On a functional equation related to competition

AbstractThe functional equation
$$f \left(\frac{x + y}{1 - xy}\right) = \frac{f\left(x\right) + f\left(y\right)} {1 + f\left(x\right) f\left(y\right)}, \quad xy < 1,$$fx+y1-xy=fx+fy1+fxfy,xy<1,(introduced by the first author in a competition model) is considered. The main result says that a function $${f : \mathbb{R} \rightarrow \mathbb{R}}$$f:R→R satisfies this equation if, and only if, $${f = {\rm tanh} \circ \, \alpha \circ {\rm tan}^{-1}}$$f=tanh∘α∘tan-1 , where $${\alpha : \mathbb{R} \rightarrow \mathbb{R}}$$α:R→R is an additive function.


Introduction
Motivated by a model of competition coming from cloud physics, the firstnamed author [1,2] introduced the following functional equation , (x, y) ∈ R 2 , xy = 1.
Applying a uniqueness result [3] for a related equation in a single variable, the form of solutions under some special regularity conditions was established (cf. Remark 5).
In Sect. 2 we present properties of solutions of this equation which in a natural way lead to the consideration of Eq. (1) with the domain restricted to the set (x, y) ∈ R 2 : xy < 1 . We prove, among other things, that if f : R → R satisfies this equation and f (y 0 ) = 1 or f (y 0 ) = −1 for some y 0 ∈ R, then f is a constant function (Proposition 1). Moreover f (0) is either 0 or −1 or 1. If f (0) = 0 then f is an odd function. In Sect. 3 we prove that the function f = tanh • α • tan −1 , where α : R → R is an arbitrary additive function, is the general solution. As a corollary we obtain that, under some weak regularity 302 P. Kahlig and J. Matkowski AEM conditions, every solution must be of the form f (x) = tanh c tan −1 (x) (x ∈ R) for some c ∈ R.

Properties of solutions of the functional equation
Since the function (x, y) → x+y 1−xy occurring in Eq. (1) is not defined on the set (x, y) ∈ R 2 : xy = 1 , instead of Eq. (1), it is natural to consider the following two functional equations on a restricted domain: and Remark 1. Since {x ∈ R : xy < 1 for some y ∈ R} = R, it is reasonable to ask for solutions of the type f : R → R of Eq. (2), that are defined on the whole R.
Note that this problem makes no sense in the case of Eq. (3), as no point (x, y) with x = 0 satisfies the condition xy > 1. Moreover, the domain of Eq. Note also that if f : R → R satisfies (1) then, clearly, it satisfies (2), and its restriction to R\ {0} satisfies (3).
We begin with the following: Proof. The result (i) is obvious.

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If y 0 > 0 we get Since the range of the function in particular f (x) = 1 for all x ∈ (0, ∞). Taking y 0 arbitrarily close to 0 from the right, we obtain which was to be shown. If y 0 < 0, the argument is similar. We omit an analogous proof of (iii). Proof. Assume that f satisfies Eq. (2) or Eq. (3), is continuous at the point x 0 , and there exists a sequence (y n ) such that lim n→∞ y n = 0 and lim n→∞ |f which implies that f (x 0 ) = 1 or f (x 0 ) = −1. In view of Proposition 1 the function f would be constant, contradicting the assumption.
In the sequel we shall deal with the functional equation (2). Indeed, setting x = y = 0 in (2) we get Hence, making use of Proposition 1, we obtain: Therefore in the sequel we are mainly interested in the solutions f : R → R of Eq. (2) such that f (0) = 0.
Indeed, for y = −x we have xy = −x 2 < 1 and, setting y = −x in Eq. (2), we obtain Indeed, in view of Corollary 1, we have f (0) = 0. The remaining part of this remark one gets immediately by setting y = y 0 in (2).
From Proposition 1 and Corollary 1 we obtain the following: . Then the following conditions are pairwise equivalent: Remark 5. Setting y = x in Eq. (2) we obtain the following functional equation in a single variable which is used in [1,2]. Assume that f : R → R satisfies Eq. (1). Replacing y in (1) by y+z 1−yz we obtain whence, setting z = y = x, we obtain the following functional equation in a single variable Vol. 87 (2014)

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By induction, this procedure and Eq. (1) lead to the following infinite system of functional equations of n variables x 1 , . . . , x n , where n ∈ N, n ≥ 2, and x 1 , . . . , x n ∈ R are such that 2j k=1 x i k = 1 (here n+1 2 denotes the largest integer not greater than n+1 2 ). Setting here x 1 , . . . , x n = x we obtain for f the system of iterative functional equations , for all n ∈ N, n ≥ 2, and x ∈ R such that [ n+1 2 ] j=1 (−1) j n 2j x 2j = 1. Proof. We have the identity x + y 1 − xy = tan tan −1 x + tan −1 y , xy < 1.