Abstract
Translativity of the beta-type function \(B_{f}:I^{2}\rightarrow \left( 0,\infty \right) \),
where f is a single variable function defined either on \(I={\mathbb {R}}\) or \(I=[ 0,\infty )\), or \(I=\left( 0,\infty \right) \), is considered. In each of these three cases a complete solution is given.
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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
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 \(\alpha :\left( 0,\infty \right) \rightarrow {\mathbb {R}}\) 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
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
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
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 \( t>0\),
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:
-
(i)
if F is \(\alpha \)-translative, then \(\alpha \) is an additive function;
-
(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
and
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 \(t>0,\)
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:
-
(i)
the beta-type function \(B_{f}:{\mathbb {R}}^{2}\rightarrow \left( 0,\infty \right) \) is \(\alpha \)-translative;
-
(ii)
\(\alpha \equiv 0\) and \(\frac{f}{f\left( 0\right) }\) is the exponential function, that is
$$\begin{aligned} f\left( 0\right) f\left( x+y\right) =f\left( x\right) f\left( y\right) , \quad x,y\in {\mathbb {R}}; \end{aligned}$$ -
(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:
Hence, in particular, for all \(x,y\in (0,\infty )\), we get
Changing the roles of x and y in the resulting equality, we hence also get, for all \(x,y\in (0,\infty )\),
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
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.
then \(\alpha \equiv 0\), that is
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
Setting here \(y=0\) gives
whence, putting \(b:=f\left( 0\right) \) and \(s:=x+t\), we get
Replacing here s by \(s+t\) with \(t\le s\), we get
Taking t such that \(t\le s\le 2t,\) in view of (4.2) we also have
Since, taking \(s=t\) in (4.2) gives
we hence get
From (4.3) and (4.4) we obtain
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:
-
(i)
the beta-type function \(B_{f}:[ 0,\infty ) ^{2}\rightarrow \left( 0,\infty \right) \) is \(\alpha \)-translative;
-
(ii)
\(\alpha \equiv 0\) and \(\frac{f}{f\left( 0\right) }\) is an exponential function, that is
$$\begin{aligned} f\left( 0\right) f\left( x+y\right) =f\left( x\right) f\left( y\right) ,\quad x,y\ge 0; \end{aligned}$$ -
(iii)
the beta-type function \(B_{f}\) is constant.
Proof
Assume (i). By Lemma 1 we have \(\alpha \equiv 0\) and
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
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.
then \(\alpha \equiv 0\), that is
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
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
Setting \(y=x\) in (5.2) we get
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
Changing the roles of x and t in (5.4) gives us
whence, by (5.4) and (5.5), we obtain that \(g(x)-x\) is constant. It follows that, for some \(b\ge 0\),
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
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
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:
-
(i)
the beta-type function \(B_{f}:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right) \) is \(\alpha \)-translative;
-
(ii)
\(\alpha \equiv 0\) and there exists \(C>0\) such that the function \( \frac{f}{C}\) is an exponential function, that is
$$\begin{aligned} Cf\left( x+y\right) =f\left( x\right) f\left( y\right) ,\quad x,y>0; \end{aligned}$$ -
(iii)
the function \(B_{f}\) is constant.
Proof
Assume (i). Then, based on Lemma 2, we have
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 \(b\in {\mathbb {R}}\),
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:
and
where \(b>0\) is arbitrarily fixed. Replacing x in (5.11) by b and 2b, respectively, we immediately get
and
These two formulas, together with the following one, obtained by replacing x by b in (5.12):
allow us to write
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
and
which, using (5.13), we can write as
Now, comparing the right-hand sides of (5.14) and (5.15), we obtain
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
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
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 \(b\in {\mathbb {R}}\),
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
Then
-
(i)
\(\alpha \) is additive;
-
(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) -
(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
Note that for every \(n\in {\mathbb {N}}\)
Indeed, if \(x,y>-n\) then, from the definition of \(G_{n}\), (6.1), and the additivity of \(\alpha \) (see Remark 2), we have
Moreover from (6.1), for every n and \(x,y>0\),
It follows that the function \(G:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}\) given by (6.2), that is
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
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
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 \(\alpha \equiv 0\).
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Małolepszy, T., Matkowski, J. Translativity of beta-type functions. Aequat. Math. 97, 121–132 (2023). https://doi.org/10.1007/s00010-022-00907-0
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DOI: https://doi.org/10.1007/s00010-022-00907-0