On continuous on rays solutions of a composite-type equation

Let X be a real linear space. We characterize solutions $${f, g : X \rightarrow \mathbb{R}}$$f,g:X→R of the equation f(x + g(x)y) = f(x)f(y), where f is continuous on rays. Our result refers to papers by Brzdȩk (Acta Math Hungar 101:281–291, 2003), Chudziak (Aequat Math, doi:10.1007/s00010-013-0228-4, 2013) and Jabłońska (J Math Anal Appl 375:223–229, 2011).


Introduction
The Go lab-Schinzel equation f (x + f (x)y) = f (x)f (y) was introduced in [5] in connection with looking for subgroups of the centroaffine group of the field. This equation and its generalizations seem to be very important composite-type functional equations because of their applications, especially in algebra, in the theory of geometric objects, as well as in differential equations appearing in meteorology and fluid mechanics (the extensive bibliography concerning the Go lab-Schinzel-type functional equations and their applications can be found in [2] and [8]).
In 2012 in [7] it was proved that solutions of the pexiderized Go lab-Schinzel equation, i.e. the equation with four unknown functions, can be described by solutions of the following equation:

E. Jab lońska AEM
For the first time the equation (1.1) was studied in 2006 by J. Chudziak [3]. Among others, he determined all real solutions of this equation under the assumption that g is continuous. Seven years later, in [9], real solutions of (1.1) were studied under the assumption that the function f is continuous. It turned out that the following theorem holds (which is essential in our considerations): and r > 0 such that f and g have one of the forms: Consequently, we obtain that the continuity of each nonconstant the function f implies the continuity of g provided f, g satisfy (1.1).
Here we prove that if f, g satisfy (1.1), then the continuity on rays of the nonconstant function f implies the continuity on rays of g.
For the first time solutions of (1.1) in the class of continuous on rays functions f and g were studied in [6]. J. Chudziak [4] generalized this result, assuming only the continuity on rays of g. We also show, how to obtain forms of solutions of (1.1) using Chudziak's result from [4] under the assumption that f is continuous on rays.
Our paper also refers to the paper [1], where J. Brzdȩk considered the equation under the assumptions that f is continuous on rays, M is continuous and H is symmetric.
In the whole paper we use the following notation: Moreover, X is a real linear space.

Preliminary lemmas
First we recall two lemmas on basic properties of solutions of (1.1), which will be useful in the sequel.
Let us recall that a function k : X → R is continuous on rays if and only if It means that 0 ∈ f (X) and, according to Lemma 2.1(ii), 0 ∈ g(X). Thus g = 1, which gives a contradiction.

The main result
Proof. Let f and g satisfy (1.1) and f be continuous on rays. Clearly, according to (1.1), if f is constant, then f = 1 or f = 0 and (i) holds.
So, assume that f is not constant and g = c for some c ∈ R. Clearly, by (1.1), c = 0. Suppose that c = 1. Then, in view of (1.1), for x ∈ X. According to Lemma 2.1(ii) 0 ∈ f (X), so f = 1, which contradicts the assumption. Thus g = 1 and, using (1.1), we find that f is a nonconstant exponential function. Hence, according to [ First we prove a property of f, g, which is essential in the whole proof: 4) or there is r > 0 such that Next, we will show that g is continuous on rays (then we will be able to apply Chudziak's result from [4]).
(3.6) So, assume that n i=1 RA ⊂ A for some n ≥ 1. Then, by (1.1) and the first step of induction, for every z i ∈ A and a i ∈ R, where i ∈ {1, . . . , n + 1}, we obtain It means that n+1 i=1 RA ⊂ A and hence, by mathematical induction, (3.6) holds and A is a linear subspace of X. Now, from Lemma 2.2 and Lemma 2.3 we have A = B. Hence, since (3.4) holds for each x ∈ A, we obtain that g x = 1 for each x ∈ A. Moreover, if x ∈ X \ A = X \ B, then, in view of Theorem 1.1, g x is given by one of conditions (1.3)-(1.5). It means that g is continuous on rays.
Case 2. Now, we consider the case where there is x 0 ∈ A such that g x0 , f x0 are given by (3.5). Then f (x 0 ) = f x0 (1) = 1, g(x 0 ) = g x0 (1) = −1 and, for each x ∈ F , Hence, according to Lemma 2.1(iii), Thus, by the property (A), one of conditions (3.4)-(3.5) holds for x ∈ F ; i.e. for x ∈ F either or there is r > 0 such that First we show that (3.8) holds for each x ∈ F . Suppose for contradiction that there is x ∈ F such that (3.7) holds. Then Now, we prove that if f y , g y fulfill (3.10) for some y ∈ X \ {0}, then g y = 1. So, let f y = 1 and g y (R) ⊂ {−1, 1} for y ∈ X \ {0} and suppose for contradiction that there is α ∈ R with g y (α) = −1. Hence, by Lemma 2.1(ii), for each x ∈ F we have But f x is given by (3.11) or (3.12) for x ∈ F \ A, so and whence c x = 0, which is a contradiction.
In this way we obtain that either f x = 1 and g x = 1, or g x , f x satisfy one of conditions (3.11), (3.12) for x ∈ X \ {0}. It means that g is continuous on rays.
In both cases the continuity on rays of g was proved. Now, we can apply Chudziak's result; by [4,Proposition 3] f and g are given by one of the following two forms: where L : X → R is a nontrivial linear functional and ψ : R → R is a nonconstant multiplicative function, or with a nontrivial linear functional L : X → R and a nonconstant multiplicative function ψ : [0, ∞) → [0, ∞).