Translativity of beta-type functions

Translativity of the beta-type function Bf:I2→0,∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$B_{f}:I^{2}\rightarrow \left( 0,\infty \right) $$\end{document}, Bfx,y:=fxfyfx+y,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} B_{f}\left( x,y\right) :=\frac{f\left( x\right) f\left( y\right) }{f\left( x+y\right) }, \end{aligned}$$\end{document}where f is a single variable function defined either on I=R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$I={\mathbb {R}}$$\end{document} or I=[0,∞)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$I=[ 0,\infty )$$\end{document}, or I=0,∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$I=\left( 0,\infty \right) $$\end{document}, is considered. In each of these three cases a complete solution is given.


Introduction
For a positive function f on an interval I that is closed under addition, the bivariable function B f : I 2 → (0, ∞) defined by is called a beta-type function and f is called its generator. Some basic properties of beta-type functions, like homogeneity or their relation with means, were studied in [3,4], respectively. In this note we examine the conditions under which the beta-type function is (positively) translative, i.e. that for a function α : (0, ∞) → R and all x, y ∈ I and t ∈ (0, ∞) (Definition 2). Theorem 1 in section 3 gives a full description of such functions in the simplest case when I = R. In a little more difficult case I = [0, ∞), considered in Sect. 4, Theorem 2 gives the solution. The case I = (0, ∞), considered in Sect. 5, turned out to be the most challenging; in particular, the methods used in the 2. Preliminaries Definition 1. Let I ⊂ R be an interval that is closed under addition, and f : I → (0, ∞) be an arbitrary function. Then B f : is called a beta-type function in I, and f is referred to as its generator [2].
For an interval I ⊂ R and t ∈ R put t + I := {t + x : x ∈ I} and −I : (2.1) Definition 3. Let I ⊂ R be an interval such that t + I ⊂ I for every t < 0. A function F : I 2 → R is said to be negatively translative, if there is a function α : (−∞, 0) → R such that  Due to this remark, in the sequel we consider only positively translative function. In addition, we use the term α-translative function for a positively translative function with the function α. (ii) if F is nonnegative and α-translative, then α (t) = at (t > 0) for some real a ≥ 0 (called the order of translativity of F ).
Proof. (i) For all x, y ∈ I and s, t > 0, using (2.1) we have and (ii) From (i), by induction, we have, for all n ∈ N, x, y ∈ I, and t > 0, If α (t) < 0 for some t > 0, then fixing arbitrarily x, y ∈ I and then choosing n ∈ N such that F (x, y) + nα (t) < 0, we would get a contradiction with the nonnegativity of F (x + nt, y + nt). Thus α is nonnegative, and being additive, it must be of the form α (t) = at (t > 0) for some a ≥ 0 ([1, p. 33]; see also [5, p. 145]).

Translativity of beta-type functions in R 2
In this section we prove the following Theorem 1. Let f : R → (0, ∞) and α : (0, ∞) → R. The following conditions are pairwise equivalent: Proof. Assume (i). Remark 2 then implies that α(t) = at for some nonnegative a, so the translativity Eq. (2.1), by Definition 1 of B f , takes the following form: Hence, in particular, for all x, y ∈ (0, ∞), we get Changing the roles of x and y in the resulting equality, we hence also get, for all x, y ∈ (0, ∞), These two equalities imply that a (y − x) = 0 for all x, y ∈ (0, ∞) and, in consequence, that a = 0, whence α(t) = 0 for all t > 0. Now, take arbitrary x, y ∈ R. There exists a sufficiently large t > 0 for which x + t and y + t are positive. Therefore, from (3.1) and (3.2) we obtain which proves (ii).
Since the implications (ii) =⇒ (iii) and (iii) =⇒ (i) are easy to verify, the proof is complete.
Remark 3. The exponential function in (ii) need not be continuous.

Translativity of beta-type functions in
Proof. From (4.1), the definition of B f and Remark 2, we easily conclude that α (t) = at for some real a and all t > 0, whence Setting here y = 0 gives Replacing here s by s + t with t ≤ s, we get Vol. 97 (2023) Translativity of beta-type functions 125 Taking t such that t ≤ s ≤ 2t, in view of (4.2) we also have Since, taking s = t in (4.2) gives we hence get
The main result of this section reads as follows: The following conditions are pairwise equivalent: (iii) the beta-type function B f is constant.
Proof. Assume (i). By Lemma 1 we have α ≡ 0 and Take arbitrary x, y ∈ [0, ∞). Assuming that y ≤ x, which can be done without any loss of generality, and applying the above equality with t = y, we get which proves (ii). As the remaining two implications (ii) =⇒ (iii) and (iii) =⇒ (i) are simple to check, the proof is complete. AEM 2 We begin with an analogue of Lemma 1.

Lemma 2. Assume that
Proof. In view of Remark 2, α (t) = at for some nonnegative a and for all t > 0. Assume, on the contrary, that a > 0 and that there exists a function f : (0, ∞) → (0, ∞) satisfying (5.1), i.e. that Writing this equality in the form we see that, without any loss of generality, we can assume that a = 1, i.e. that Replacing here x, y, t, respectively by 2x, 2y, 2t, we obtain which shows that the function g (x) : Changing the roles of x and t in (5.4) gives us x, x, t > 0, (5.5) whence, by (5.4) and (5.5), we obtain that g(x) − x is constant. It follows that, for some b ≥ 0, Vol. 97 (2023) Translativity of beta-type functions 127 Hence, using (5.3), we have which is equivalent to On the other hand, squaring both sides of (5.2), gives us From (5.6) and (5.7) we obtain, for all x, y, t > 0, To finish the proof, we consider two cases separately: b > 0 and b = 0. In the first case equality (5.8), which must hold for all x, y, t > 0, in particular must also be valid for x = b, y = 3b, t = b. Putting these values to (5.8) we get 19 56 However, because the quadratic polynomial 19 56 p 2 − 2bp + 23 7 b 2 of the variable p does not have real roots, we get a desired contradiction. Now, if b = 0, equality (5.8) simplifies to (x + t) (y + t) (x + y) 2t (x + t) (y + t) x + y + 2t − t 2 = 0, but once again, there does not exist a real-valued function f defined on (0, ∞), satisfying this equality for all x, y > 0 (to see that, set, for instance, x = t = 1, y = 2 and repeat the reasoning from the former case).
Now we are ready to give a complete solution to the translativity problem of B f in (0, ∞) 2 .