1 Introduction

For a positive function f on an interval I that is closed under addition, the bivariable function \(B_{f}:I^{2}\rightarrow \left( 0,\infty \right) \) defined by

$$\begin{aligned} B_{f}\left( x,y\right) :=\frac{f\left( x\right) f\left( y\right) }{f\left( x+y\right) }, \end{aligned}$$

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

$$\begin{aligned} B_{f}\left( x+t,y+t\right) =B_{f}\left( x,y\right) +\alpha \left( t\right) \end{aligned}$$

for a function \(\alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\) and all \(x,y\in I\) and \(t\in \left( 0,\infty \right) \) (Definition 2). Theorem 1 in section 3 gives a full description of such functions in the simplest case when \(I={\mathbb {R}}\). In a little more difficult case \(I=[ 0,\infty )\), considered in Sect. 4, Theorem 2 gives the solution. The case \(I=\left( 0,\infty \right) \), considered in Sect. 5, turned out to be the most challenging; in particular, the methods used in the previous cases do not work. A complete solution (Theorem 3) required some more sophisticated arguments.

Trying to find a simpler and more universal approach, we prove, in particular, that every translative function acting on \(\left( 0,\infty \right) ^{2}\) has a unique translative extension acting on \({\mathbb {R}}^{2}\) (Theorem 4). We observe that the extension of the translative beta-type function is also of this type if and only if \(\alpha \equiv 0\).

2 Preliminaries

Definition 1

Let \(I\subset {\mathbb {R}}\) be an interval that is closed under addition, and \( f:I\rightarrow \left( 0,\infty \right) \) be an arbitrary function. Then \( B_{f}:I^{2}\rightarrow \left( 0,\infty \right) \) given by

$$\begin{aligned} B_{f}\left( x,y\right) :=\frac{f\left( x\right) f\left( y\right) }{f\left( x+y\right) },\quad x,y\in I, \end{aligned}$$

is called a beta-type function in I, and f is referred to as its generator [2].

For an interval \(I\subset {\mathbb {R}}\) and \(t\in {\mathbb {R}}\) put \( t+I:=\left\{ t+x:x\in I\right\} \) and \(-I:=\left\{ -x:x\in I\right\} \).

Definition 2

Let \(I\subset {\mathbb {R}}\) be an interval such that \(t+I\subset I\) for every \(t>0\). A function \(F:I^{2}\rightarrow {\mathbb {R}}\) is said to be positively translative, if there is a function \(\alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\) such that

$$\begin{aligned} F\left( x+t,y+t\right) =F\left( x,y\right) +\alpha \left( t\right) , \quad x,y\in I, \quad t>0. \end{aligned}$$
(2.1)

Definition 3

Let \(I\subset {\mathbb {R}}\) be an interval such that \(t+I\subset I\) for every \(t<0\). A function \(F:I^{2}\rightarrow {\mathbb {R}}\) is said to be negatively translative, if there is a function \(\alpha :\left( -\infty ,0\right) \rightarrow {\mathbb {R}}\) such that

$$\begin{aligned} F\left( x+t,y+t\right) =F\left( x,y\right) +\alpha \left( t\right) ,\quad x,y\in I, \quad t<0. \end{aligned}$$

Remark 1

A function \(F:I^{2}\rightarrow {\mathbb {R}}\) is negatively translative if, and only if, the function \({\bar{F}}:\left( -I\right) ^{2}\rightarrow {\mathbb {R}}\) defined by \({\bar{F}}\left( x,y\right) :=F\left( -x,-y\right) \) is positively translative.

Indeed, if F is negatively translative with a function \(\alpha :\left( -\infty ,0\right) \rightarrow {\mathbb {R}}\) then, setting \({\bar{\alpha }}:\left( 0,\infty \right) \rightarrow {\mathbb {R}}\) by \({\bar{\alpha }}\left( t\right) :=\alpha \left( -t\right) \), we have for all \(x,y\in \left( -I\right) \) and \( t>0\),

$$\begin{aligned} {\bar{F}}\left( x+t,y+t\right)= & {} F\left( -x+\left( -t\right) ,-y+\left( -t\right) \right) =F\left( -x,-y\right) +\alpha \left( -t\right) \\= & {} {\bar{F}} \left( x,y\right) +{\bar{\alpha }}\left( t\right) , \end{aligned}$$

so \({\bar{F}}\) is positively translative (with the function \({\bar{\alpha }}\)). Similarly one can show the converse implication.

Due to this remark, in the sequel we consider only positively translative function. In addition, we use the term \(\alpha \)-translative function for a positively translative function with the function \(\alpha \).

Remark 2

Under the conditions of Definition 2:

  1. (i)

    if F is \(\alpha \)-translative, then \(\alpha \) is an additive function;

  2. (ii)

    if F is nonnegative and \(\alpha \)-translative, then \(\alpha \left( t\right) =at\) (\(t>0\)) for some real \(a\ge 0\) (called the order of translativity of F).

Proof

(i) For all \(x,y\in I\) and \(s,t>0,\) using (2.1) we have

$$\begin{aligned} F\left( x+s+t,y+s+t\right)= & {} F\left( \left( x+s\right) +t,\left( y+s\right) +t\right) \\ =F\left( x+s,y+s\right) +\alpha \left( t\right)= & {} F\left( x,y\right) +\alpha \left( s\right) +\alpha \left( t\right) , \end{aligned}$$

and

$$\begin{aligned} F\left( x+s+t,y+s+t\right) =F\left( x,y\right) +\alpha \left( s+t\right) , \end{aligned}$$

whence \(\alpha \left( s+t\right) =\alpha \left( s\right) +\alpha \left( t\right) \), so \(\alpha \) is additive in \(\left( 0,\infty \right) \).

(ii) From (i), by induction, we have, for all \(n\in {\mathbb {N}}\), \(x,y\in I\), and \(t>0,\)

$$\begin{aligned} F\left( x+nt,y+nt\right) =F\left( x,y\right) +n\alpha \left( t\right) . \end{aligned}$$

If \(\alpha \left( t\right) <0\) for some \(t>0\), then fixing arbitrarily \( x,y\in I\) and then choosing \(n\in {\mathbb {N}}\) such that \(F\left( x,y\right) +n\alpha \left( t\right) <0\), we would get a contradiction with the nonnegativity of \(F\left( x+nt,y+nt\right) \). Thus \(\alpha \) is nonnegative, and being additive, it must be of the form \(\alpha \left( t\right) =at\,\ \)(\( t>0\)) for some \(a\ge 0\) ([1, p. 33]; see also [5, p. 145]). \(\square \)

3 Translativity of beta-type functions in \({\mathbb {R}}^{2}\)

In this section we prove the following

Theorem 1

Let \(f:{\mathbb {R}}\rightarrow \left( 0,\infty \right) \) and \(\ \alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\). The following conditions are pairwise equivalent:

  1. (i)

    the beta-type function \(B_{f}:{\mathbb {R}}^{2}\rightarrow \left( 0,\infty \right) \) is \(\alpha \)-translative;

  2. (ii)

    \(\alpha \equiv 0\) and \(\frac{f}{f\left( 0\right) }\) is the exponential function, that is

    $$\begin{aligned} f\left( 0\right) f\left( x+y\right) =f\left( x\right) f\left( y\right) , \quad x,y\in {\mathbb {R}}; \end{aligned}$$
  3. (iii)

    the beta-type function \(B_{f}\) is constant.

Proof

Assume (i). Remark 2 then implies that \(\alpha (t)=at\) for some nonnegative a, so the translativity Eq. (2.1), by Definition 1 of \(B_{f}\), takes the following form:

$$\begin{aligned} \frac{f\left( x+t\right) f\left( y+t\right) }{f\left( x+y+2t\right) }=\frac{ f\left( x\right) f\left( y\right) }{f\left( x+y\right) }+at ,\quad x,y\in {\mathbb {R}},\quad t>0. \end{aligned}$$
(3.1)

Hence, in particular, for all \(x,y\in (0,\infty )\), we get

$$\begin{aligned} \frac{f\left( x\right) f\left( y\right) }{f\left( x+y\right) }= & {} \frac{ f\left( \left( x-y\right) +y\right) f\left( 0+y\right) }{f\left( \left( \left( x-y\right) +y\right) +\left( 0+y\right) \right) }=\frac{f\left( x-y\right) f\left( 0\right) }{f\left( x-y\right) }+ay \\= & {} f\left( 0\right) +ay. \end{aligned}$$

Changing the roles of x and y in the resulting equality, we hence also get, for all \(x,y\in (0,\infty )\),

$$\begin{aligned} \frac{f\left( x\right) f\left( y\right) }{f\left( x+y\right) }=f\left( 0\right) +ax. \end{aligned}$$
(3.2)

These two equalities imply that \(a \left( y-x\right) =0\) for all \(x,y\in (0,\infty )\) and, in consequence, that \(a=0\), whence \(\alpha (t)=0\) for all \(t>0\).

Now, take arbitrary \(x,y\in {\mathbb {R}}\). 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

$$\begin{aligned} \frac{f\left( x\right) f\left( y\right) }{f\left( x+y\right) }=\frac{f\left( x+t\right) f\left( y+t\right) }{f\left( x+y+2t\right) }=f(0), \end{aligned}$$

which proves (ii).

Since the implications \(\left( ii\right) \Longrightarrow \left( iii\right) \) and \(\left( iii\right) \Longrightarrow \left( i\right) \) are easy to verify, the proof is complete. \(\square \)

Remark 3

The exponential function in (ii) need not be continuous.

4 Translativity of beta-type functions in \([ 0,\infty ) ^{2}\)

Lemma 1

Assume that \(f:[ 0,\infty ) \rightarrow \left( 0,\infty \right) \) and \(\ \alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\). If \(B_{f}: [ 0,\infty ) ^{2}\rightarrow \left( 0,\infty \right) \) is \(\alpha \)-translative, i.e.

$$\begin{aligned} B_{f}\left( x+t,y+t\right) =B_{f}\left( x,y\right) +\alpha \left( t\right) , \quad x,y\ge 0,\quad t>0, \end{aligned}$$
(4.1)

then \(\alpha \equiv 0\), that is

$$\begin{aligned} B_{f}\left( x+t,y+t\right) =B_{f}\left( x,y\right) ,\quad x,y\ge 0,\quad t>0. \end{aligned}$$

Proof

From (4.1), the definition of \(B_{f}\) and Remark 2, we easily conclude that \( \alpha \left( t\right) =at\) for some real a and all \(t>0\), whence

$$\begin{aligned} \frac{f\left( x+t\right) f\left( y+t\right) }{f\left( x+y+2t\right) }=\frac{ f\left( x\right) f\left( y\right) }{f\left( x+y\right) }+at, \quad x,y\ge 0,\quad t>0. \end{aligned}$$

Setting here \(y=0\) gives

$$\begin{aligned} \frac{f\left( x+t\right) f\left( t\right) }{f\left( x+2t\right) }=f\left( 0\right) +at, \quad x\ge 0, \quad t>0, \end{aligned}$$

whence, putting \(b:=f\left( 0\right) \) and \(s:=x+t\), we get

$$\begin{aligned} \frac{f\left( s\right) f\left( t\right) }{f\left( s+t\right) }=b+at,\quad 0< t\le s. \end{aligned}$$
(4.2)

Replacing here s by \(s+t\) with \(t\le s\), we get

$$\begin{aligned} f\left( s+2t\right) =f\left( s+t\right) \frac{f\left( t\right) }{b+at} =f\left( s\right) \left( \frac{f\left( t\right) }{b+at}\right) ^{2}, \quad 0< t\le s. \end{aligned}$$
(4.3)

Taking t such that \(t\le s\le 2t,\) in view of (4.2) we also have

$$\begin{aligned} f\left( 2t+s\right) =f\left( 2t\right) \frac{f\left( s\right) }{b+as}. \end{aligned}$$

Since, taking \(s=t\) in (4.2) gives

$$\begin{aligned} f\left( 2t\right) =\frac{\left( f\left( t\right) \right) ^{2}}{b+at}, \end{aligned}$$

we hence get

$$\begin{aligned} f\left( 2t+s\right) =\frac{\left( f\left( t\right) \right) ^{2}}{b+at}\frac{ f\left( s\right) }{b+as},\quad t\le s\le 2t. \end{aligned}$$
(4.4)

From (4.3) and (4.4) we obtain

$$\begin{aligned} f\left( s\right) \left( \frac{f\left( t\right) }{b+at}\right) ^{2}=\frac{ \left( f\left( t\right) \right) ^{2}}{b+at}\frac{f\left( s\right) }{b+as}, \quad 0< t\le s\le 2t, \end{aligned}$$

which reduces to \(a\left( s-t\right) =0\). This proves that \(a=0.\) \(\square \)

The main result of this section reads as follows:

Theorem 2

Let \(f:[ 0,\infty ) \rightarrow \left( 0,\infty \right) \) and \(\alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\). The following conditions are pairwise equivalent:

  1. (i)

    the beta-type function \(B_{f}:[ 0,\infty ) ^{2}\rightarrow \left( 0,\infty \right) \) is \(\alpha \)-translative;

  2. (ii)

    \(\alpha \equiv 0\) and \(\frac{f}{f\left( 0\right) }\) is an exponential function, that is

    $$\begin{aligned} f\left( 0\right) f\left( x+y\right) =f\left( x\right) f\left( y\right) ,\quad x,y\ge 0; \end{aligned}$$
  3. (iii)

    the beta-type function \(B_{f}\) is constant.

Proof

Assume (i). By Lemma 1 we have \(\alpha \equiv 0\) and

$$\begin{aligned} \frac{f\left( x+t\right) f\left( y+t\right) }{f\left( x+y+2t\right) }=\frac{ f\left( x\right) f\left( y\right) }{f\left( x+y\right) },\quad x,y\ge 0,\quad t>0. \end{aligned}$$

Take arbitrary \(x,y\in [ 0,\infty ).\) Assuming that \(y\le x\), which can be done without any loss of generality, and applying the above equality with \(t=y\), we get

$$\begin{aligned} \frac{f\left( x\right) f\left( y\right) }{f\left( x+y\right) }=\frac{f\left( \left( x-y\right) +y\right) f\left( 0+y\right) }{f\left( \left( \left( x-y\right) +y\right) +\left( 0+y\right) \right) }=\frac{f\left( x-y\right) f\left( 0\right) }{f\left( x-y\right) }=f\left( 0\right) , \end{aligned}$$

which proves \(\left( ii\right) \).

As the remaining two implications \(\left( ii\right) \Longrightarrow \left( iii\right) \) and \(\left( iii\right) \Longrightarrow \left( i\right) \) are simple to check, the proof is complete. \(\square \)

5 Translativity of beta-type functions in \(\left( 0,\infty \right) ^{2}\)

We begin with an analogue of Lemma 1.

Lemma 2

Assume that \(f:\left( 0,\infty \right) \rightarrow \left( 0,\infty \right) \) and \(\ \alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\). If \(B_{f}:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right) \) is \(\alpha \)-translative, i.e.

$$\begin{aligned} B_{f}\left( x+t,y+t\right) =B_{f}\left( x,y\right) +\alpha \left( t\right) , \quad x,y>0,\quad t>0, \end{aligned}$$
(5.1)

then \(\alpha \equiv 0\), that is

$$\begin{aligned} B_{f}\left( x+t,y+t\right) =B_{f}\left( x,y\right) ,\quad x,y>0,\quad t>0. \end{aligned}$$

Proof

In view of Remark 2, \(\alpha \left( t\right) =at\) for some nonnegative a and for all \(t>0.\) Assume, on the contrary, that \(a >0\) and that there exists a function \(f:\left( 0,\infty \right) \rightarrow \left( 0,\infty \right) \) satisfying (5.1), i.e. that

$$\begin{aligned} \frac{f\left( x+t\right) f\left( y+t\right) }{f\left( x+y+2t\right) }=\frac{ f\left( x\right) f\left( y\right) }{f\left( x+y\right) }+at,\quad x,y,t>0. \end{aligned}$$

Writing this equality in the form

$$\begin{aligned} \frac{\frac{f\left( x+t\right) }{a}\frac{f\left( y+t\right) }{a}}{\frac{ f\left( x+y+2t\right) }{a}}=\frac{\frac{f\left( x\right) }{a}\frac{f\left( y\right) }{a}}{\frac{f\left( x+y\right) }{a}}+t,\quad x,y,t>0 , \end{aligned}$$

we see that, without any loss of generality, we can assume that \(a=1\), i.e. that

$$\begin{aligned} \frac{f\left( x+t\right) f\left( y+t\right) }{f\left( x+y+2t\right) }=\frac{ f\left( x\right) f\left( y\right) }{f\left( x+y\right) }+t,\quad x,y,t>0. \end{aligned}$$
(5.2)

Replacing here x, y, t,  respectively by 2x, 2y, 2t, we obtain

$$\begin{aligned} \frac{f(2(x+t))f(2(y+t))}{f(2(x+y+2t))}=\frac{f(2x)f(2y)}{f(2(x+y))}+2t,\quad x,y,t>0. \end{aligned}$$
(5.3)

Setting \(y=x\) in (5.2) we get

$$\begin{aligned} \frac{\left( f\left( x+t\right) \right) ^{2}}{f\left( 2\left( x+t\right) \right) }=\frac{\left( f\left( x\right) \right) ^{2}}{f\left( 2x\right) }+t,\quad x,t>0, \end{aligned}$$

which shows that the function \(g\left( x\right) :=\frac{\left( f\left( x\right) \right) ^{2}}{f\left( 2x\right) }\) satisfies the functional equation

$$\begin{aligned} g\left( x+t\right) =g\left( x\right) +t,\quad x,t>0. \end{aligned}$$
(5.4)

Changing the roles of x and t in (5.4) gives us

$$\begin{aligned} g(t+x)=g(t)+x,\quad x,t>0, \end{aligned}$$
(5.5)

whence, by (5.4) and (5.5), we obtain that \(g(x)-x\) is constant. It follows that, for some \(b\ge 0\),

$$\begin{aligned} \frac{\left( f\left( x\right) \right) ^{2}}{f\left( 2x\right) }=x+b,\quad x>0. \end{aligned}$$

Hence, using (5.3), we have

$$\begin{aligned} \frac{\frac{(f(x+t))^{2}}{x+t+b}\frac{(f(y+t))^{2}}{y+t+b}}{\frac{ (f(x+y+2t))^{2}}{x+y+2t+b}}=\frac{\frac{(f(x))^{2}}{x+b}\frac{(f(y))^{2}}{y+b }}{\frac{(f(x+y))^{2}}{x+y+b}}+2t,\quad x,y,t>0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \left( \frac{f(x+t)f(y+t)}{f(x+y+2t)}\right) ^{2}= & {} \frac{(x+t+b)(y+t+b)(x+y+b) }{(x+y+2t+b)(x+b)(y+b)}\left( \frac{f(x)f(y)}{f(x+y)}\right) ^{2}\nonumber \\&+\frac{2t\left( x+t+b\right) \left( y+t+b\right) }{x+y+2t+b},\quad x,y,t>0. \end{aligned}$$
(5.6)

On the other hand, squaring both sides of (5.2), gives us

$$\begin{aligned} \left( \frac{f(x+t)f(y+t)}{f(x+y+2t)}\right) ^{2}=\left( \frac{f(x)f(y)}{ f(x+y)}\right) ^{2}+2t\frac{f(x)f(y)}{f(x+y)}+t^{2}. \end{aligned}$$
(5.7)

From (5.6) and (5.7) we obtain, for all \(x,y,t>0\),

$$\begin{aligned} \left( \frac{\left( x+t+b\right) \left( y+t+b\right) \left( x+y+b\right) }{ \left( x+y+2t+b\right) \left( x+b\right) \left( y+b\right) }-1\right) \left( \frac{f(x)f(y)}{f(x+y)}\right) ^{2} \end{aligned}$$
$$\begin{aligned} -2t\frac{f(x)f(y)}{f(x+y)}+\frac{2t\left( x+t+b\right) \left( y+t+b\right) }{ x+y+2t+b}-t^{2}=0. \end{aligned}$$
(5.8)

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

$$\begin{aligned} \frac{19}{56}\left( \frac{f\left( b\right) f\left( 3b\right) }{f\left( 4b\right) }\right) ^{2}-2b\left( \frac{f\left( b\right) f\left( 3b\right) }{ f\left( 4b\right) }\right) +\frac{23}{7}b^{2}=0. \end{aligned}$$

However, because the quadratic polynomial \(\frac{19}{56}p^{2}-2bp+\frac{23}{7 }b^{2}\) of the variable p does not have real roots, we get a desired contradiction. Now, if \(b=0\), equality (5.8) simplifies to

$$\begin{aligned}&\left( \frac{\left( x+t\right) \left( y+t\right) \left( x+y\right) }{ \left( x+y+2t\right) xy }-1\right) \left( \frac{f(x)f(y)}{f(x+y)}\right) ^{2}-2t\frac{f(x)f(y)}{f(x+y)} \\&\quad +\frac{2t\left( x+t\right) \left( y+t\right) }{ x+y+2t}-t^{2}=0, \end{aligned}$$

but once again, there does not exist a real-valued function f defined on \((0,\infty )\), 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). \(\square \)

Now we are ready to give a complete solution to the translativity problem of \(B_{f}\) in \((0,\infty )^{2}\).

Theorem 3

Let \(f:\left( 0,\infty \right) \rightarrow \left( 0,\infty \right) \) and \(\ \alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\). The following conditions are equivalent:

  1. (i)

    the beta-type function \(B_{f}:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right) \) is \(\alpha \)-translative;

  2. (ii)

    \(\alpha \equiv 0\) and there exists \(C>0\) such that the function \( \frac{f}{C}\) is an exponential function, that is

    $$\begin{aligned} Cf\left( x+y\right) =f\left( x\right) f\left( y\right) ,\quad x,y>0; \end{aligned}$$
  3. (iii)

    the function \(B_{f}\) is constant.

Proof

Assume (i). Then, based on Lemma 2, we have

$$\begin{aligned} \frac{f(x+t)f(y+t)}{f(x+y+2t)}=\frac{f(x)f(y)}{f(x+y)},\quad x,y,t>0. \end{aligned}$$
(5.9)

It is easy to observe that (5.9) implies that the related beta-type function \( B_{f}\) for arbitrary \(b\in {\mathbb {R}}\) must be constant on the set \(\mathcal { D}_{b}=\{(x,y)\in (0,\infty )^{2}:y=x+b\}\). For a given real number b we denote this constant value by \(C_{b}\). Hence (5.9) can be equivalently written as the following infinite system of equalities, indexed by \(b\in {\mathbb {R}}\),

$$\begin{aligned} \frac{f(x)f(x+b)}{f(2x+b)}=C_{b},\quad x>\max \{0,-b\}. \end{aligned}$$
(5.10)

Now we prove that in fact \(C_{b}=C_{0}\) for each \(b\in {\mathbb {R}}\). First we consider the case \(b>0\). By (5.10) the function f satisfies the following two equalities:

$$\begin{aligned} f(x)f(x)=C_{0}f(2x),\quad x>0, \end{aligned}$$
(5.11)

and

$$\begin{aligned} f(x)f(x+b)=C_{b}f(2x+b),\quad x>0, \end{aligned}$$
(5.12)

where \(b>0\) is arbitrarily fixed. Replacing x in (5.11) by b and 2b, respectively, we immediately get

$$\begin{aligned} f(b)=\frac{\left( f\left( \frac{1}{2}b\right) \right) ^{2}}{C_{0}} \end{aligned}$$

and

$$\begin{aligned} f(2b)=\frac{(f(b))^{2}}{C_{0}}=\frac{\left( f\left( \frac{1}{2}b\right) \right) ^{4}}{C_{0}^{3}}. \end{aligned}$$
(5.13)

These two formulas, together with the following one, obtained by replacing x by b in (5.12):

$$\begin{aligned} f(b)f(2b)={C_{b}}f(3b) \end{aligned}$$

allow us to write

$$\begin{aligned} f(3b)=\frac{f(b)f(2b)}{C_{b}}=\frac{\left( f\left( \frac{1}{2}b\right) \right) ^{6}}{C_{b}C_{0}^{4}}. \end{aligned}$$
(5.14)

On the other hand, after putting \(x=\frac{3}{2}b\) in (5.11) and \(x=\frac{1}{2}b \) in (5.12), we get

$$\begin{aligned} f(3b)=\frac{\left( f\left( \frac{3}{2}b\right) \right) ^{2}}{C_{0}} \end{aligned}$$

and

$$\begin{aligned} f\left( \frac{3}{2}b\right) =\frac{C_{b}f(2b)}{f\left( \frac{1}{2}b\right) }, \end{aligned}$$

which, using (5.13), we can write as

$$\begin{aligned} f(3b)=\frac{C_{b}^{2}}{C_{0}^{7}}\left( f\left( \frac{1}{2}b\right) \right) ^{6}. \end{aligned}$$
(5.15)

Now, comparing the right-hand sides of (5.14) and (5.15), we obtain

$$\begin{aligned} \frac{1}{C_{b}C_{0}^{4}}=\frac{C_{b}^{2}}{C_{0}^{7}}, \end{aligned}$$

whence, by the positivity of \(C_{0}\) and \(C_{b}\), we get \(C_{b}=C_{0},\) that is the desired equality.

The proof in the case \(b<0\) runs quite similarly. The main difference is that we are now looking for two different formulas for \(f(-3b)\) expressed in terms of only \(f\left( -\frac{1}{2}b\right) \). The first formula of this kind has the form

$$\begin{aligned} f(-3b)=\frac{\left( f\left( -\frac{1}{2}b\right) \right) ^{6}}{C_{b}C_{0}^{4} }, \end{aligned}$$
(5.16)

which can be obtained by putting \(x=-\frac{3}{2}b\) and \(x=-b\) in (5.11) and \(x=-\frac{3}{2}b\) in (5.12). In turn, replacing \(x=-\frac{1}{2}b\) in (5.11) and \(x=-2b\) in (5.12) leads to the second formula

$$\begin{aligned} f(-3b)=\frac{C_{b}^{2}}{C_{0}^{7}}\left( f\left( -\frac{1}{2}b\right) \right) ^{6}. \end{aligned}$$
(5.17)

Finally, equalities (5.16) and (5.17) imply that also in the case of arbitrary \( b<0\) we have \(C_{b}=C_{0}\). This shows that, for all \(b\in {\mathbb {R}}\),

$$\begin{aligned} \frac{f(x)f(y)}{f(x+y)}=C_{0},\ \ \ (x,y)\in {\mathcal {D}}_{b}. \end{aligned}$$

Because \(\bigcup \limits _{b}{\mathcal {D}}_{b}=(0,\infty )^{2}\), the proof of (ii) is completed.

Implications \(\left( ii\right) \Longrightarrow \left( iii\right) \) and \( \left( iii\right) \Longrightarrow \left( i\right) \) are obvious. \(\square \)

6 Extension theorem for translative functions

To look at the problem of translativity of beta-type functions from a broader perspective, we begin with the following

Theorem 4

Assume that the functions \(F:\left( 0,\infty \right) ^{2}\rightarrow {\mathbb {R}}\) and \(\alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\) satisfy the equation

$$\begin{aligned} F\left( x+t,y+t\right) =F\left( x,y\right) +\alpha \left( t\right) ,\quad x,y,t>0. \end{aligned}$$
(6.1)

Then

  1. (i)

    \(\alpha \) is additive;

  2. (ii)

    the function \(G:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}\) given by

    $$\begin{aligned} G\left( x,y\right) :=F\left( x+n,y+n\right) -\alpha \left( n\right) \quad \text {if} \quad \left( x,y\right) \in \left( -n,+\infty \right) ^{2},\quad n\in {\mathbb {N}},\nonumber \\ \end{aligned}$$
    (6.2)

    is well defined, and it is a unique function such that

    $$\begin{aligned} G\left( x+t,y+t\right) =G\left( x,y\right) +\alpha \left( t\right) ,\quad x,y\in {\mathbb {R}},\quad t>0, \end{aligned}$$
    (6.3)

    and

    $$\begin{aligned} G\left( x,y\right) =F\left( x,y\right) ,\quad x,y>0; \end{aligned}$$
    (6.4)
  3. (iii)

    if \(\alpha \equiv 0\) and F is nonnegative (positive), then so is G.

Proof

To prove (i) we can argue as in Remark 2.

(ii) For every \(n\in {\mathbb {N}}\) define \(G_{n}:\left( -n,\infty \right) ^{2}\rightarrow {\mathbb {R}}\) by

$$\begin{aligned} G_{n}\left( x,y\right) :=F\left( x+n,y+n\right) -\alpha \left( n\right) ,\quad x,y>-n. \end{aligned}$$

Note that for every \(n\in {\mathbb {N}}\)

$$\begin{aligned} G_{n+1}\left( x,y\right) =G_{n}\left( x,y\right) ,\quad x,y>-n. \end{aligned}$$

Indeed, if \(x,y>-n\) then, from the definition of \(G_{n}\), (6.1), and the additivity of \(\alpha \) (see Remark 2), we have

$$\begin{aligned} G_{n+1}\left( x,y\right)= & {} F\left( x+n+1,y+n+1\right) -\alpha \left( n+1\right) \\= & {} F\left( \left( x+n\right) +1,\left( y+n\right) +1\right) -\alpha \left( n+1\right) \\= & {} \left( F\left( x+n,y+n\right) +\alpha \left( 1\right) \right) -\left( \alpha \left( n\right) +\alpha \left( 1\right) \right) \\= & {} F\left( x+n,y+n\right) -\alpha \left( n\right) =G_{n}\left( x,y\right) . \end{aligned}$$

Moreover from (6.1), for every n and \(x,y>0\),

$$\begin{aligned} G_{n}\left( x,y\right) :=F\left( x+n,y+n\right) -\alpha \left( n\right) = \left( F\left( x,y\right) +\alpha \left( n\right) \right) -\alpha \left( n\right) =F\left( x,y\right) . \end{aligned}$$

It follows that the function \(G:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}\) given by (6.2), that is

$$\begin{aligned} G\left( x,y\right) :=G_{n}\left( x,y\right) \quad \text {if} \quad \left( x,y\right) \in \left( -n,\infty \right) ^{2}, \end{aligned}$$

is well defined and satisfies condition (6.4).

To see that G satisfies (6.3), take arbitrary \(x,y\in {\mathbb {R}}\) and \(t>0\). Choosing \(n\in {\mathbb {N}}\) such that \(x,y\in \left( -n,\infty \right) \), making use of the definition of G and applying property (6.1) of F, we have

$$\begin{aligned} G\left( x+t,y+t\right)= & {} G_{n}\left( x+t,y+t\right) =F\left( x+t+n,y+t+n\right) -\alpha \left( n\right) \\= & {} \left( F\left( x+n,y+n\right) +\alpha \left( t\right) \right) -\alpha \left( n\right) \\= & {} \left( F\left( x+n,y+n\right) -\alpha \left( n\right) \right) +\alpha \left( t\right) =G_{n}\left( x,y\right) +\alpha \left( t\right) \\= & {} G\left( x,y\right) +\alpha \left( t\right) , \end{aligned}$$

which proves (24). Since the uniqueness of G is obvious, result (ii) is proved.

To prove (iii), assume \(\alpha \equiv 0\) and the nonnegativity (positivity) of F. From (6.3) and (6.4) we have

$$\begin{aligned} G\left( x,y\right) =F\left( x+t,y+t\right) ,\quad t+x>0, \quad t+y>0, \end{aligned}$$

so G is nonnegative (positive). \(\square \)

Remark 4

The counterpart of this result holds true on replacing the interval \(\left( 0,\infty \right) \) by \(\left( c,\infty \right) \,\), where \(c\ge 0\) is arbitrarily fixed.

As a simple consequence of Theorems 4 and 1 we obtain that for a \(\alpha \)-translative beta-type function \(B_{f}\) defined on \( \left( 0,\infty \right) ^{2}\) its unique \(\alpha \)-translative extension G is a beta-type function on \({\mathbb {R}}^{2}\), provided that \(\alpha \equiv 0\).