Semigroup-valued solutions of some composite equations

AbstractLet X be a linear space over the field K of real or complex numbers and (S, °) be a semigroup. We determine all solutions of the functional equation
$$f(x+g(x)y)=f(x)\circ f(y)\quad \text{for}\quad x,y\in X$$f(x+g(x)y)=f(x)∘f(y)forx,y∈Xin the class of pairs of functions (f,g) such that f : X → S and g : X → K satisfies some regularity assumptions. Several consequences of this result are presented.


Introduction
Let X be a linear space over the field K of real or complex numbers. The solutions f : X → K of the Go lab-Schinzel functional equation f (x + f (x)y) = f (x)f (y) for x, y ∈ X, (1) have been intensively studied in the last half-century. Equation (1) is one of the most important equations of a composite type and plays a prominent role e.g. in the determination of substructures of various algebraical structures [1, pp. 311-319], [3,4]. The solutions of (1) and its further generalizations, namely where n is a nonnegative integer and t is a nonzero real number; and where M : K → K, have been considered under various regularity assumptions e.g. in [2][3][4] and [19][20][21]. In the real case the functional equation where • is a binary operation on R satisfying some additional conditions (commutativity, associativity etc.), was studied in [5,7,16,25] and [26]. For more information concerning (1)- (4) and their further applications (e.g. to mathematical meteorology and fluid dynamics) we refer to the survey paper [6]. Various aspects of stability problems for the Go lab-Schinzel functional equations were considered in [8][9][10][11][12][13]15,17] and [18]. In the case where (S, •) is an arbitrary semigroup, the general solution of the equation in the class of pairs (f, g) such that f : R → S and g : R → R is continuous, was determined in [14]. The functional equation f (x + g(x)y) = f (x)f (y) for x, y ∈ X was considered in [22] under the assumption that f and g, mapping a real linear space X into R, are continuous on rays.
In the present paper we generalize substantially the results from [14] and [22] in various directions. Namely, we determine the general solution of the equation in the case where X is a linear space over the field K of real or complex numbers, (S, •) is an arbitrary semigroup, f : X → S and g : X → K satisfies some regularity assumptions. Several consequences of this result are presented, as well. In particular, applying our main result and using a natural correspondence between (5) and the pexiderized version of the Go lab-Schinzel equation, that is we obtain a generalization of the results in [23].
In what follows B(x, r) denotes the open ball (in K) with a center at x ∈ K and a radius r > 0. Let us recall [24, p. 596] that given a nonempty subset A of X, we say that a ∈ A is an algebraically interior point of A, provided, for By int a A we denote a set of all algebraically interior points of A. If f : X → R and x ∈ X then a function f x : R → R is given by where g is an arbitrary function and f ≡ s ∈ E(S), is a solution of (5).
Next, we present a result describing degenerate solutions of (5), i.e. such solutions (f, g) that either f or g is constant. (5) if and only if one of the following conditions is valid: Proof. It is clear that if one of the conditions (i) − (iii) holds, then (f, g) is a degenerate solution of (5). So, assume that (f, g) is a degenerate solution of (5). If f is constant, then according to Remark 1, we get (i) with s := f (0). Now, assume that f is nonconstant and g is constant, say g ≡ c. If c = 0, then (ii) holds with S 0 := f (X). The case where c = 1 leads to (iii). Suppose that c / ∈ {0, 1}. Then, in view of (5), This yields a contradiction, because f is nonconstant.
From now on we will deal only with the non-degenerate solutions of (5), i.e. with such solutions (f, g) that neither f nor g is constant.
The following result plays a crucial role in our considerations.

Proposition 2.
Let X be a linear space over the field K of real or complex numbers, (S, •) be a semigroup, f : X → S and g : X → K. Assume that (f, g) is a non-degenerate solution of (5). Then each of the following regularity conditions: or there exists a nontrivial R-linear functional L : X → R such that Proof. In view of (5), for every x, y, z ∈ X, we have Thus, as • is associative, we get Since f is nonconstant, this means that g(x + g(x)y) = 0. In this way we have proved that, for every x, y ∈ X, it holds that So, if (C 1 ) is valid, according to [17,Theorem 1], we obtain that Hence, applying [4, Theorem 3], we get the assertion. Now, assume that (C 2 ) holds. We show that 0 ∈ g(X). Suppose that 0 ∈ g(X). Then, by (5), we get (5) and (11), we obtain Vol. 88 (2014) Semigroup-valued solutions of some composite equations 187 Next, as (f, g) is a non-degenerate solution of (5), there is Then, in view of (12), f (a) = f (0). Furthermore, let r x > 0 be such that We claim that there is r > 0 such that Suppose that (14) does not hold. Then there exists a sequence (t n ) of elements of K converging to 0 and such that f (t n x) = f (0) for n ∈ N. Then, by (12), g(t n x) = 1 for n ∈ N and so, in view of (5), for every n ∈ N, we get On the other hand, from (13) it follows that a + t n x ∈ G 1 for sufficiently large n ∈ N. Hence, by (12), f (a + t n x) = f (0) for sufficiently large n ∈ N, which yields a contradiction. In this way we have proved (14). Since the Therefore, making use of (12), we get Now, we show by induction that for every b ∈ B(0, r) and n ∈ N it holds that Note that by (5), (14) and (15), for every b ∈ B(0, r), we have Thus, (16) is valid for n = 1. Next, fix n ∈ N and assume that (16) holds for every b ∈ B(0, r). Then, in view of (12), g((k + bn)x) = 1, so applying (5), (14) and (16), for every b ∈ B(0, r), we obtain In this way we have proved that (16) holds for every b ∈ B(0, r) and n ∈ N. Note however that, taking n 0 ∈ N with k n0 ∈ B(0, r), we have This yields a contradiction and shows that 0 ∈ g(X). Since by (C 2 ) applying [17, Theorem 1], from (9) we deduce (10). Thus, according to [4, Theorem 3], we get the assertion.

Main results
The next theorem is the main result of the paper. and such that g is of the form (7) and (ii) there exist a nontrivial R-linear functional L : X → R, functions a : X → S and φ : [0, ∞) → S satisfying (17), and l ∈ Z L (φ([0, ∞))) ∩ E(S) such that g is of the form (8) and Vol. 88 (2014) Semigroup-valued solutions of some composite equations 189 Proof. Assume that a pair (f, g) is a non-degenerate solution of (5). Then, according to Proposition 2, g is either of the form (7) or (8). In the first case, let us fix x 0 ∈ X \ ker L and define the functions Π 1 , Π 2 : X → X by and Then Π 1 and Π 2 are linear and Π 1 (x) + Π 2 (x) = x for x ∈ X. Moreover L(Π 1 (x)) = 0 for x ∈ X, which yields that Therefore, taking a : X → S of the form and φ : K → S of the form in view of (5), we obtain Moreover, for every x ∈ X, we get Hence by (25) and (27), we obtain Thus, taking (28) into account, we conclude that (20) holds. It remains to show (19). To this end note that, in view of (26), for every x ∈ X and t ∈ K, we have 190 J. Chudziak AEM Hence, by (5), we obtain So, taking (27) and (28) into account, we get (19). In this way we have proved that (i) is valid.
Since the converse is easy to check, the proof is completed.
In the case of a commutative semigroup, a description of the solutions of (5) is significantly simpler. Namely, we have the following result. and z ∈ Z(ψ([0, ∞))) such that ψ = z, g is of the form (8) and Proof. Assume that a pair (f, g) is a non-degenerate solution of (5). Then one of the conditions (i), (ii) of Theorem 1 holds. In the case of (i), (19) and the commutativity of • imply that a(2x) Thus, in view of (17), for every x ∈ X, we get Example 1. Let X be a real linear space of dimension at least 2 and let S = R 2 be endowed with the following binary operation (x 1 , y 1 ) • (x 2 , y 2 ) = (x 1 y 1 , x 1 y 2 + y 1 ) for (x 1 , y 1 ), (x 2 , y 2 ) ∈ S.
Define the functions a : X → S and φ : R → S by a(x) = (1, x) for x ∈ X and φ(t) = (t, 0) for t ∈ R, respectively. Then an easy calculation shows that (17)- (19) hold. So, taking a nontrivial R-linear functional L : X → R and applying Theorem 1(i), we conclude that a pair of functions (f, g), where g : X → R is of the form (7) and f : X → S is given by satisfies (5). Note also that f is injective and, as dim X ≥ 2, L is not. Thus, f can not be represented in the form f (x) = ψ(L(x) + 1) for x ∈ X with some ψ : R → S.
Next, we present the result concerning the solutions of (5) in the case where (S, •) is a group. Proof. Suppose that (f, g) is a non-degenerate solution of (5). Then one of the conditions (i) or (ii) of Theorem 1 holds. In the first case, putting t = 0 in (18) and considering the fact that (S, •) is a group, we conclude that φ is constant. Furthermore, applying (19) with t = 0, we get φ(0) • a(x) = a(0) • φ(0) for x ∈ X, so also a is constant. Hence, in view of (20), f is constant, which yields a contradiction. In the case where condition (ii) of Theorem 1 holds, the same arguments yield a contradiction (note that in this case φ = e, Z L (φ([0, ∞))) ∩ E(S) = {e} and so l = e).
In this way we have proved that Eq. (5) has only degenerate solutions. Furthermore, as (S, •) is a group, we have E(S) = {e}. Moreover, the only subsemigroup S 0 of (S, •) such that u • v = u for u, v ∈ S 0 , is S 0 = {e}. Therefore, applying Proposition 1, we obtain the assertion.
We complete the paper with two results concerning (6), which generalize some results in [23]. satisfies equation (6) if and only if there exist x 0 ∈ X, s, t ∈ S, k ∈ K \ {0} and functions f : X → S, g : X → K such that a pair (f, g) satisfies (5) and Proof. Assume that a quadruple (F, G, H, K) satisfies Eq. (6). By the assumption, there is x 0 ∈ X such that F (x 0 ) • p = p • F (x 0 ) = e for some p ∈ S. In (6) taking y = 0 and next x = x 0 , we get and respectively. Since, in view of (42), using the commutativity of •, we derive that So considering (42), we get In a similar way, using (43), we obtain Furthermore, by (6) and (44), for every x, y ∈ X, we get Hence, in view of (42) and (43), we obtain Note also that G(x 0 ) = 0. Otherwise, as G is nonconstant, In (45) putting y = y 1 with G(y 1 ) = 0, we would have that F is constant, which yields a contradiction. Now, let and