Schinzel type Page 7 of 15 28 Since A g \ {0} = ∅, this implies that y 0 = 0. Thus, replacing in (24) x by ty 0 , we obtain g(ty 0 ) = 1 − (1 − g(y 0 ))t whenever ty 0 ∈ A g .
In particular, we have Hence, putting we get x 0 = 0 and Moreover, in view of (24)-(25), we have Therefore, taking we conclude that g is of the form (22). Since {0, y 0 } ⊆ A g , it follows from (25) and (28) that Note that, in view of (26), we have x 0 ∈ A g . Thus, taking into account (28), we get G ⊆ R \ {0}. We show that G is a non-trivial subgroup of the group (R \ {0}, ·) and φ : G → S, given by is a homomorphism of G into (S, •). To this end, fix ω, ξ ∈ G. Then ω = 1 − s and ξ = 1 − t with some s, t ∈ R such that sx 0 ∈ A g and tx 0 ∈ A g . Hence, applying Lemma 2.2(ii), in view of (26), we obtain Furthermore, using (26) again, we get Hence, according to Lemma 2.2 (ii), we have − s 1−s x 0 ∈ A g and so In this way we have proved that G is a subgroup of the group (R \ {0}, ·). Note also that, in view of (29), G is non-trivial. Furthermore, making use of (6), (26) and (30), we get Therefore, φ is a homomorphism of G into (S, •). Moreover, taking into account (27)-(28) and (30), in view of Lemma 2.1, we obtain that there exists a z ∈ E(S) such that (21) and (23) hold. Note that, if φ(t z ) = z for some t z ∈ G then, making use of (21), we get Since f is a non-constant function, this contradicts (23). Thus, (20) holds. Suppose that φ is periodic. We claim that To this end, fix an x ∈ X. If x ∈ Span{x 0 } then x − t 0 x 0 ∈ Span{x 0 } and x + t 0 x 0 ∈ Span{x 0 }. Hence, in view of (23), we get If x = tx 0 for some t ∈ R with 1 − t ∈ G then, by (13), we have 1 − t − t 0 ∈ G and 1−t+t 0 ∈ G. Thus, applying (23) again, we obtain (32). Finally, if x = tx 0 for some t ∈ R with 1 − t ∈ G then, in view of (13), we get 1 − t + t 0 ∈ G and 1 − t − t 0 ∈ G. Hence, making use of (14) and (23), we conclude that In this way we have proved (31). Since f is non-periodic and t 0 x 0 = 0, this yields a contradiction. Therefore, φ is non-periodic and so, (d) is valid. A standard computations show that, if one of the possibilities (a)-(d) holds, then the pair (f, g) satisfies equation (6). Furthermore, it is obvious that in the cases (a)-(b) f is non-periodic.
Corollary 2.5. Assume that f : X → R and g : X → R. If the pair (f, g) satisfies equation (7) and f is non-periodic then one of the following possibilities holds: (i) there exists an a ∈ R \ {0} such that g is of the form (15) and (ii) g = 1 and there exist an injective additive function a : X → R such that (iii) there exist a non-trivial proper subgroup A of the group (X, +), a nontrivial subgroup A 0 of A and an injective additive function a : A → R such that g is of the form (18) and (iv) there exist an x 0 ∈ X \ {0}, a non-trivial subgroup G of the group (R \ {0}, ·) and a non-periodic multiplicative function φ : G → R \ {0} such that g is of the form (22) and Conversely, if one of the possibilities (i)-(iv) holds, then the pair (f, g) satisfies equation (7) and f is non-periodic.
Proof. Assume that the pair (f, g) satisfies equation (7) and f is non-periodic. Then, applying Theorem 2.4, with (S, •) being the multiplicative semigroup of real numbers, we conclude that one of the possibilities (a)-(d) holds. Note that in this case, we have E(S) = {0, 1} and so (s, z) = (1, 0) is the only pair of elements of E(S) such that s = z and s • z = z. Hence, (a) leads to (i).
In the case (b), we have f (x) = 0 for x ∈ X. In fact, if f (y) = 0 for some y ∈ X, then which contradicts the injectivity of f . Hence, Thus, defining a function a : X → R by a(x) = ln f (x) for x ∈ X, we conclude that a is injective and additive. Moreover, f is of the form (38) and so (ii) is valid.
Hence, A 0 is a subgroup of A and (39) holds. Note also that, as A is a nontrivial subgroup of the group (X, +), in view of (39), A 0 is non-trivial. Let a : A → R be given by a(x) = ln |ψ(x)| for x ∈ A. Then a is an additive function and Therefore, as z = 0, taking into account (19), we conclude that f is of the form (40). In order to show that a is injective, suppose that a(x 0 ) = 0 for some x 0 ∈ A. Then a(2x 0 ) = 0. Moreover, as A 0 is a subgroup of A, in view of (39), we have 2x 0 ∈ A 0 . Thus, applying (42), we get ψ(2x 0 ) = ψ(0) = 1. Since ψ is injective this implies that 2x 0 = 0 and so x 0 = 0. Hence, a is injective. In this way we have proved that (iii) holds.
The converse is a direct consequence of Theorem 2.4.