Reducing the polynomial-like iterative equations order and a generalized Zolt\'an Boros' problem

We present a technique for reducing the order of polynomial-like iterative equations; in particular, we answer a question asked by Wenmeng Zhang and Weinian Zhang. Our method involves the asymptotic behaviour of the sequence of consecutive iterates of the unknown function at a given point. As an application we solve a problem of Zolt\'an Boros posed during the 50th ISFE (2012).


Introduction
Suppose I ⊂ R to be a non-degenerated interval and let g : I → I be a function. Assuming g to be continuous we are interested in lowering the order of the equation where N is a positive integer, the coefficients a 0 , a 1 , . . . , a N are real and a 0 = 0; here and throughout this paper g n stands for the n-th iterate of g. Note that a 0 = 0 implies that a n = 0 for some n = 1, . . . , N. From now on we assume that a N = 0. It turns out that continuous solutions to (1.1) deeply depend on the roots of its characteristic equation (1.2) a N r N + . . . + a 1 r + a 0 = 0, which usually is obtained by assuming that g has the form g(x) = rx. Up to now the case N = 2 is the only non-trivial one which has been completely solved (see [12]). In fact, the problem still remains open even for N = 3 (see [7]). These difficulties follow from the non-linearity of the operator g → g n . Nonetheless, a lot of investigation was done in this matter; see a survey on functional equations with superpositions of the unknown function [1, section 3] and a survey on iterative equations of polynomial type [19]. One of methods for finding solutions to equation (1.1), and also to its non-homogenous counterpart, where zero on the right-hand side is replaced by an arbitrary continuous function, is based on lowering its order. First such results on the whole real line were obtained in [8] in the case where all roots of the characteristic equation are real and satisfy some special conditions. Further research in this direction was done in [15,17,20], but still most cases remain unsolved. For some exploration of equation (1.1) on intervals see [6,11,13,16,18] and on half-lines see [4].
Let us note that (1.2) also can be considered as the characteristic equation of the recurrence relation (1.3) a N x m+N + . . . + a 1 x m+1 + a 0 x m = 0 which might be obtained by choosing x 0 ∈ I arbitrarily and putting x m = g(x m−1 ) for all m ∈ N. Such an approach we will examine in the present paper. It can be easily observed that if a polynomial b M r M + . . . + b 1 r + b 0 divides a polynomial a N r N + . . . + a 1 r + a 0 and a function g satisfies the equation then it satisfies also (1.1); for a simple proof see [9]. The main objective of this paper is to give particular conditions under which a converse holds, i.e., we want to find some conditions guaranteeing that if a continuous function g satisfies (1.1), then there is a divisor of a N r N + . . . + a 1 r + a 0 such that g is a solution to the corresponding iterative equation of lower order. Of course, it is not possible in each case. For instance, we may consider the equation g 2 (x) − 2g(x) + x = 0 whose continuous solution (on the whole real line) is The paper is organized as follows. Section 2 contains basic properties of solutions to considered equations and preliminary information on linear homogenous recurrence relations. In Section 3 we prove theorems on eliminating non-real roots from the characteristic equation which show that, in new crucial cases, equation (1.1) is equivalent to an equation of lower order. In particular, we generalize results from [20] and give an answer to the question posed in the last section of that paper. In Section 4 we obtain similar results as in the previous section, but we eliminate real roots of opposite sign. These theorems allow us to solve, in Section 5, a problem of Zoltán Boros. Namely, for a given integer n ≥ 3 we determine all continuous self-mapping functions f acting on an interval and satisfying (1.5) f n (x) = [f (x)] n x n−1 . For n = 2 the equation above can be solved using the mentioned result of Nabeya or [10]. In the last section we propose some further problems.

Preliminaries
As we mentioned before we assume that a 0 = 0. Proof. Choose x, y ∈ I and suppose that g(x) = g(y). Then The foregoing lemma directly implies that every continuous solution to (1.1) is strictly monotone. It means that the sequence (x m ) m∈N 0 given by x 0 ∈ I and x m = g(x m−1 ) for all m ∈ N is either monotone (in the case of increasing g) or anti-monotone (in the case of decreasing g). By anti-monotone we mean that the expression (−1) m (x m − x m−1 ) does not change its sign when m runs through N 0 .
In the case where g is bijective h = g −1 is also a self-mapping function acting on I. Putting g −N (x) in place of x in (1.1) we obtain the equation which is called the dual equation. It is worth mentioning that if r 1 , . . . , r N are roots of (1.2), then the roots of the characteristic equation of (2.1) are r −1 1 , . . . , r −1 N . Whence if a bijective function g satisfies (1.1) and its inverse satisfies an iterative equation of a lower order, then the order of (1.1) can be reduced. Moreover, a root from the characteristic equation can be eliminated if and only if its inverse can be eliminated from the characteristic equation of the dual equation. For the theory of linear recurrence relations see, for instance, [5, §3.2]. We will recall only the most significant theorem in this matter. In order to do this and simplify the writing we introduce the following notation. For a given polynomial c K r K + . . . + c 1 r + c 0 we denote by R(c K , . . . , c 0 ) the set {(r 1 , k 1 ), . . . , (r p , k p )} of all pairs of pairwise distinct roots r 1 , . . . , r p and their multiplicities k 1 , . . . , k p , respectively. Here and throughout the paper by a polynomial we mean a polynomial with real coefficients. Note that in the introduced notation k 1 + . . . + k p equals the degree of c K r K + . . . + c 1 r + c 0 . Moreover, (µ, k), (µ, k) ∈ R(c K , . . . , c 0 ) forces µ to be non-real.

3) if and only if it is given by
where A k is a polynomial whose degree equals at most l k − 1 for k = 1, . . . , p and B j , C j are polynomials whose degrees equal at most k j − 1, with φ j being an argument of µ j , for j = 1, . . . , q.

Eliminating non-real roots
We will make use of the following lemma whose proof can be found in [14].
Lemma 3.1. Let the sequence (x m ) m∈N 0 be given by x m = N n=1 (a n cos mφ n + b n sin mφ n ) with a n , b n ∈ R and φ n ∈ (0, 2π) for n = 1, . . . , N. If lim inf m→∞ x m ≥ 0, then x m = 0 for all m ∈ N 0 . Lemma 3.2. Suppose λ 1 , . . . , λ N to be complex numbers such that λ n = r(cos φ n + i sin φ n ) with r > 0 and φ n ∈ (0, π) for n = 1, . . . , N. Morever, let F : N 0 → R be a function such that lim m→∞ F (m) r m = 0 and let A n , B n be polynomials for n = 1, . . . , N. If the sequence (x m ) m∈N 0 given by Proof. Assume that k is the greatest number from degrees of the polynomials A 1 , . . . , A N and B 1 , . . . , B N . We may write A n (m) = k 's are possibly zeros. We first consider the case where (x m ) m∈N 0 is monotone. By considering the sequence (−x m ) m∈N 0 if necessary, we may asssume that x m ≥ 0 for all but finitely many m ∈ N 0 . It means that the inequality holds for all but finitely many m ∈ N 0 . Consequently, we obtain In view of Lemma 3.1 It means that we can eliminate terms with m k from N n=1 (A n (m) cos mφ n +B n (m) sin mφ n ) and now we can assume that the greatest number from degrees of the polynomials A n 's and B n 's is equal to l ≤ k − 1. The same procedure shows that the terms with m l also can be eliminated. Continuing in this fashion, we obtain Secondly, we consider the case where (x m ) m∈N 0 is anti-monotone. By considering the sequence (−x m ) m∈N 0 if necessary, we may asssume that (−1) m (x m − x m−1 ) ≥ 0 for all m ∈ N. It implies that the inequality holds for all m ∈ N and, as previously, Since we have cos(2m − 1)φ n = cos φ n cos 2mφ n + sin φ n sin 2mφ n and sin(2m − 1)φ n = cos φ n sin 2mφ n − sin φ n cos 2mφ n , limit (3.2) is of the form as in Lemma 3.1. Therefore, and, consequently, the equality   l 1 ) . . . , (λ p , l p ), (µ 1 , k 1 ), (µ 1 , k 1 ), . . . , (µ q , k q ), (µ q , k q )}.    where C is a constant. Since a constant has no effect on the monotonicity or the antimonotonicity of a sequence, we can apply Lemma 3.2 and argue as before. As an immediate corollary from the theorems above we obtain a result which for I = R was also proven in [15].

Eliminating roots of opposite sign
Below we are going to prove that the order of polynomial-like equation (1.1) also can be lowered when the minimal and the maximal (with respect to the absolute value) root of its characteristic equation are real and of opposite sign. However, it is necessary to distinguish two cases-when the unknown function monotonically increases and when it monotonically decreases. Proof. Fix x ∈ R and define a sequence (x m ) m∈N 0 by (3.4 for all m ∈ N 0 , where F represents the part of the solution to (1.3) for which the roots r 1 , λ 1 , . . . , λ p are responsible and A is a polynomial of degree at most k 2 − 1. Since where a k 2 −1 stands for the coefficient at m k 2 −1 in the polynomial A, the monotonicity of (x m ) m∈N 0 forces a k 2 −1 = 0. Continuing in this fashion, we eliminate all non-zero terms from A and obtain A ≡ 0. It means that x m = F (m) for all m ∈ N 0 and, again by Theorem 2.3, it safisfies the relation b M x m+M + . . . + b 1 x m+1 + b 0 x m = 0, which ends the proof. where j is the index of the negative number from r 1 and r 2 , in the case where where j is the index of the negative number from r 1 and r 2 , in the case where the positive root from r 1 and r 2 equals 1.
Proof. Fix x ∈ R and define a sequence (x m ) m∈N 0 by (3.4). By Lemma 2.2 this sequence satisfies equation (1.3). Firstly, we consider the case where r 1 = 1 = r 2 . By replacing equation (1.1) with the dual equation if necessary, we may assume that r 2 > 0. Since g is decreasing, (x m ) m∈N 0 is anti-monotone and of the form where F represents the part of the solution to (1.3) for which the roots r 1 , λ 1 , . . . , λ p are responsible and A is a polynomial of degree at most k 2 − 1. If it were A ≡ 0, we would have which is a contradiction to the anti-monotonicity of (x m ) m∈N 0 . Consequently, x m = F (m) for all m ∈ N 0 and, similarly as in the previous proof, (1.4) is satisfied with b j 's as in assertion (i). Secondly, we consider the case where either r 1 = 1 or r 2 = 1. By replacing equation (1.1) with the dual equation if necessary, we may assume that r 2 = 1. Again (x m ) m∈N 0 is anti-monotone and of the form where F represents the part of the solution to (1.3) for which the roots r 1 , λ 1 , . . . , λ p are responsible and A is a polynomial of degree at most k 2 − 1. If the polynomial A(m + 1) − A(m) were not constant, then denoting by k its degree, we would have where a k stands for the coefficient at m k in A(m + 1) − A(m), which is a contradiction to the anti-monotonicity of (x m ) m∈N 0 . Consequently, A is constant and (1.4) holds with b j 's as in assertion (ii).
Reasoning in the same manner as in the proof of Theorem 3.6 and making use of the proof above, one can show that a single root 1 does not prevent the elimination of other roots. More precisely, we have the following result.  We finish this part of the present paper by observing that in each of the mentioned cases the order of equation (1.1) can be esentially lowered. The key property allowing us to use one of the results above is the monotonicity that is implied by the continuity of solutions to considered equation.

A Zoltán Boros' problem
In [3] the authors solved the original problem posed by Zoltán Boros (see [2]) during the 50th International Symposium on Functional Equations (2012). Namely, we determined all continuous solutions to equation (1.5) for n = 3. Let us recall this result.  Conversely, if I ⊂ R is an interval and g : I → I is a solution to (5.2), then the formula f = exp •g • log defines a function acting from exp I into itself such that (1.5) holds for every x ∈ exp I. it follows that 1 is a root of equation (5.3) of multiplicity 2. We will need two lemmas on the behaviour of the roots of this equation. (ii) If n is odd, then the function f has a local maximum at the point x = −1 and a local minimum at the point x = 1. Hence there exists a unique point r 0 = 1 such that f (r 0 ) = 0. Clearly, r 0 < −1. Since f ′ (r 0 ) = 0, we conclude that r 0 is a single root of equation (5.3).