The archetypal equation and its solutions attaining the global extremum

Let (Ω,F,P)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\Omega , \mathcal {F}, \mathbb {P})$$\end{document} be a probability space and let α,β:F→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha , \beta : \mathcal {F} \rightarrow ~\mathbb {R}$$\end{document} be random variables. We provide sufficient conditions under which every bounded continuous solution φ:R→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi : \mathbb {R} \rightarrow \mathbb {R}$$\end{document} of the equation φ(x)=∫Ωφα(ω)(x-β(ω))P(dω)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ \varphi (x) = \int _{ \Omega } \varphi \left( \alpha (\omega ) (x-\beta (\omega ))\right) \mathbb {P}(d\omega )$$\end{document} is constant. We also show that any non-constant bounded continuous solution of the above equation has to be oscillating at infinity.


Introduction
The paper concerns the linear functional equation of infinite order where μ is a given Borel probability measure on R 2 . We can think about equation (1.1) also in the language of random variables. Given a probability space (Ω, F, P) let (α, β) : Ω → R 2 be a fixed random vector with a distribution μ. Then the equation is equivalent to (1.1). The study of equation (1.1) was initiated by Derfel [5] in 1989. He considered equation (1.2) under the additional assumption α > 0 a.s. and he noticed 506 M. Sudzik AEM that the behaviour of solutions of (1.1) crucially depends on the value of the integral ln |a|μ(da, db). (1.3) He proved that, under some additional technical assumptions, if K ∈ (−∞, 0) then equation (1.1) has only trivial, i.e. constant, solutions in the class of bounded continuous functions. If K ∈ (0, ∞) and α > 0 a.s., then he constructed a non-trivial bounded continuous solution. The details of this theorem and its proof are included in [4], Theorem 1.1.
Many deep results connected with equation (1.1) were obtained and collected by Bogachev et al. [3] and [4]. In those papers equation (1.1) was named by the authors as the archetypal equation since it is a rich source of many famous functional and differential-functional equations. For instance, one can obtain a balanced version of the pantograph equation; details can be found in [4]. It is worth noting that functional equations of a more general form than (1.2) were also considered in the literature; an interested reader, can find them e.g. among papers written by Polish mathematicians from the Silesian University: Baron, Kapica and Morawiec (see, for instance, [2] and [8]). Moreover, a lot of interesting remarks connected with the archetypal equation, its special cases and equations of similar forms can be found in [1] and [6].
Here we can confine ourselves to (1.1) since equations with linearly transformed arguments are of prominent interest in applications. We have to add that the main results obtained in this paper hold under the condition P(α < 0) > 0. (1.4) In this case we still do not have extensive knowledge about the existence of non-constant solutions of the archetypal equation in the class of bounded continuous functions. It is worth noting that every absolutely continuous solution of (1.1) whose derivative belongs to L 1 (R) is constant when (1.4) holds (see Theorem 4.5 from [3] [4]). Each of them is uniformly continuous. It was also proved in [4] that if |α| = 1 a.s. and K ≤ 0, then every bounded and uniformly continuous solution of (1.1) is constant. The problem of the existence or non-existence of non-trivial solutions of (1.1) when (1.4) is true and K > 0 is still open today.
In the present paper we will prove that, under some mild technical assumption imposed on the measure μ, every bounded continuous solution of (1.1) reaching the global extreme value must be constant. In the proof we make use of the method in [9]. In the third section we prove some results connected with the asymptotical behaviour of bounded continuous solutions of the archetypal equation.

Solutions attaining the global extreme value
Let us begin with a simple observation.
Remark 2.1. Note that each constant function is a solution of the archetypal equation since μ is a probability measure. Furthermore, the linear combination of solutions of (1.1) is still its solution. Hence, studying bounded solutions ϕ : R → R of (1.1) which attain their global extremum we can assume without loss of generality that min ϕ(R) = 0 and sup ϕ(R) ≤ 1.
By the support of a Borel measure μ on R n , where n ∈ N, we mean the set suppμ of all x ∈ R n such that each neighborhood of x is of positive measure μ. Observe that suppμ is a closed subset of R n . In what follows μ stands for an arbitrary Borel probability measure on R 2 .
We prove the following technical lemma which will be our basic tool in the proof of the main theorems.
We will also prove a lemma which asserts that some subsets are dense in R. The statement is obvious but it can help to draw attention to the crucial properties of sets which are considered in the proof of Theorem 2.4. We will also present a simple proof of the lemma for the sake of completeness.
Proof. We will consider the case when t is positive. We fix a, b ∈ R such that a < b and define the interval I = (a, b). We have to show that I ∩ D is nonempty. Let n 0 ∈ N be such that (2.1) Let s 0 be the greatest number of Z for which Hence, by the definition of the number s 0 , we have Moreover, using the definition of s 0 and inequality (2.1) we get Therefore u n0 + s0t+t v n 0 ∈ I. It means that D is dense in R. If t is negative then the proof is similar, so we omit it. Proof. The pair (1, b) will be denoted by (a 0 , b 0 ) from this point. Take any bounded continuous solution ϕ : R → R of (1.1) attaining its global extremum. In view of Remark 2.1 we can assume that ϕ : R → [0, 1] and there exists x 0 ∈ R such that ϕ(x 0 ) = 0. Since ϕ is continuous, it is sufficient to show that ϕ takes value 0 on some dense subset D of the real line. The proof is split into the three general cases and in each of them we will describe how to obtain the desired dense set. We shall examine each of the following situations: Furthermore we distinguish two complementary subcases for both case I and case II.

I.A.
There exist a ∈ (R \ Q) ∩ (−∞, 0) and b ∈ R such that (a, b) ∈ suppμ. I.B. If (a, b) ∈ suppμ and a ∈ R \ Q, then a is positive. II.A. There exist a ∈ (Q \ Z) ∩ (−∞, 0) and b ∈ R such that (a, b) ∈ suppμ. II.B. If (a, b) ∈ suppμ and a ∈ Q \ Z, then a is positive. In summary, we obtain the five cases (I.A, I.B, II.A, II.B and III) and note that the measure μ fulfills at least one of them. We shall prove that in each one the function ϕ is constant. I.A. Assume that there exists (a, b) ∈ suppμ such that a ∈ (−∞, 0) \ Q. Put (a 1 , b 1 ) := (a, b) and remember that a 0 = 1 and b 0 = 0. Fix also arbitrary k, l ∈ N. If we apply Lemma 2.2 with (c i , d i ) = (a 0 , b 0 ), where i = 1, 2, . . . , k, then we obtain the equality ϕ(x 0 − kb 0 ) = 0 since a 0 = 1. Applying equality (1.1) for x 0 − kb 0 one can deduce that The Kronecker density theorem [7, Chapter XXIII] asserts that the set is also dense as b 0 = 0. This implies that ϕ is constant and the proof is complete in this subcase.
I.B. This case means that if (a, b) ∈ suppμ and a is negative, then a ∈ Q. Since μ ((−∞, 0) × R) > 0 there exists (a, b) ∈ suppμ with a < 0. Put (a 1 , b 1 ) := (a, b). Then a 1 ∈ Q. Take also (a 2 , b 2 ) ∈ suppμ with a 2 ∈ R \ Q. Then a 2 > 0. As before we obtain the equality ϕ (a 1 (x 0 − b 1 ) − a 1 kb 0 ) = 0, where k ∈ N is fixed. In the next step, applying equality (1.1) to a 1 ( Fix an arbitrary l ∈ N. Finally, we apply Lemma 2.2 with (c i , d i ) = (a 0 , b 0 ) for i = 1, . . . , l to the point a 1 a 2 (x 0 − b 1 ) − a 2 b 2 − a 1 a 2 kb 0 . As a result we have Note that a 1 a 2 is a negative irrational number, so again using the Kronecker density theorem we have constructed a dense subset D of R while k, l runs through N. The proof is complete in this case.
II. Now we assume that suppμ ∩ ((Q \ Z) × R) = ∅. Since we are done in case I we may assume that suppμ ⊆ Q × R. Let A ⊆ Q be the smallest set such that suppμ ⊆ A × R, i.e. For every i ∈ I we also define a Borel measure μ i by the equality Note that for any i ∈ I we have μ i (R) ∈ (0, 1) and i∈I μ i (R) = 1. Moreover, equation (1.1) can be rewritten in the form

II.A.
In this case there exists i ∈ I such that a i is negative and noninteger. First of all remember that a 0 = 1 and b 0 = 0. Without loss of generality we may assume that i = 1. In other words a 1 ∈ (−∞, 0) ∩ (Q \ Z). Then there exist coprimes q, q 0 ∈ Z such that a 1 = q/q 0 and q < 0. The definitions of q, q 0 and a 1 imply that q 0 ≥ 2. We fix also any b 1 ∈ R for which (a 1 , b 1 ) ∈ suppμ. Define the sequence (D n ) n∈N of sets putting Since b 0 = 0 and q 0 > 1, the density of the set D follows from Lemma 2.3.
Our goal is to prove that the function ϕ vanishes on D. We shall use mathematical induction with respect to n. In the first step we show that ϕ |D1 = 0, i.e.
Fix any k, l ∈ N ∪ {0}. In the beginning note that we can obtain the equality Note that the expression q 0 l+qk runs through the whole Z, while k, l ∈ N∪{0}, since q and q 0 are coprime and of different signs. Therefore equality (2.3) implies that Vol. 95 (2021) The archetypal equation and its... 511 Let n ∈ N be fixed and assume that ϕ |Dn = 0. We prove that ϕ |Dn+1 = 0. Take any s ∈ Z, put and observe that r ∈ D n+1 . Since q and q 0 are coprime and of different signs, we can find k, l ∈ N ∪ {0} such that q n+1 0 l + qk = s. If we use the induction hypothesis and equality (2.2), then we will get 0 = ϕ a n and thus, as ϕ is non-negative, In particular, for j = 1 we have The function ϕ is non-negative and continuous. Hence Taking b = b 1 in the above equality we get Thus, using the representation a 1 = q/q 0 , we come to the equality ϕ a n+1 Recall that a 0 = 1 and the numbers k and l were chosen so that the sum , then we will get ϕ a n+1 The above equality means that ϕ(r) = 0. Since r was an arbitrary element of D n+1 , we have proven that Consequently, by mathematical induction, we get ϕ |D = 0 and the proof has been completed in case II.A.
II.B. In this case A \ Z = ∅ and each negative element of A is an integer. Recall that we have a 0 = 1 and b 0 = 0. We know that μ ((−∞, 0) × R) > 0. This condition implies that we can choose i ∈ I such that a i ∈ (−∞, 0). All such numbers are, in view of the assumptions of case II.B, integers. Without loss of generality we may assume that i = 1. Then a 1 ∈ Z. Take an arbitrary Take any b 2 ∈ suppμ 2 . Then a 2 must be positive. We also define a parameter m ∈ N such that a 1 a m 2 ∈ Q \ Z since it may happen that a 1 a 2 ∈ Z. There exist coprimes q, q 0 ∈ Z for which a 1 a m 2 = q/q 0 and q < 0. Then q 0 ≥ 2. The proof in this case is analogous to that of the previous subcase. This time a sequence of sets (D n ) n∈N will be defined by and again we put We have b 0 = 0 and q 0 > 1. Define a sequence (u n ) n∈N by Then, in view of Lemma 2.3, we will get the density of D. Again we shall use mathematical induction to prove that ϕ is constant on D. First of all we are going to show that ϕ |D1 = 0. As previously we start with the equality Vol. 95 (2021) The archetypal equation and its... 513 Since a 1 a m 2 = q/q 0 , we have . . , l, to the point from the above equality, then we will obtain Note that q and q 0 are coprime and they have different signs which implies that the expression qk + q 0 l can take any integer value when k, l run through N ∪ {0}. Hence Fix n ∈ N and assume that ϕ |Dn = 0. We will show the equality ϕ |Dn+1 = 0. Take any s ∈ Z and put Note that r is a member of the set D n+1 . We show that ϕ(r) = 0. As in the proof of case II.A we fix k, l ∈ N ∪ {0} such that lq n+1 0 + kq = s. Such a choice is possible since q and q 0 are coprime and q < 0 < q 0 . Moreover, we have since the argument from the above equality belongs to D n . Applying equality (2.2) to the point from the last equality, for every i ∈ I and b ∈ suppμ i we get In particular for i = 1 and b = b 1 , we have Since a 1 a m 2 = q/q 0 , the last equality can be rewritten as Recall that a 0 = 1. If we use Lemma 2.2 with (c j , d j ) = (a 0 , b 0 ), where j = 1, 2, . . . , l, to the number from the above equality, then we will get We have chosen the numbers k and l in such a way that lq n+1 i.e. ϕ(r) = 0. This equality implies that ϕ |Dn+1 = 0 since r was taken arbitrarily. In view of the mathematical induction we have ϕ |D = 0 and the proof is complete in case II.B. III. We shall consider the last case when suppμ ⊂ Z × R, that is A ⊆ Z. Recall that a 0 = 1 and b 0 = 0. Since μ (((−∞, 0) \ {−1}) × R) > 0 we can choose i ∈ I such that a i is a negative integer different from −1. Without loss of generality we may assume that i = 1. Fix any b 1 ∈ suppμ 1 . We define Note that D n contains as a subset the set We have b 0 = 0 and a 2 1 > 1. Lemma 2.3 asserts that the set Hence also D is dense. Moreover, now we will show that the sequence (D n ) n∈N∪{0} has the following properties: P1. If r ∈ D n for some n ∈ N and i ∈ I \ {1}, then P2. If r ∈ D n for some n ∈ N, then a 1 (r − b) ∈ D n−1 for every b ∈ suppμ 1 .
Fix n ∈ N and take any r ∈ D n . Let p ∈ N and (e 1 , f 1 ), . . . , (e p , f p ) ∈ suppμ be such that a 1 is one of e 1 , . . . , e p and with some s ∈ Z. Take an arbitrary i ∈ I and b ∈ suppμ i . Then (a i , b) ∈ suppμ and where e p+1 := a i and f p+1 :

M. Sudzik AEM
Now we are going to show that the function ϕ is constant on D. As before we will use mathematical induction and start with showing that ϕ |D0 = 0. Let us recall that f 1 ), . . . , (e p , f p ) ∈ suppμ and ∃ i0∈{1,2,...,p} e i0 = a 1 .
Since e i0 is a negative integer and k, l ∈ N ∪ {0}, the expression l + e i0 k can attain any integer value. In particular, we can find k, l ∈ N ∪ {0} such that l + e i0 k = s. This means that equality (2.4) holds. Hence ϕ |D0 = 0. Now assume that ϕ |Dn = 0 for some n ∈ N∪{0}. We prove that ϕ |Dn+1 = 0. Let r ∈ D n+1 be fixed and put J := I \ {1}. Then property P2 asserts that a 1 (r − b) ∈ D n for every b ∈ suppμ 1 . Hence using equality (2.2) and the inductive hypothesis we get The function ϕ is bounded above by 1. Therefore If we use equation (2.2) for every i ∈ J, then we will get Property P1 asserts that a i (r − b 1 ) ∈ D n+1 for every i ∈ J and b 1 ∈ suppμ i . Moreover, property P2 implies that a 1 (a i (r − b 1 ) − b 2 ) ∈ D n for every i ∈ J and (a i , b 1 ), (a 1 , b 2 ) ∈ suppμ. Since ϕ |Dn = 0, we have ϕ (a 1 (a i (r − b 1 ) − b 2 )) = 0 for every i ∈ J and (a i , b 1 ), (a 1 , b 2 ) ∈ suppμ.
Hence and from the fact that ϕ is bounded above by 1 we have for all i ∈ I the inequalities If we put the above estimations into the equality In a similar way, using P1 and P2 several times and taking into account that the function ϕ vanishes on D n and the fact that ϕ satisfies equation (2.2), one can inductively show that where q ∈ N and μ ⊗ (i 1 ,...,iq ) denotes the product of μ i1 , . . . , μ iq . Therefore, since all values of ϕ lie in [0, 1], we have and thus ϕ(r) = 0 because of the condition μ 1 (R) ∈ (0, 1). Consequently, we get ϕ(x) = 0 for every x ∈ D n+1 . Summarizing we see that ϕ vanishes on D.
If we apply Lemma 2.2 with (c i , d i ) = (a σ(i) , b σ(i) ) for i = 1, . . . , 2s to the point x 0 , we will get by t for simplicity. Then the last equality can be rewritten as If we again use Lemma 2.2 with (c i , d i ) = (a σ(i) , b σ(i) ) for i = 1, . . . , 2s to the point x 0 + t, then we will get ϕ(x 0 + 2t) = 0. By the induction one can prove that Without loss of generality we may assume that |a 1 | < 1. Let k ∈ N ∪ {0} and n ∈ N be taken arbitrarily. If we use Lemma 2.2 with (c i , d i ) = (a 1 , b 1 ) for every i = 1, 2, . . . , n to the points x 0 + kt and y 0 − kt, then we obtain the equalities ϕ ⎛ ⎝ a n 1 x 0 − n j=1 a j 1 b 1 + a n 1 kt ⎞ ⎠ = 0 (2.9) and ϕ ⎛ ⎝ a n respectively. We know that t = 0, that is either positive, or negative. We consider the first case. Conditions (2.7) and (2.8) imply that the set of zeros of the function ϕ is unbounded both from above and from below. Therefore for every n ∈ N we can choose a zero u n ∈ R of the function ϕ such that a n 1 u n − n j=1 a j 1 b 1 < 0.
Vol. 95 (2021) The archetypal equation and its... 521 We define also a sequence (v n ) n∈N of zeros of the function ϕ fulfilling the opposite inequalities a n Equalities (2.9) and (2.10) imply that for every n ∈ N and k ∈ N ∪ {0} we have ϕ ⎛ ⎝ a n 1 u n − n j=1 a j 1 b 1 + a n 1 kt Then ϕ |E∪F = 0. Lemma 2.8 asserts that E and −F are dense in (0, +∞). Hence E ∪ F is dense in R, and thus ϕ is constant. If t is negative, the proof is similar and we omit it.

The asymptotics of solutions
In this part we study the asymptotic behaviour of bounded continuous solutions of (1.1) which do not necessarily attain the global extremes. Vol. 95 (2021) The archetypal equation and its... 523 Proof. We will show equalities (3.1) only. Choose a sequence (x n ) n∈N of reals such that If it is bounded, then one can find its subsequence (y n ) n∈N which is convergent to some x 0 ∈ R. Then, by the continuity of ϕ, we have ϕ(x 0 ) = inf ϕ(R). Hence, by Theorem 2.4, we know that ϕ is constant and equalities (3.1) hold. So we may assume that (x n ) n∈N is unbounded. Then we can choose a subsequence (y n ) n∈N such that either y n → −∞, or y n → +∞. Consider, for instance, the first possibility. Then lim n→+∞ ϕ(y n ) = inf ϕ(R), and thus lim inf Since, in view of Theorem 4.2 from [3], we have lim inf we come to (3.1).
From this theorem the following result can be immediately deduced. We also have the analogous results generated by Theorem 2.9.  The above corollaries can be compared with a similar theorem by Bogachev, Derfel and Molchanov (see [3,Theorem 4
Corollaries 3.2 and 3.4 cannot be used here for the same reasons as in the previous example. Therefore, in view of Theorem BDM, every bounded continuous solution ϕ : R → R of the equation which has at least one of the limits lim x→−∞ ϕ(x) and lim x→+∞ ϕ(x) must be constant.
In conclusion, the presented theorems show that non-constant solutions of the archetypal equation in the case μ ((−∞, 0) × R) > 0 have to be oscillating functions. The question of the existence of bounded continuous solutions, which are not constant, is still an open problem.
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