Abstract
We determine the solutions of conditional composite type functional equations related to some problems in mathematical meteorology and fluid mechanics.
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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
where r is a positive real constant. It turns out that Eq. (1) by some simple substitutions can be reduced to the equation
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
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
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
and
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
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
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
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
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
Furthermore, for every \(i\in \{1,2,3,4,5,6\}\), we have
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
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
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
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
Thus, letting \(t\rightarrow 0^+\) and setting \(\delta _j:=\beta _j^2\gamma _j\) for \(j\in \{1,\ldots ,n\}\), we get
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
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
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
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
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
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
Moreover, for every \(x\in \mathcal {C}\), there exists a \(c(x)\in \mathbb {R}\) such that
In view of (24), for every \(x\in \mathcal {C}\) and \(\alpha \), \(t\in [0,\infty )\), we get
which implies that
We claim that
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
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
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
Then, taking into account (11) and (15), in view of Lemma 2.3, we obtain
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
Hence
In the same way, we get
Moreover, from (12)–(14) and (16) it follows that
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
for every \(x,y\in \bigcup _{i=3}^6 \mathcal {C}_{(i)}\) and sufficiently small \(t\in V_x\cap V_y\). Hence
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
Moreover, according to Lemma 2.2, we get \(r>0\) whenever \(\bigcup _{i=4}^6 \mathcal {C}_{(i)} \ne \emptyset \). Furthermore, taking
in view of (35), we obtain
Thus, for every \(x \in \mathcal {C}\) and \(\alpha \in (0,\infty )\), we get
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
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
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
Thus, in view of (43), we obtain
On the other hand, making use of (6) and (43), for every \(x\in \mathcal {C}_{(1)}\) and \(y\in \mathcal {C}\), we get
Hence
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
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
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
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
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
Hence, applying (6), we get
Moreover, according to (47), we obtain
Thus, making use of (43), we get
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
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,
Thus, in view of (6), we obtain
On the other hand, we get
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
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
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
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
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
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
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
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 \(x\in Lin \; \mathcal {C}\).
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Acknowledgments
The authors wish to express their thanks to the referee for valuable comments that helped to improve the layout of the results. The research was supported, in part, by ESF project CZ.1.07/2.3.00/20.0002. The work of the first author was partially supported by the Centre for Innovation and Transfer of Natural Sciences and Engineering Knowledge.
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Chudziak, J., Kočan, Z. Functional equations of the Goła̧b-Schinzel type on a cone. Monatsh Math 178, 521–537 (2015). https://doi.org/10.1007/s00605-015-0784-x
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DOI: https://doi.org/10.1007/s00605-015-0784-x