Functional equations of the Goła̧b-Schinzel type on a cone

Abstract

We determine the solutions of conditional composite type functional equations related to some problems in mathematical meteorology and fluid mechanics.

1 Introduction

Dealing with some problems in mathematical meteorology and fluid mechanics (e.g. evaporation of cloud droplets and water discharging from a reservoir) Kahlig and Matkowski [15] noted that the corresponding nonlinear differential equations exhibit some symmetries which can be expressed by the functional equation

$$\begin{aligned} F\left( x+ys(x)^{r}\right) =s(x)F(y) \;\; \text{ for } \;\; x,y \in [0, \infty ), \end{aligned}$$

(1)

where r is a positive real constant. It turns out that Eq. (1) by some simple substitutions can be reduced to the equation

$$\begin{aligned} f(x+f(x)y) = f(x)f(y)\quad \text{ for } \;\; x,y \in [0, \infty ), \end{aligned}$$

(2)

which is the Goła̧b-Schinzel equation on a restricted domain. More details on the Goła̧b-Schinzel type functional equation can be found in [3]. Continuous solutions $f:\mathbb {R}\rightarrow \mathbb {R}$ of Eq. (2) have been determined in [1]. Applying the results of [1], P. Kahlig and J. Matkowski [15] have determined the solutions of (1) in the class of pairs of functions (F, s), where $F:[0,\infty )\rightarrow [0,\infty )$ is continuous (or monotone) and $s:[0,\infty )\rightarrow \mathbb {R}$. Inspired by this fact several authors have considered conditional versions of the Goła̧b-Schinzel equation. In particular the solutions of the equation

$$\begin{aligned} f(x+f(x)y) = f(x)f(y) \;\;\; \text{ whenever } \;\;\; x,y, x+f(x)y\ge 0 \end{aligned}$$

has been considered in [19]. For further results concerning conditional Goła̧b-Schinzel type functional equations we refer to [2, 4, 17]–[18] and [20]–[21].

In a series of papers [7, 9] and [12]–[14] the following generalization of the Goła̧b-Schinzel equation

$$\begin{aligned} F(x+G(x)y)=F(x)F(y) \end{aligned}$$

(3)

has been studied in various settings. Further going pexiderization of the Goła̧b-Schinzel equation have been investigated in [5]–[6] and [10]–[11].

In a recent paper [8] the results of [1] and [19] have been generalized. More precisely, the continuous solutions of the equations

$$\begin{aligned} f(x+g(x)y) = f(x)f(y) \;\;\; \text{ whenever } \;\;\; x,y, x+g(x)y\ge 0 \end{aligned}$$

(4)

and

$$\begin{aligned} f(x+g(x)y) = f(x)f(y) \;\;\; \text{ for } \;\;\; x,y\in [0,\infty ) \end{aligned}$$

(5)

have been determined.

In the present paper, applying the results of [8], we deal with a similar problem, but in a much more general setting. Namely, given a real linear space X and a convex cone $\mathcal {C}$ in X, that is a nonempty subset of X such that $\alpha x+\beta y\in \mathcal {C}$ for $x,y\in \mathcal {C}$ and $\alpha ,\beta \in [0,\infty )$, we determine the solutions of the equation

$$\begin{aligned} f(x+g(x)y)=f(x)f(y) \;\;\; \text{ whenever } \;\;\; x,y,x+g(x)y\in \mathcal {C} \end{aligned}$$

(6)

in the class of pairs (f, g) of functions continuous on rays mapping $\mathcal {C}$ into $\mathbb {R}$. Roughly speaking, our main result says that the pair (f, g) of functions continuous on rays satisfies Eq. (6) if and only if there exist continuous functions $\tilde{f}, \tilde{g}:\mathbb {R}\rightarrow \mathbb {R}$ and a nontrivial linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that the pair $(\tilde{f},\tilde{g})$ satisfies Eq. (4), $f=\tilde{f}\circ L_{|\mathcal {C}}$ and either $g=\tilde{g}\circ L_{|\mathcal {C}}$, or $L(\mathcal {C})=(-\infty ,0]$ and, for every $x\in \mathcal {C}$, $g(x)=(\tilde{g}\circ L)(x)$ whenever $1+L(x)>0$ and $g(x)\ge 0$, otherwise. Furthermore we show that the similar assertion holds for the solution of the equation

$$\begin{aligned} F(x+ G(x)y) = F(x)F(y) \;\; \text{ for } \;\; x,y \in \mathcal {C} \end{aligned}$$

(7)

in the class of pairs (F, G) of functions continuous on rays mapping X into $\mathbb {R}$. Let us recall that a function $F:X\rightarrow \mathbb {R}$ is said to be continuous on rays provided, for every $x\in X$, the function $F_x:\mathbb {R}\rightarrow \mathbb {R}$ given by $F_x(t)=F(tx)$ for $t\in \mathbb {R}$, is continuous. A function $f:\mathcal {C}\rightarrow \mathbb {R}$ is said to be continuous on rays provided, for every $x\in \mathcal {C}$, the function $f_x:[0,\infty )\rightarrow \mathbb {R}$ given by $f_x(t)=f(tx)$ for $t\in [0,\infty )$, is continuous.

In what follows X stands for a real linear space and $\mathcal {C}$ denotes a convex cone in X. These assumptions will not be repeated.

2 Preliminary results

The following result will play a crucial role in our considerations [cf. [8, Theorem 3.1]]. So, we provide it here for reader’s convenience.

Theorem 2.1

Assume that $f,g:[0,\infty ) \rightarrow \mathbb {R}$ are continuous functions. Then the pair (f, g) satisfies Eq. (4) if and only if one of the subsequent possibilities holds:

• (i) $f(x)=0 \;\; \text{ for } \;\; x\in [0,\infty )$ and g is arbitrary;

• (ii) $f(x)=1 \;\; \text{ for } \;\; x\in [0,\infty )$ and g is arbitrary;

• (iii) $g(x)=1 \;\; \text{ for } \;\; x\in [0,\infty )$ and there exists a $c\in \mathbb {R}{\setminus }\{0\}$ such that

  $$\begin{aligned} f(x)=e^{cx} \;\;\; \text{ for } \;\;\; x\in [0,\infty ); \end{aligned}$$

• (iv) There exist a $c\in (0,\infty )$ and an $r\in \mathbb {R}{\setminus }\{0\}$ such that

  $$\begin{aligned} g(x) = 1 + cx \;\;\; \text{ for } \;\;\; x\in [0,\infty ) \end{aligned}$$

  (8)

  and

  $$\begin{aligned} f(x) = (1 + cx)^r \;\;\; \text{ for } \;\;\; x\in [0,\infty ); \end{aligned}$$

• (v) There exist a $c\in (-\infty ,0)$ and an $r\in (0,\infty )$ such that g is of the form (8) and either

  $$\begin{aligned} f(x) = |1 + cx|^r \;\;\; \text{ for } \;\;\; x\in [0,\infty ) \end{aligned}$$

  (9)

  or

  $$\begin{aligned} f(x) = |1 + cx|^r \cdot \mathrm{sgn}\,(1+cx)\;\;\; \text{ for } \;\;\; x\in [0,\infty ); \end{aligned}$$

  (10)

• (vi) There exist a $c\in (-\infty ,0)$, an $r\in (0,\infty )$ and a continuous function $\phi :\left[ -\frac{1}{c},\infty \right) \rightarrow [0,\infty )$ such that $\phi \left( -\frac{1}{c}\right) =0$,

  $$\begin{aligned} g(x) = \left\{ \begin{array}{ll} 1 + cx &{}\quad \text{ for } \;\; x \in [0, -\frac{1}{c})\\ \phi (x) &{}\quad \text{ for } \;\; x\in [-\frac{1}{c},\infty ) \end{array} \right. \end{aligned}$$

  and

  $$\begin{aligned} f(x)=(\max \{1 + cx,0\})^{r} \;\;\; \text{ for } \;\;\; x\in [0,\infty ). \end{aligned}$$

Lemma 2.1

Assume that $f,g:\mathcal {C}\rightarrow \mathbb {R}$ and the pair (f, g) satisfies Eq. (6). Then either $f=0$ or $f(0)=1$.

Proof

Applying (6) with $x=y=0$, we get $f(0)\in \{0,1\}$. Furthermore, if $f(0)=0$ then putting in (6) $y=0$ we obtain that $f=0$. $\square$

Lemma 2.2

Assume that functions $f,g:\mathcal {C}\rightarrow \mathbb {R}$ are continuous on rays, $f\ne 0$ and the pair (f, g) satisfies Eq. (6). Then, for every $x\in \mathcal {C}$, one of the following possibilities holds:

• $(C_1)$ $f_x(t)=1 \;\; \text{ for } \;\; t\in [0,\infty )$;

• $(C_2)$ $g_x(t)=1 \;\; \text{ for } \;\; t\in [0,\infty )$ and there exists a $c(x)\in \mathbb {R}{\setminus }\{0\}$ such that

  $$\begin{aligned} f_x(t)=e^{c(x)t} \;\; \text{ for } \;\; t\in [0,\infty ); \end{aligned}$$

• $(C_3)$ there exist a $c(x)\in (0,\infty )$ and an $r(x)\in \mathbb {R}{\setminus }\{0\}$ such that

  $$\begin{aligned} g_x(t)=1+c(x)t \;\;\; \text{ for } \;\;\; t\in [0,\infty ) \end{aligned}$$

  (11)

  and

  $$\begin{aligned} f_x(t)=(1+c(x)t)^{r(x)} \;\;\; \text{ for } \;\;\; t\in [0,\infty ); \end{aligned}$$

  (12)

• $(C_4)$ there exist a $c(x)\in (-\infty ,0)$ and an $r(x)\in (0,\infty )$ such that $g_x$ is of the form (11) and

  $$\begin{aligned} f_x(t)=|1+c(x)t|^{r(x)} \;\;\; \text{ for } \;\;\; t\in [0,\infty ); \end{aligned}$$

  (13)

• $(C_5)$ there exist a $c(x)\in (-\infty ,0)$ and an $r(x)\in (0,\infty )$ such that $g_x$ is of the form (11) and

  $$\begin{aligned} f_x(t)=|1+c(x)t|^{r(x)} \cdot \mathrm{sgn}\,(1+c(x)t)\;\;\; \text{ for } \;\;\; t\in [0,\infty ); \end{aligned}$$

  (14)

• $(C_6)$ there exist a $c(x)\in (-\infty ,0)$, an $r(x)\in (0,\infty )$ and a continuous function $\phi ^{(x)}:\left[ -\frac{1}{c(x)},\infty \right) \rightarrow [0,\infty )$ such that $\phi ^{(x)}\left( -\frac{1}{c(x)}\right) =0$,

  $$\begin{aligned} g_x(t) = \left\{ \begin{array}{l@{\quad }l} 1 + c(x)t &{} \text{ for } \;\; t \in \left[ 0,-\frac{1}{c(x)}\right) ,\\ \phi ^{(x)}(t) &{} \text{ for } \;\; t\in \left[ -\frac{1}{c(x)},\infty \right) \end{array} \right. \end{aligned}$$

  (15)

  and

  $$\begin{aligned} f_{x}(t)=(\max \{1 + c(x)t,0\})^{r(x)} \;\;\; \text{ for } \;\;\; t\in [0,\infty ). \end{aligned}$$

  (16)

Proof

According to Lemma 2.1, we get $f(0)=1$. Moreover, in view of (6), for every $x\in \mathcal {C}$ and $s,t\in [0, \infty )$ such that $sx+g(sx)tx\in \mathcal {C}$, we obtain

$$\begin{aligned} f_x(s+g_x(s)t)=f(sx+g(sx)tx)=f(sx)f(tx)=f_x(s)f_x(t). \end{aligned}$$

Furthermore, for every $x\in \mathcal {C}$ and $s,t\in [0,\infty )$, we have $sx+g(sx)tx=(s+g_x(s)t)x$, that is $sx+g(sx)tx\in \mathcal {C}$ provided $s+g_x(s)t\ge 0$. Therefore, for every $x\in \mathcal {C}$, the pair $(f_x,g_x)$ satisfies Eq. (4). Since, for every $x\in \mathcal {C}$, $f_x$ and $g_x$ are continuous and $f_x(0)=f(0)=1$, applying Theorem 2.1, we get the assertion. $\square$

Remark 2.1

In the sequel, for every $i\in \{1,2,3,4,5,6\}$, we set

$$\begin{aligned} \mathcal {C}_{(i)}:=\{x\in \mathcal {C}|(C_i) \; \text{ holds } \text{ for } \; x\}. \end{aligned}$$

According to Lemma 2.2, if $f\ne 0$ then $0\in \mathcal {C}_{(1)}$, $\bigcup _{i=1}^{6}\mathcal {C}_{(i)}=\mathcal {C}$ and

$$\begin{aligned} \mathcal {C}_{(i)}\cap \mathcal {C}_{(j)}=\emptyset \;\;\, \text{ for } \;\,\; i,j\in \{1,2,3,4,5,6\}, i\ne j. \end{aligned}$$

(17)

Furthermore, for every $i\in \{1,2,3,4,5,6\}$, we have

$$\begin{aligned} \lambda x\in \mathcal {C}_{(i)} \;\,\; \text{ whenever } \;\,\; x\in \mathcal {C}_{(i)} \;\; \text{ and } \;\; \lambda \in (0,\infty ). \end{aligned}$$

(18)

Note also that $g_x(0)=1$ for $x\in \bigcup _{i=2}^{6}\mathcal {C}_{(i)}$, so $g(0)=1$ provided $\bigcup _{i=2}^{6}\mathcal {C}_{(i)}\ne \emptyset$.

Lemma 2.3

Assume that functions $f,g:\mathcal {C}\rightarrow \mathbb {R}$ are continuous on rays, f is nonconstant and the pair (f, g) satisfies Eq. (6). Then

$$\begin{aligned} g_{x}(t) > 0 \quad \text{ for } \quad x \in \mathcal {C}_{(1)}, t \in [0, \infty ). \end{aligned}$$

(19)

Proof

Since f is nonconstant, according to Remark 2.1, we have $\bigcup _{i=2}^{6}\mathcal {C}_{(i)}\ne \emptyset$ and so $g(0)=1$. Suppose that $g_{x}(t)\le 0$ for some $x \in \mathcal {C}_{(1)}$ and $t \in [0, \infty )$. As $g_x$ is continuous and $g_{x}(0)=g(0)=1$, this means that $t>0$ and $g(t_0x)=g_{x}(t_0)=0$ for some $t_0\in (0,t]$. Hence $t_{0}x+g(t_{0}x)y=t_{0}x\in \mathcal {C}$ for $y \in \mathcal {C}$ and so, in view of (6), we obtain

$$\begin{aligned} 1=f_{x}(t_0)=f(t_{0}x)=f(t_{0}x+g(t_{0}x)y)=f(t_{0}x)f(y)=f_{x}(t_{0})f(y)=f(y) \end{aligned}$$

for $y\in \mathcal {C}$, which yields a contradiction.

In the proof of the next proposition we will need the following lemma.

Lemma 2.4

Assume that $n\in \mathbb {N}$, $\gamma _i\in \mathbb {R}$ for $i\in \{0,\ldots ,n\}$, $\beta _i\in \mathbb {R}{\setminus }\{0\}$ for $i\in \{1,\ldots ,n\}$ and $\beta _j\ne \beta _k$ for $j,k\in \{1,\ldots ,n\}$, $j\ne k$. Suppose that

$$\begin{aligned} \sum _{j=1}^{n}\gamma _j\ln (1+\beta _jt)=\gamma _0 t \;\;\; \text{ for } \;\;\; t\in [0,\varepsilon ) \end{aligned}$$

(20)

where $\varepsilon \in (0,\infty )$ is fixed. Then $\gamma _i=0$ for $i\in \{0,\ldots ,n\}$.

Proof

Differentiating (20) with respect to t, for every $k\in \{2,\ldots ,n+1\}$, we obtain

$$\begin{aligned} \sum _{j=1}^{n}\frac{\beta _j^k\gamma _j}{(1+\beta _jt)^k}=0 \;\;\; \text{ for } \;\;\; t\in (0,\varepsilon ). \end{aligned}$$

Thus, letting $t\rightarrow 0^+$ and setting $\delta _j:=\beta _j^2\gamma _j$ for $j\in \{1,\ldots ,n\}$, we get

$$\begin{aligned} \sum _{j=1}^{n}\beta _j^k\delta _j=0 \;\;\; \text{ for } \;\;\; k\in \{0,\ldots ,n-1\}. \end{aligned}$$

The determinant of this system of n linear equations with variables $\delta _1,\ldots ,\delta _n$ is the Vandermonde determinant. Hence, as $\beta _j\ne \beta _k$ for $j,k\in \{1,\ldots ,n\}$, $j\ne k$, it is different from 0. Thus $\delta _i=0$ for $i\in \{1,\ldots ,n\}$. Since $\beta _i\ne 0$ for $i\in \{1,\ldots ,n\}$, this means that $\gamma _i=0$ for $i\in \{1,\ldots ,n\}$ which, in view of (20), gives $\gamma _0=0$.

Proposition 2.1

Assume that functions $f,g:\mathcal {C}\rightarrow \mathbb {R}$ are continuous on rays and the pair (f, g) satisfies Eq. (6). If $\mathcal {C}_{(2)}\ne \emptyset$ then $g=1$ and there exists a nontrivial linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $f(x)=e^{L(x)}$ for $x\in \mathcal {C}$.

Proof

Assume that $\mathcal {C}_{(2)}\ne \emptyset$. First we show that $\mathcal {C}=\mathcal {C}_{(1)}\cup \mathcal {C}_{(2)}$. Suppose that this is not true. Since $\mathcal {C}_{(2)}\ne \emptyset$, we have $f\ne 0$ and so, according to Remark 2.1, we get $\bigcup _{i=3}^{6}\mathcal {C}_{(i)}\ne \emptyset$. Let $y\in \bigcup _{i=3}^{6}\mathcal {C}_{(i)}$. Then there exist $c(y),r(y)\in \mathbb {R}{\setminus }\{0\}$ such that

$$\begin{aligned} f_y(t)=(1+c(y)t)^{r(y)} \;\;\; \text{ for } \text{ sufficiently } \text{ small } \;\;\; t\ge 0. \end{aligned}$$

(21)

Fix an $x\in \mathcal {C}_{(2)}$. Then $tx+g(tx)ty=tx+g_x(t)ty=tx+ty\in \mathcal {C}$ for $t\in [0,\infty )$ and so, taking into account (6) and (21), we get

$$\begin{aligned} f_{x+y}(t)&=f(tx+ty)=f(tx+g(tx)ty)=f(tx)f(ty)\nonumber \\&=f_x(t)f_y(t)=e^{c(x)t}(1+c(y)t)^{r(y)} \;\;\; \text{ for } \text{ sufficiently } \text{ small } \;\;\; t\ge 0.\quad \end{aligned}$$

(22)

If $x+y\in \mathcal {C}_{(1)}\cup \mathcal {C}_{(2)}$ then $f_{x+y}(t)=e^{c(x+y)t}$ for $t\in [0,\infty )$ with some $c(x+y)\in \mathbb {R}$ and so from (22) we derive that

$$\begin{aligned} r(y)\ln (1+c(y)t)=(c(x+y)-c(x))t \;\;\; \text{ for } \text{ sufficiently } \text{ small } \;\;\; t\ge 0. \end{aligned}$$

Thus, as $c(y)\ne 0$, according to Lemma 2.4, we get $r(y)=0$, which yields a contradiction. If $x+y\in \bigcup _{i=3}^{6}\mathcal {C}_{(i)}$ then there exist $c(x+y),r(x+y)\in \mathbb {R}{\setminus }\{0\}$ such that $f_{x+y}(t)=(1+c(x+y)t)^{r(x+y)}$ for sufficiently small $t\ge 0$. Hence, in view of (22), we get

$$\begin{aligned}&r(x+y)\ln (1+c(x+y)t)-r(y)\ln (1+c(y)t)\\&\quad =c(x)t \;\;\; \text{ for } \text{ sufficiently } \text{ small } \;\;\; t\ge 0. \end{aligned}$$

Therefore, as $c(x+y)\ne 0$ and $c(y)\ne 0$, in the case where $c(x+y)\ne c(y)$, applying Lemma 2.4, we conclude that $r(x+y)=r(y)=c(x)=0$, which yields a contradiction. If $c(x+y)=c(y)$ then we have

$$\begin{aligned} (r(x+y)-r(y))\ln (1+c(y)t)=c(x)t \;\;\; \text{ for } \text{ sufficiently } \text{ small } \;\;\; t\ge 0. \end{aligned}$$

Thus, as $c(y)\ne 0$, according to Lemma 2.4, we obtain that $c(x)=0$ which again gives a contradiction.

In this way we have proved that $\mathcal {C}=\mathcal {C}_{(1)}\cup \mathcal {C}_{(2)}$. Therefore, according to Lemma 2.3, we get $g_x(t)>0$ for $x\in \mathcal {C}$, $t\in [0,\infty )$ and so from (6) we derive that

$$\begin{aligned} f(x+g(x)y)=f(x)f(y) \;\;\; \text{ for } \;\;\; x,y\in \mathcal {C}. \end{aligned}$$

(23)

Moreover, for every $x\in \mathcal {C}$, there exists a $c(x)\in \mathbb {R}$ such that

$$\begin{aligned} f_x(t)=e^{c(x)t} \;\;\; \text{ for } \;\;\; t\in [0,\infty ). \end{aligned}$$

(24)

In view of (24), for every $x\in \mathcal {C}$ and $\alpha$, $t\in [0,\infty )$, we get

$$\begin{aligned} e^{c(\alpha x)t}=f_{\alpha x}(t)=f_x(\alpha t)=e^{c(x)\alpha t}, \end{aligned}$$

which implies that

$$\begin{aligned} c(\alpha x)=\alpha c(x) \;\; \text{ for } \;\; x\in \mathcal {C},\alpha \in [0,\infty ). \end{aligned}$$

(25)

We claim that

$$\begin{aligned} c(x+y)=c(x)+c(y) \;\; \text{ for } \;\; x,y\in \mathcal {C}. \end{aligned}$$

(26)

Fix $x,y\in \mathcal {C}$. If $x\in \mathcal {C}_{(1)}$ then $c(x)=0$ whence, in view of (23) and (24), we obtain

$$\begin{aligned} e^{c(x+y)t}&=f_{x+y}(t)=f(tx+ty)=f\left( ty+g(ty)\frac{tx}{g(ty)}\right) \\&=f(ty)f\left( \frac{tx}{g(ty)}\right) =f_y(t)f_x\left( \frac{t}{g(ty)}\right) =f_y(t)=e^{c(y)t} \;\; \text{ for } \;\; t\in [0,\infty ). \end{aligned}$$

Hence $c(x+y)=c(y)=c(x)+c(y)$. If $x\in \mathcal {C}_{(2)}$ then from (23) it follows that $f(x+y)=f(x)f(y)$. Since, in view of (24), $f(x+y)=f_{x+y}(1)=e^{c(x+y)}$, $f(x)=f_{x}(1)=e^{c(x)}$ and $f(y)=f_{y}(1)=e^{c(y)}$, this implies that $c(x+y)=c(x)+c(y)$. Thus (26) is proved.

Now, applying [16, Theorem 4.4.1, p. 88], from (25) and (26) we derive that there exists a linear functional $L:\text{ Lin }\;\mathcal {C}\rightarrow \mathbb {R}$ such that

$$\begin{aligned} c(x) = L(x) \quad \text{ for } \quad x \in \mathcal {C}. \end{aligned}$$

(27)

Hence, according to (24), $f(x)=f_x(1)=e^{L(x)} \;\; \text{ for } \;\; x\in \mathcal {C}$. Furthermore, as $\mathcal {C}_{(2)} \ne \emptyset$, we get that $L_{|\mathcal {C}}$ is not identically zero and so L is nontrivial. Finally, inserting into (23) f of the above form, we obtain $(g(x)-1)L(y)=0$ for $x,y\in \mathcal {C}$. Since $L|_{\mathcal {C}}$ is not identically zero, this means that $g=1$. $\square$

Proposition 2.2

Assume that functions $f,g:\mathcal {C}\rightarrow \mathbb {R}$ are continuous on rays, f is nonconstant and the pair (f, g) satisfies Eq. (6). If $\mathcal {C}_{(2)}=\emptyset$ then $\bigcup _{i=3}^6 \mathcal {C}_{(i)} \ne \emptyset$ and there exist an $r\in \mathbb {R}{\setminus }\{0\}$ and a nontrivial linear functional $L:\text{ Lin }\;\mathcal {C}\rightarrow \mathbb {R}$ such that $r>0$ whenever $\bigcup _{i=4}^6 \mathcal {C}_{(i)} \ne \emptyset$, $\mathcal {C}_{(1)}=\mathcal {C}\cap \ker \;L$ and:

• $(\alpha )$ If $x \in \mathcal {C}_{(3)}$ then $L(x)>0$, $f(x)=(1+L(x))^{r}$ and

  $$\begin{aligned} g(x)=1+L(x); \end{aligned}$$

  (28)

• $(\beta )$ If $x\in \mathcal {C}_{(4)}$ then $L(x)<0$, (28) holds and $f(x)=|1+L(x)|^{r}$;

• $(\gamma )$ If $x \in \mathcal {C}_{(5)}$ then $L(x)<0$, (28) holds and $f(x) =|1+L(x)|^{r} \mathrm{sgn}\,(1+L(x))$;

• $(\delta )$ If $x \in \mathcal {C}_{(6)}$ then $L(x)<0$,

  $$\begin{aligned} f(x)=(\max \{1+L(x),0\})^{r}, \end{aligned}$$

  (29)
  $$\begin{aligned} g(x)=1+L(x) \;\;\; \text{ whenever } \;\;\; 1+L(x)>0, \end{aligned}$$

  (30)

  and

  $$\begin{aligned} g(x)\ge 0\;\;\; \text{ whenever } \;\;\; 1+L(x)\le 0. \end{aligned}$$

  (31)

Proof

Assume that $\mathcal {C}_{(2)}=\emptyset$. Since f is nonconstant, we have $\mathcal {C}_{(1)}\ne \mathcal {C}$ and so, according to Remark 2.1, we get $\bigcup _{i=3}^{6}\mathcal {C}_{(i)}\ne \emptyset$. Let

$$\begin{aligned} V_{x} = \left\{ \begin{array}{ll} [0, \infty ) &{}\quad \text{ for } \;\; x \in \mathcal {C}_{(1)}\cup \mathcal {C}_{(3)},\\ \left[ 0, -\frac{1}{c(x)}\right) &{} \quad \text{ for } \;\; x \in \mathcal {C}_{(4)}\cup \mathcal {C}_{(5)}\cup \mathcal {C}_{(6)}. \end{array} \right. \end{aligned}$$

Then, taking into account (11) and (15), in view of Lemma 2.3, we obtain

$$\begin{aligned} g(tx)=g_{x}(t)>0 \quad \text{ for } \quad x\in \mathcal {C}, t\in V_{x}. \end{aligned}$$

(32)

Thus, for every $x,y\in \mathcal {C}$ and $t\in V_{x}$, we have $\frac{t}{g(tx)}y\in \mathcal {C}$ and $tx+g(tx)\frac{t}{g(tx)}y =tx+ty\in \mathcal {C}$. Therefore, making use of (6), for every $x,y\in \mathcal {C}$ and $t\in V_{x}$, we obtain

$$\begin{aligned} f(tx+ty)=f\left( tx+g(tx)\frac{t}{g(tx)}y\right) =f(tx)f\left( \frac{t}{g(tx)}y\right) . \end{aligned}$$

Hence

$$\begin{aligned} f_{x+y}(t) = f_{x}(t) f_{y}\left( \frac{t}{g_{x}(t)}\right) \quad \text{ for } \quad x,y \in \mathcal {C}, t \in V_{x}. \end{aligned}$$

(33)

In the same way, we get

$$\begin{aligned} f_{y+x}(t) = f_{y}(t) f_{x}\left( \frac{t}{g_{y}(t)}\right) \quad \text{ for } \quad x,y \in \mathcal {C}, t \in V_{y}. \end{aligned}$$

(34)

Moreover, from (12)–(14) and (16) it follows that

$$\begin{aligned} f_{x}(t) = (1 + c(x)t)^{r(x)} \quad \text{ for } \quad x\in \bigcup _{i=3}^{6} \mathcal {C}_{(i)}, t \in V_{x}. \end{aligned}$$

(35)

Note also that, for every $x,y\in \bigcup _{i=3}^6 \mathcal {C}_{(i)}$ and sufficiently small $t\in V_x\cap V_y$, we have $\frac{t}{g_x(t)}=\frac{t}{1+c(x)t}\in V_y$ and $\frac{t}{g_y(t)}=\frac{t}{1+c(y)t}\in V_x$. Therefore, taking into account (33)–(35), we conclude that

$$\begin{aligned} (1+c(x)t)^{r(x)}\left( 1+\frac{c(y)t}{1+c(x)t}\right) ^{r(y)}=(1+ c(y)t)^{r(y)}\left( 1+\frac{c(x)t}{1+c(y)t}\right) ^{r(x)} \end{aligned}$$

for every $x,y\in \bigcup _{i=3}^6 \mathcal {C}_{(i)}$ and sufficiently small $t\in V_x\cap V_y$. Hence

$$\begin{aligned} \left( 1+\frac{c(x)c(y)t^2}{1+(c(x)+c(y))t}\right) ^{r(x)-r(y)}=1 \end{aligned}$$

for every $x,y\in \bigcup _{i=3}^6 \mathcal {C}_{(i)}$ and sufficiently small $t\in V_x\cap V_y$. Since $c(x)c(y)\ne 0$ for $x,y\in \bigcup _{i=3}^6 \mathcal {C}_{(i)}$, this implies that $r(x)=r(y)$ for $x,y\in \bigcup _{i=3}^6 \mathcal {C}_{(i)}$. Thus there exists an $r\in \mathbb {R} {\setminus } \{0\}$ such that

$$\begin{aligned} r(x) = r \quad \text{ for } \quad x \in \bigcup _{i=3}^{6} \mathcal {C}_{(i)}. \end{aligned}$$

Moreover, according to Lemma 2.2, we get $r>0$ whenever $\bigcup _{i=4}^6 \mathcal {C}_{(i)} \ne \emptyset$. Furthermore, taking

$$\begin{aligned} c(x) = 0 \quad \text{ for } \quad x \in \mathcal {C}_{(1)}, \end{aligned}$$

(36)

in view of (35), we obtain

$$\begin{aligned} f_{x}(t) = (1+c(x)t)^{r} \quad \text{ for } \quad x \in \mathcal {C}, t \in V_{x}. \end{aligned}$$

(37)

Thus, for every $x \in \mathcal {C}$ and $\alpha \in (0,\infty )$, we get

$$\begin{aligned} (1+c(\alpha x)t)^{r}=f_{\alpha x}(t)=f_{x}(\alpha t)= (1+c(x)\alpha t)^{r} \;\; \text{ for } \;\; t\in \frac{1}{\alpha } V_{x} \cap V_{\alpha x}, \end{aligned}$$

which implies (25). Next we show that (26) is valid. Fix $x,y \in \mathcal {C}$. If $y \in \mathcal {C}_{(1)}$ then, taking into account (33), we get $f_{x+y}(t)=f_{x}(t)$ for $t\in V_{x}$. Therefore, in view of (37), we obtain $(1+c(x+y)t)^{r}=(1+c(x)t)^{r}$ for $t \in V_{x} \cap V_{x+y}$. Hence $c(x+y)=c(x)$ and so, according to (36), we get $c(x+y)=c(x)+c(y)$. In the case where $x\in \mathcal {C}_{(1)}$, taking into account (34), we obtain the same assertion. Assume that $x,y \in \bigcup _{i=3}^{6} \mathcal {C}_{(i)}$. Since, as we have already noted, $\frac{t}{g_{x}(t)}=\frac{t}{1+c(x)t}\in V_y$ for sufficiently small $t\in V_x$, making use of (33) and (37), we obtain

$$\begin{aligned} (1+c(x+y)t)^{r} = (1+c(x)t)^{r} \left( 1+ \frac{c(y)t}{1+c(x)t}\right) ^{r}=(1+ (c(x) + c(y))t)^{r} \end{aligned}$$

for sufficiently small $t\in V_x\cap V_{x+y}$. Thus $c(x+y)=c(x)+c(y)$ and so (26) is proved. Therefore, applying [16, Theorem 4.4.1, p. 88], from (25) and (26) we deduce that there exists a linear functional $L: \text{ Lin }\; \mathcal {C}\rightarrow \mathbb {R}$ such that (27) holds. Since $\bigcup _{i=3}^{6} \mathcal {C}_{(i)} \ne \emptyset$, L is nontrivial. Furthermore, for every $x\in \mathcal {C}$, we have $f(x)=f_x(1)$ and $g(x)=g_x(1)$, so taking into account (27) and applying Lemma 2.2, we obtain $(\alpha )$–$(\delta )$. From $(\alpha )$–$(\delta )$ it follows that $\mathcal {C}\cap \ker L\subseteq \mathcal {C}_{(1)}$. On the other hand, (27) and (36) imply that $\mathcal {C}_{(1)}\subseteq \mathcal {C}\cap \ker L$. Thus $\mathcal {C}_{(1)}=\mathcal {C}\cap \ker L$. $\square$

3 Main results

The next theorem is the main result of the paper.

Theorem 3.1

Assume that functions $f,g:\mathcal {C}\rightarrow \mathbb {R}$ are continuous on rays. Then the pair (f, g) satisfies Eq. (6) if and only if one of the following possibilities holds:

• (a) $f=0$ and g is arbitrary;

• (b) $f=1$ and g is arbitrary;

• (c) $g=1$ and there exists a nontrivial linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $f(x)=e^{L(x)}$ for $x\in \mathcal {C}$;

• (d) There exist an $r\in \mathbb {R}{\setminus }\{0\}$ and a nontrivial linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $r>0$ whenever $L(\mathcal {C})\cap (-\infty ,0)\ne \emptyset$,

  $$\begin{aligned} g(x)=\max \{L(x)+1,0\} \;\; \text{ for } \;\; x\in \mathcal {C} \end{aligned}$$

  (38)

  and

  $$\begin{aligned} f(x)=(\max \{1+L(x),0\})^{r} \;\;\; \text{ for } \;\;\; x\in \mathcal {C}; \end{aligned}$$

  (39)

• (e) There exist an $r\in (0, \infty )$ and a linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $L(\mathcal {C})\cap (-\infty ,0)\ne \emptyset$,

  $$\begin{aligned} g(x)=1+L(x) \;\; \text{ for } \;\; x\in \mathcal {C}, \end{aligned}$$

  (40)

  and either

  $$\begin{aligned} f(x)=|1+L(x)|^{r} \;\; \text{ for } \;\; x\in \mathcal {C} \end{aligned}$$

  (41)

  or

  $$\begin{aligned} f(x)=|1+L(x)|^{r}\text{ sgn }(1+L(x)) \;\; \text{ for } \;\; x\in \mathcal {C}; \end{aligned}$$

  (42)

• (f) There exist an $r \in (0, \infty )$ and a linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $L(\mathcal {C})=(-\infty ,0]$, f is of the form (39) and, for every $x\in \mathcal {C}$, (30)–(31) hold.

Proof

A straightforward calculation shows that if one of the possibilities (a)–(f) holds then the pair (f, g) satisfies Eq. (6).

Assume that the pair (f, g) satisfies Eq. (6). If f is constant then, in view of Lemma 2.1, $f=0$ or $f=1$ and so, either (a) or (b) holds. Furthermore, if $\mathcal {C}_{(2)}\ne \emptyset$ then, applying Proposition 2.1, we get (c). So, assume that f is nonconstant and $\mathcal {C}_{(2)}=\emptyset$. Then, according to Proposition 2.2, $\bigcup _{i=3}^6 \mathcal {C}_{(i)} \ne \emptyset$ and there exist an $r\in \mathbb {R}{\setminus }\{0\}$ and a nontrivial linear functional $L:\text{ Lin }\;\mathcal {C}\rightarrow \mathbb {R}$ such that $(\alpha )$–$(\delta )$ hold, $r>0$ whenever $\bigcup _{i=4}^6 \mathcal {C}_{(i)} \ne \emptyset$ and $\mathcal {C}_{(1)}=\mathcal {C}\cap \text{ ker }L$. In particular, for every $x\in \mathcal {C}$, we obtain

$$\begin{aligned} f(x)=(1+L(x))^{r} \;\;\; \text{ whenever } \;\;\; 1+L(x)>0. \end{aligned}$$

(43)

Note also that, according to Lemma 2.3, for every $x\in \mathcal {C}_{(1)}$ and $y\in \mathcal {C}$, we get $x+g(x)ty\in \mathcal {C}$ for $t\in [0,\infty )$. Moreover, we have

$$\begin{aligned} 1+L(x+g(x)ty)=1+g(x)tL(y)>0 \;\; \text{ for } \text{ sufficiently } \text{ small } \;\; t\in [0,\infty ). \end{aligned}$$

Thus, in view of (43), we obtain

$$\begin{aligned} f(x+g(x)ty)=(1+g(x)tL(y))^r \;\; \text{ for } \text{ sufficiently } \text{ small } \;\; t\in [0,\infty ). \end{aligned}$$

On the other hand, making use of (6) and (43), for every $x\in \mathcal {C}_{(1)}$ and $y\in \mathcal {C}$, we get

$$\begin{aligned}&f(x+g(x)ty)=f(x)f(ty)=f(ty)\\&\quad =(1+tL(y))^r \;\; \text{ for } \text{ sufficiently } \text{ small } \;\; t\in [0,\infty ). \end{aligned}$$

Hence

$$\begin{aligned} g(x)=1 \;\; \text{ for } \;\; x\in \mathcal {C}_{(1)}. \end{aligned}$$

(44)

Therefore, taking into account $(\alpha )$–$(\delta )$, we conclude that:

• If $\mathcal {C}_{(4)}=\mathcal {C}_{(5)}=\mathcal {C}_{(6)}=\emptyset$ then (d) holds;

• If $\mathcal {C}_{(4)}\ne \emptyset$ and $\mathcal {C}_{(5)}=\mathcal {C}_{(6)}=\emptyset$, or $\mathcal {C}_{(5)}\ne \emptyset$ and $\mathcal {C}_{(4)}=\mathcal {C}_{(6)}=\emptyset$ then (e) holds;

• If $\mathcal {C}_{(3)}=\mathcal {C}_{(4)}=\mathcal {C}_{(5)}=\emptyset$ and $\mathcal {C}_{(6)}\ne \emptyset$ then (f) holds.

Suppose that $\mathcal {C}_{(3)}\ne \emptyset$, $\mathcal {C}_{(6)}\ne \emptyset$ and $\mathcal {C}_{(4)}=\mathcal {C}_{(5)}=\emptyset$. We claim that in this case (d) is valid. In view of (29)–(30), in order to show this, it is enough to prove that $g(x)=0$ for every $x\in \mathcal {C}$ with $1+L(x)\le 0$. Fix an $x\in \mathcal {C}$ such that $1+L(x)\le 0$. Then $x\in \mathcal {C}_{(6)}$ and so, in view of (31), we obtain that $g(x)\ge 0$. Thus, taking a $y\in \mathcal {C}_{(3)}$, we get $x+g(x)sy\in \mathcal {C}$ for $s\in [0,\infty )$. Hence, making use of (6) and (29), we obtain

$$\begin{aligned} f(x+g(x)sy)=f(x)f(sy)=0\;\;\; \text{ for } \;\;\; s\in [0,\infty ). \end{aligned}$$

(45)

On the other hand, according to Proposition 2.2, we have $f(z)\ge 1$ for $z\in \mathcal {C}_{(1)}\cup \mathcal {C}_{(3)}$. Thus from (45) we derive that $x+g(x)sy\not \in \mathcal {C}_{(1)}\cup \mathcal {C}_{(3)}$ for $s\in [0,\infty )$ and so, as $\mathcal {C}_{(4)}=\mathcal {C}_{(5)}=\emptyset$, we get $x+g(x)sy\in \mathcal {C}_{(6)}$ for $s\in [0,\infty )$. Therefore, taking into account (29) and (45), we obtain

$$\begin{aligned} 1+L(x)+g(x)sL(y)=1+L(x+g(x)sy)\le 0 \;\;\; \text{ for } \;\;\; s\in [0,\infty ). \end{aligned}$$

Furthermore, $L(y)>0$ because $y\in \mathcal {C}_{(3)}$. Hence, we have $g(x)=0$.

Now, in order to complete the proof, it is enough to show that at most one of the sets $\mathcal {C}_{(i)}$ for $i\in \{4,5,6\}$ is nonempty. For the proof by contradiction suppose that $\mathcal {C}_{(j_1)}\ne \emptyset$ and $\mathcal {C}_{(j_2)}\ne \emptyset$ for some $j_1,j_2\in \{4,5,6\}$, $j_1\ne j_2$. Fix an $x\in \mathcal {C}_{(j_1)}$, a $y\in \mathcal {C}_{(j_2)}$ and put $\mathcal {C}(x,y):=\{\alpha x+\beta y:\alpha , \beta \in (0,\infty )\}$ and $\mathcal {\bar{C}}(x,y):=\{\alpha x+\beta y:\alpha , \beta \in [0,\infty ), \alpha +\beta >0\}$. Obviously $\emptyset \ne \mathcal {C}(x,y)\subset \mathcal {\bar{C}}(x,y)$. Moreover, as $L(x),L(y)<0$, we have $L(z)<0$ for $z\in \mathcal {\bar{C}}(x,y)$, that is

$$\begin{aligned} \mathcal {\bar{C}}(x,y)\subseteq \mathcal {C}_{(4)}\cup \mathcal {C}_{(5)}\cup \mathcal {C}_{(6)}. \end{aligned}$$

(46)

Thus one of the following two cases is possible:

• 1. $\mathcal {C}(x,y)\cap \mathcal {C}_{(k)}\ne \emptyset$ for some $k\in \{4,5\}$,

• 2. $\mathcal {C}(x,y)\subseteq \mathcal {C}_{(6)}$.

Case 1 Let $z\in \mathcal {C}(x,y)\cap \mathcal {C}_{(k)}$. Then $L(z)<0$ and so $\lambda _n:=-\left( 1+\frac{1}{n}\right) \frac{1}{L(z)}>0$ for $n\in \mathbb {N}$. Put $z_n:=\lambda _n z$ for $n\in \mathbb {N}$. According to (18), $z_n\in \mathcal {C}_{(k)}$ for $n\in \mathbb {N}$, which in view of (28), gives

$$\begin{aligned} g(z_n)=1+L(z_n)=1+\lambda _nL(z)=-\frac{1}{n}\quad \text{ for } \;\; n\in \mathbb {N}. \end{aligned}$$

(47)

Furthermore, for every $n\in \mathbb {N}$, we have $z_n\in \mathcal {C}(x,y)$, so there exist $\alpha _n,\beta _n\in (0,\infty )$ such that $z_n=\alpha _n x+\beta _n y$. Fix an arbitrary $w\in \mathcal {\bar{C}}(x,y)$. Then $L(w)<0$ and $w=\alpha _w x+\beta _w y$ for some $\alpha _w$, $\beta _w\in [0,\infty )$ with $\alpha _w+\beta _w>0$. Thus, taking an $s\in (0,\infty )$ with $1+sL(w)<0$, in view of (47), for sufficiently large $n\in \mathbb {N}$, we obtain

$$\begin{aligned} z_n+g(z_n)sw=z_n-\frac{1}{n}sw=\left( \alpha _n-\frac{1}{n}s\alpha _w\right) x+\left( \beta _n-\frac{1}{n}s\beta _w\right) y\in \mathcal {C}(x,y)\subset \mathcal {C}. \end{aligned}$$

Hence, applying (6), we get

$$\begin{aligned} f(z_n+g(z_n)sw)=f(z_n)f(sw) \;\; \text{ for } \text{ sufficiently } \text{ large } \;\; n\in \mathbb {N}. \end{aligned}$$

(48)

Moreover, according to (47), we obtain

$$\begin{aligned} 1+L(z_n+g(z_n)sw)=1+L(z_n)-\frac{1}{n}sL(w)=-\frac{1}{n}(1+sL(w))>0 \quad \text{ for } \;\; n\in \mathbb {N}. \end{aligned}$$

Thus, making use of (43), we get

$$\begin{aligned} f(z_n+g(z_n)sw)=\left( -\frac{1}{n}(1+sL(w))\right) ^r>0\quad \text{ for } \;\; n\in \mathbb {N}. \end{aligned}$$

(49)

On the other hand, in view of (46), $w\in \mathcal {C}_{(l)}$ for some $l\in \{4,5,6\}$. Hence, as $1+sL(w)<0$ and $1+L(z_n)=-\frac{1}{n}$ for $n\in \mathbb {N}$, applying Proposition 2.2, for every $n\in \mathbb {N}$, we obtain

$$\begin{aligned} f(z_n)f(sw)= \left\{ \begin{array}{l@{\quad }l} -|\frac{1}{n}(1+sL(w))|^r &{} \text{ whenever } \;\; l\in \{4,5\},\; l\ne k,\\ 0 &{} \text{ whenever } \;\; l=6. \end{array} \right. \end{aligned}$$

Thus, taking into account (48) and (49), we get that $k=l$. Hence $w\in \mathcal {C}_{(k)}$. Since $w\in \mathcal {\bar{C}}(x,y)$ is fixed arbitrarily, this implies that $\mathcal {\bar{C}}(x,y)\subseteq \mathcal {C}_{(k)}$. In particular $x,y\in \mathcal {C}_{(k)}$ and so $\mathcal {C}_{(j_1)}\cap \mathcal {C}_{(k)}\ne \emptyset$ and $\mathcal {C}_{(j_2)}\cap \mathcal {C}_{(k)}\ne \emptyset$. Thus, in view of (17), we get $j_1=k=j_2$, which yields a contradiction.

Case 2 Fix a $w\in \mathcal {C}(x,y)$. Then $L(w)<0$ and so, taking $w_0:=-\frac{w}{2L(w)}$, we get $w_{0}\in \mathcal {C}(x,y)$ and $1+L(w_0)=\frac{1}{2}>0$. Hence, making use of (29) and (30), we obtain $f(w_0)=\frac{1}{2^r}$ and $g(w_0)=\frac{1}{2}$, respectively. Let $s\in (0,\infty )$ be such that $1+sL(w)<0$. Suppose that $j_1\in \{4,5\}$. Then, as $x\in \mathcal {C}_{(j_1)}$, according to Proposition 2.2, we have $f_x(s)\ne 0$. Moreover,

$$\begin{aligned} w_0+g(w_0)sx=w_0+\frac{s}{2}x\in \mathcal {C}(x,w_0)\subset \mathcal {C}(x,y)\subseteq \mathcal {C}_{(6)}\subset \mathcal {C}. \end{aligned}$$

Thus, in view of (6), we obtain

$$\begin{aligned} f(w_0+g(w_0)sx)=f(w_0)f(sx)=\frac{1}{2^r}f_{x}(s)\ne 0. \end{aligned}$$

On the other hand, we get

$$\begin{aligned} 1+L(w_0+g(w_0)sx)=1+L(w_0)+g(w_0)sL(x)=\frac{1}{2}(1+sL(x))<0, \end{aligned}$$

so making use of (29), we conclude that $f(w_0+g(w_0)sx)=0$. This is a contradiction. Consequently $j_1=6$. In the same way we obtain that $j_2=6$, which yields a contradiction. $\square$

Remark 3.1

From Theorems 2.1 and 3.1 it follows that if the functions $f,g:\mathcal {C}\rightarrow \mathbb {R}$ are continuous on rays, then the pair (f, g) satisfies Eq. (6) if and only if there exist continuous functions $\tilde{f}, \tilde{g}:\mathbb {R}\rightarrow \mathbb {R}$ and a nontrivial linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that the pair $(\tilde{f},\tilde{g})$ satisfies Eq. (4), $f=\tilde{f}\circ L_{|\mathcal {C}}$ and either $g=\tilde{g}\circ L_{|\mathcal {C}}$, or $L(\mathcal {C})=(-\infty ,0]$ and, for every $x\in \mathcal {C}$, $g(x)=(\tilde{g}\circ L)(x)$ whenever $1+L(x)>0$ and $g(x)\ge 0$, otherwise.

From Theorem 3.1 we derive the following result, which generalizes the main result in [19].

Corollary 3.1

Assume that a function $f:\mathcal {C}\rightarrow \mathbb {R}$ is continuous on rays. Then f satisfies equation

$$\begin{aligned} f(x+f(x)y)=f(x)f(y) \;\;\; \text{ whenever } \;\;\; x,y, x+f(x)y\in \mathcal {C} \end{aligned}$$

(50)

if and only if $f=0$ or $f=1$ or there exists a nontrivial linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that either $f(x)=1+L(x)$ for $x\in \mathcal {C}$, or $f(x)=\max \{1+L(x),0\}$ for $x\in \mathcal {C}$.

Corollary 3.2

If a function $f:\mathcal {C} \rightarrow \mathbb {R}$ is continuous on rays and satisfies Eq. (50), then there exists a function $F:X\rightarrow \mathbb {R}$ continuous on rays and satisfying equation

$$\begin{aligned} F(x+F(x)y)=F(x)F(y) \;\;\; \text{ for } \;\;\; x,y\in X \end{aligned}$$

such that $f(x)=F(x)$ for $x \in \mathcal {C}$.

Remark 3.2

The counterpart of Corollary 3.2 does not hold for Eq. (6) even in the case $X=\mathbb {R}$ and $\mathcal {C}=[0,\infty )$ (cf. [8]).

Now, we determine the solutions of Eq. (7).

Theorem 3.2

Assume that functions $F,G:X\rightarrow \mathbb {R}$ are continuous on rays. Let $G^{-}:=\{x\in \mathcal {C}|G(x)<0\}$ and $G^{-}-\mathcal {C}:=\{x-y|(x,y)\in G^{-}\times \mathcal {C}\}$. Then the pair (F, G) satisfies Eq. (7) if and only if one of the following possibilities holds:

• (i) G is arbitrary and $F(x)=0$ for $x\in \mathcal {C}\cup (G^{-}-\mathcal {C})$;

• (ii) G is arbitrary and $F(x)=1$ for $x\in \mathcal {C}\cup (G^{-}-\mathcal {C})$;

• (iii) $G(x)=1$ for $x\in \mathcal {C}$ and there exists a nontrivial linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $F(x)=e^{L(x)}$ for $x\in \mathcal {C}$;

• (iv) There exist an $r\in \mathbb {R}{\setminus }\{0\}$ and a linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $r>0$ whenever $L(\mathcal {C})\cap (-\infty ,0)\ne \emptyset$,

  $$\begin{aligned} G(x)=\max \{1+L(x),0\} \quad \text{ for } \;\; x\in \mathcal {C} \end{aligned}$$

  (51)

  and

  $$\begin{aligned} F(x)=(\max \{1+L(x),0\})^r \quad \text{ for } \;\; x\in \mathcal {C}; \end{aligned}$$

  (52)

• (v) There exist an $r\in (0,\infty )$ and a linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $L(\mathcal {C})\cap (-\infty ,0)\ne \emptyset$,

  $$\begin{aligned} G(x)=1+L(x) \;\; \text{ for } \;\; x\in \mathcal {C} \end{aligned}$$

  (53)

  and either

  $$\begin{aligned} F(x)=|1+L(x)|^{r} \;\; \text{ for } \;\; x\in \text{ Lin } \; \mathcal {C} \end{aligned}$$

  (54)

  or

  $$\begin{aligned} F(x)=|1+L(x)|^{r}sgn(1+L(x)) \;\; \text{ for } \;\; x\in \text{ Lin } \; \mathcal {C}; \end{aligned}$$

  (55)

• (vi) There exist an $r\in (0,\infty )$ and a linear functional $L:\text{ Lin }\;\mathcal {C}\rightarrow \mathbb {R}$ such that $L(\mathcal {C})=(-\infty ,0]$, F is of the form (52) and, for every $x\in \mathcal {C}$,

  $$\begin{aligned} G(x)=1 + L(x) \;\;\; \text{ whenever } \;\;\; 1+L(x)>0 \end{aligned}$$

  and $G(x)\ge 0$, otherwise.

Proof

Assume that the pair (F, G) satisfies (7). Then the functions $F_{|\mathcal {C}}$ and $G_{|\mathcal {C}}$ are continuous on rays and the pair $(F_{|\mathcal {C}}, G_{|\mathcal {C}})$ satisfies Eq. (6). Hence, according to Theorem 3.1, one of the possibilities (a)–(f) holds. Note that (c) implies (iii), (d) yields (iv) and (f) leads to (vi).

If (a) holds then $F(x)=0$ for $x\in \mathcal {C}$. Moreover, if $G^{-}\ne \emptyset$ then taking a $z\in G^{-}-\mathcal {C}$, we have $z=u-y$ for some $u\in G^{-}$ and $y\in \mathcal {C}$. Thus, as $-\frac{1}{G(u)}y\in \mathcal {C}$, making use of (7), we obtain

$$\begin{aligned} F(z)=F(u-y)=F\left( u+ G(u)\left( \frac{-1}{G(u)}y\right) \right) = F(u)F\left( \frac{-1}{G(u)}y\right) =0. \end{aligned}$$

Hence $F(x)=0$ for $x \in G^{-}-\mathcal {C}$ and so (i) is valid. The similar arguments show that (b) leads to (ii).

Finally consider the case where (e) holds. Then (40) implies (53). Suppose that (41) holds. Then

$$\begin{aligned} F(x)=|1+L(x)|^r \;\;\; \text{ for } \;\;\; x\in \mathcal {C}. \end{aligned}$$

(56)

Furthermore, taking a $z\in \text{ Lin }\;\mathcal {C}$, we have $z=x-y$ for some $x,y\in \mathcal {C}$. If $1+L(x)<0$ then $-\frac{y}{1+L(x)}\in \mathcal {C}$ and so, applying (7), (53) and (56), we obtain

$$\begin{aligned} F(z)&=F(x-y)=F\left( x+(1+L(x))\left( \frac{-y}{1+L(x)}\right) \right) \\&=F\left( x+G(x)\left( \frac{-y}{1+L(x)}\right) \right) =F(x)F\left( \frac{-y}{1+L(x)}\right) \\&=|1+L(x)|^r\left| 1-\frac{L(y)}{1+L(x)}\right| ^r =|1+L(x)-L(y)|^r\\&=|1+L(x-y)|^r=|1+L(z)|^r. \end{aligned}$$

If $1+L(x)=0$ then taking $x^{\prime }=2x$ and $y^{\prime }=x+y$, we have $x^{\prime },y^{\prime }\in \mathcal {C}$, $z=x-y=x^{\prime }-y^{\prime }$ and $1+L(x^{\prime })=1+2L(x)=-1<0$. Therefore, arguing as in the previous case, we get that $F(z)=|1+L(z)|^r$. If $1+L(x)>0$ then taking an $x_0\in \mathcal {C}$ with $L(x_0)<0$, $\lambda :=-\frac{2}{L(x_0)}(1+L(x))$, $\tilde{x}:=x+\lambda x_0$ and $\tilde{y}:=y+\lambda x_0$, we get $\lambda >0$, $\tilde{x}, \tilde{y}\in \mathcal {C}$, $z=x-y=\tilde{x}-\tilde{y}$ and

$$\begin{aligned} 1+L(\tilde{x})= & {} 1+L(x+\lambda x_0)=1+L(x)-\frac{2}{L(x_0)}(1+L(x))L(x_0)\\= & {} -(1+L(x))<0. \end{aligned}$$

Therefore, making use of (7), (53) and (56), as previously we conclude that $F(z)=|1+L(z)|^r$. In this way we have proved that (41) implies (54). The similar arguments show that (42) leads to (55). Thus, (v) is valid.

The converse is easy to check. $\square$

We conclude the paper with the result describing the solutions of the equation

$$\begin{aligned} F(x+F(x)y)=F(x)F(y) \;\;\; \text{ for } \;\;\; x,y\in \mathcal {C} \end{aligned}$$

(57)

in the class of functions $F:X\rightarrow \mathbb {R}$ continuous on rays. It generalizes the main result in [1].

Corollary 3.3

Assume that a function $F:X\rightarrow \mathbb {R}$ is continuous on rays. Then F satisfies Eq. (57) if and only if one of the following possibilities holds:

• (i) $F(x)=0$ for $x\in \mathcal {C}$;

• (ii) $F(x)=1$ for $x\in \mathcal {C}$;

• (iii) There exists a nontrivial linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such $F(x)=\max \{1+L(x),0\}$ for $x\in \mathcal {C}$;

• (iv) There exist a linear functional $L:\text{ Lin } \; \mathcal {C}\rightarrow \mathbb {R}$ such that $L(\mathcal {C})\cap (-\infty ,0)\ne \emptyset$ and $F(x)=1+L(x)$ for $x\in Lin \; \mathcal {C}$.