A Pexider equation containing the aritmetic mean
In this paper we determine the solutions (φ , f 1 , f 2) of the Pexider functional equation φ ( x + y 2) (f 1 (x) - f 2 (y)) = 0 , (x , y) ∈ I 1 × I 2 , where I 1 and I 2 are nonempty open subintervals. Special cases of the above equation regularly arise in problems with two-variable means. We show that, under a rather weak regularity condition, the coordinate-functions of a typical solution of the equation are constant over several subintervals of their domain. The regularity condition in question will be that the set of zeros of φ is closed. We also discuss particular solutions where this condition is not met.
Keywords: Functional equation; Pexider equation; Arithmetic mean; Primary 39B22; Secondary 26E60
The research of the author was supported in part by the ÚNKP-22-4 New National Excellence Program of the Ministry for Culture and Innovation from the source of the National Research, Development and Innovation Fund, in part by NKFIH Grant K-134191, and in part by the Eötvös Loránd Research Network (ELKH).
Introduction
Functional equations of type
1
Graph
are closely connected with various problems with two-variable means. In relation to the Matkowski–SutÔ problem [[2]], equation (1) was studied assuming that , , for some constant , and that g is of the form , where d is a derivative and c is continuous.
In paper [[3]] Z. Daróczy and V. Totik investigated a co-equation of (1) and proved that continuous differentiability of the solutions implies that they are constant. The authors also showed that there exist differentiable, nowhere monotone (and hence non-constant) solutions as well. The interested reader can also find equations with a similar spirit in [[1]].
In [[4]], as an auxiliary functional equation, a special case of (1) played an important role in solving the equality problem of Cauchy means and quasi-arithmetic means. The solutions of (1) were determined assuming that , , and that the set of zeros of is closed. Also in terms of functions and domains, equation (1) is a pexiderization of the auxiliary equation of [[4]].
Note that, in fact, for given functions and , it is not so difficult to construct to obtain a solution. Indeed, let and be any functions, be any value but fixed, and define . Then is a solution of (1) provided that whenever . The aim of this paper is to solve the equation under a weak but reasonable condition that guarantees that the solutions have regular, more precisely, constant parts. We are going to describe the solution set of functional equation (1), where stand for nonempty open intervals, the functions , , and are unknown, and the set of zeros of is assumed to be closed in D. Note that this latter condition is trivially satisfied whenever is continuous, strictly monotone, injective or, say, has finitely many zeros.
Adapted from paper [[4]], for and a given function , the set will be denoted by . For its complementary set we shall use . Whenever we write , we mean that , , and are real valued functions defined on D, , and , respectively. A triplet of functions will be called extremal if , that is, if is identically zero or there exists such that for all and . We can then make the following trivial observation.
Proposition 1
If is extremal then solves equation (1).
In fact, it is the above statement that motivates the name extremal. As we will see later, a general solution of (1), roughly speaking, is a mixture of extremal solutions, and this nice property is provided by the condition that is closed.
A nonempty family will be called a sequence of intervals, if it is countable. Then the family in question will be simply referred to as , that is, we do not indicate the index set.
For given subsets , define
Graph
In words, consists of those elements of Q that are reflections of some element of S with respect some element of P. If at least one of the sets is empty, then is empty. For brevity, when reflecting to a given point, that is, if P is of the form , then instead of we simply write . Finally, having an interval , we shall say that is a proper subset if .
Later we want to refer to some trivial properties of a set of the form , hence we formulate the following lemma.
Lemma 2
Let be any subsets.
- If , , and , then
-
Graph
- The sets and are mirror images of each other with respect to P, more precisely, we have
-
Graph
- If Q is open and at least one of the sets P and S is open, then is open.
- Finally, if P, Q, and S are intervals, so is and
-
Graph
- provided that is nonempty.
Proof
Assertions (i), (iii), and (iv) are direct consequences of the definition.
The proof of (ii) is also elementary. Set and . By definition, , hence, in view of statement (i), follows. The reverse inclusion is trivial if B is empty, thus we may assume that this is not the case. Let be any point. Then and there exist and such that . Thus , which, together with , yields that .
We will use what the next lemma states directly in our theorem about the solutions of (1). Briefly, it says that at least one of the mirror images of and for some nonempty subinterval of D is always located at one end of and , respectively. Note that the subinterval for which we are reflecting may consist of a single point.
Lemma 3
Let be a nonempty subinterval and with . Then is a nonempty open subinterval of furthermore
- if then and
- if then .
Proof
The fact that is nonempty follows from the inclusion .
To prove (i), assume that . Then, by (iv) of Lemma 2, , and hence . Then, in view of (ii) and (iv) of Lemma 2, we have
Graph
since and . Statement (ii) is similarly verified.
Solutions where the interior of Zφ is empty
In this section we characterize the solution set of (1) in the case where the interior of is empty. To do this, we need the following property of the members of the support of .
Proposition 4
Let be a solution of equation (1) such that is different from but closed in D. Then, for any , there exists a constant such that
2
Graph
Proof
Let . Then, by Lemma 3, and are nonempty open intervals in and , respectively.
Assume first that . We are going to show that any member of A has an open neighbourhood in A on which is constant. Let be any point. Then there exists , such that . Using that the sets A and are open, one can find such that
Graph
Applying equation (1) for with and for , we get that
Graph
where . Consequently, with , we have whenever . That was arbitrary, implies that if . By the same reasoning we obtain that must also be constant on B with some value . Notice that , thus follows.
Assume now that with and let . Then for some . Let be a sequence tending to p. We claim that there exists such that .
Indeed, otherwise, an infinite number of the intervals in question would always be to the left or right of the point , more precisely, there would exist a subsequence such that, for all , either
Graph
The first inequality yields that for all . Thus, particularly, follows. Taking , we get that
Graph
The second inequality leads to a similar type of contradiction. Thus for some .
In view of the first part of the proof, it follows that is constant on some open neighbourhood of . But was arbitrary, hence is identically on A for some constant . A similar argument shows that must take some on B.
All that remains is to show is that and are equal. Let and be such that , and, again, pick a sequence tending to p. Then there exists such that . Otherwise, for some subsequence , we would have either
Graph
Both inequalities contradict the fact that . Consequently, there exist and such that . Then, by equation (1), we have , which finishes the proof.
Theorem 5
Let be such that is closed in D and has an empty interior. Then solves functional equation (1) if and only if is extremal.
Proof
The sufficiency is obvious, so we address the necessity.
If is empty, then and we are done. Hence assume that this is not the case.
By Proposition 4, one can immediately obtain that any member of has an open neighbourhood in on which is constant. This yields that must be constant on any component of . An analogous statement holds with and .
Now observe that for any components and , the constant functions and take a common value. Indeed, if this were not so, then, using equation (1), one could deduce that is identically zero on , contradicting that has an empty interior. Let us denote the common value taken by the functions and by .
Now take a point arbitrarily and let be such that . We note that such y exists: otherwise would be true, which is a contradiction. Thus, by equation (1), follows, where
Graph
Consequently, . An analogous argument shows that the same is true for , that is, if .
Remark 1
In view of Lemma 3 and Proposition 4, for a solution of (1), either at least one of the functions and is constant on its domain or at the same end of their domain they are constant with the same value. Furthermore if has an empty interior or, equivalently, if is dense in D, the only possible solution is the extremal one. To have these statements, we heavily use the topological condition on . The following example demonstrates that without this topological assumption the situation may be completely different.
Let be any function,
Graph
where stands for the characteristic function of rational numbers, that is,
Graph
Then solves equation (1) and is closed in D if and only if is extremal or, equivalently, if vanishes at any rational member of D.
Indeed, let and . If or , then . If exactly one of them is rational, then must be irrational, that is . The second part of the statement is trivial in view of the inclusion .
The solution set in general
For an index and a constant , define
Graph
In words, the extended real numbers defined above can be interpreted as follows. The set is the largest open subinterval at the left end of on which is identically . We have if and only if has no subinterval of positive length whose left endpoint is and on which takes . Finally, if and only if takes on the whole interval . With obvious modifications, similar statements are true with .
Observe that, whenever is a non-extremal solution of (1), then obviously
- neither nor is constant on its domain or
- exactly one of the functions and is constant on its domain.
Indeed, if is not extremal, then neither the interior of nor the complementary set is non-empty. So if and were both constant, they would have to take the same value, which contradicts the fact that the solution is not extremal.
Theorem 6
Let be such that is closed in D. Then solves functional equation (1) if and only if one of the following statements holds true.
- The triplet is extremal.
- There exist constants and open intervals and with
-
Graph
- such that
-
Graph
• and
-
Graph
- where and are closed proper subintervals.
- There exists a constant and, for some index , there exists a sequence of disjoint nonempty open subintervals of such that
- 3
Graph
- where and with .
Proof
First we are going to prove the necessity of cases (i), (ii), and (iii). To do this, assume that solves equation (1).
If has an empty interior then, in view of Theorem 5, must be extremal, so we obtain solution (i). If has a nonempty interior, then either or . If , then again we get that is extremal. Hence, in the sequel, we suppose that is valid. Within this sub-case, two main cases can be distinguished: either the functions and are both constant on their domain or at least one of them is not constant. Since is different from D, in the first case we get that the functions and must take the same value, so we again get case (i). Therefore, suppose also that at least one of the functions and is not constant on its domain. This means that either (C1) or (C2) is satisfied.
Case 1. Assume first that neither
nor
is constant on its domain. Since is nonempty, the intersection is nonempty. Then, by Remark 1, there exist such that at least one of the inequalities
4
Graph
is met. Define the open subintervals
5
Graph
In view of the inequalities under (4), at least one of the conditions or is satisfied. Furthermore, by our condition (C1), we have , , and hold for any . Consequently, and are indeed proper closed subintervals.
For brevity, denote . The intervals and contain at least one point, hence , which shows that A is a nonempty open subinterval of D. Now all we need to prove is that the inclusion is fulfilled. We prove this indirectly.
Suppose that we have a point for which . Since , there exist and such that and that at least one of the inclusions or holds. On the other hand, in view of our condition (C1) and of the maximality of the intervals under (5), we must have and simultaneously. These inclusions yield that and . The contradiction shows that is valid. Thus we have obtained solution (ii).
Case 2. Assume now that exactly one of the functions
and
is constant on its domain. Without loss of generality, we can assume that for some whenever . Since the set is not empty there exists a nonempty open subinterval of on which takes the value . By our assumption, this interval must be different from . This forces that is bounded from below or bounded from above.
Let consist of the open subintervals of fulfilling the condition that each of them is maximal with respect to the property that is identically on it. On the one hand, the family is nonempty. On the other hand, having a non-extremal solution, does not belong to . Denote and consider the subfamily
Graph
Let us first consider the case where the diameter of each member of is less than d, that is, where the subfamily is empty. Then, for any , it follows that either
6
Graph
holds, since . Consequently at least one of the disjoint intervals and is nonempty. Moreover, by our assumption, we have , which yields that is bounded from below or bounded from above. In this subcase, define .
If the family is nonempty, then it can be at most countable. Then let .
Now, in any of the above subcases, define . We show that vanishes on the arithmetic mean .
Indirectly, assume that this is not the case. Then there exists such that . Then, on one hand, takes on . On the other hand , consequently and hence . In the case this is impossible, thus cannot be empty. Then there exists a member of the sequence such that . Hence, by the definition of , the interval cannot be a subset of , which is a contradiction. To summarize, in this third case we have obtained solution (iii).
Finally we turn to the sufficiency of (i), (ii), and (iii). In fact, the sufficiency of (i) is obvious.
To prove the sufficiency of (ii), let , , and set . If and or and , then , and hence equation (1) is trivially satisfied. Thus we can assume that this is not the case. Then we can distinguish two subcases, namely or .
If , then follows, which means that , hence equation (1) is satisfied. If and , then and hold. Observe that
Graph
Using these, we obtain that
Graph
Since A is an interval, it follows that , and hence . Case can be treated in a similar way.
Finally, to prove the sufficiency of (iii), for brevity assume that and let , , and . If for some index n, then , thus equation (1) is satisfied. Otherwise , which yields that .
As a consequence of Theorem 6, we obtain Theorem 3 of paper [[4]]. For the sake of clarity, the statement in question is formulated using the terminology of the current paper.
Corollary
Let be a nonempty open subinterval, be such that is closed in I, and . Then the pair is a solution of equation
7
Graph
if and only if either f is constant on I and is any function or there exist constants and open subintervals with
Graph
such that
8
Graph
Proof
Apply Theorem 6 for and .
Author Contributions
TK wrote the manuscript.
Funding Information
Open access funding provided by University of Debrecen.
Declarations
Conflict of interest
The authors declare no competing interests.
Publisher's Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
[
References
1
Daróczy Z, Laczkovich M. On functions taking the same value on many pairs of points. Real Anal. Exchange. 2008; 33; 2: 385-393. 2458255. 10.14321/realanalexch.33.2.0385
2
Daróczy Z, Páles Z. Gauss-composition of means and the solution of the Matkowski-Sutô problem. Publ. Math. Debrecen. 2002; 61: 157-218. 1914652. 10.5486/PMD.2002.2713
3
Daróczy Z, Totik V. Remarks on a functional equation. Acta Sci. Math. (Szeged). 2015; 81; 3–4: 527-534. 3443768. 10.14232/actasm-015-805-1
4
Kiss T, Páles Z. On a functional equation related to two-variable Cauchy means. Math. Inequal. Appl. 2019; 22; 4: 1099-1122. 4027704
]
By Tibor Kiss
Reported by Author