Means of iterates

We determine continuous bijections $f$, acting on a real interval into itself, whose $k$-fold iterate is the quasi-arithmetic mean of all its subsequent iterates from $f^0$ up to $f^n$ (where $0\le k\le n$). Namely, we prove that if at most one of the numbers $k,n$ is odd, then such functions consist of at most three affine pieces.


Introduction
Consider integers n 2, 0 k n and a non-trivial interval I ⊂ R. The following problem arises when studying polynomial-like iterative equations, means of functions or convergence of means of iterates: Determine all continuous functions F : I → I with the k-fold iterate F k being a mean of all its subsequent iterates up to F n ; more precisely, find all self-mappings F ∈ C(I) such that for every x ∈ I, where M : I n+1 → I is a mean and C(I) stands for the family of all continuous real functions acting on I.
In this paper we concentrate on equation (1) in the case where M is a quasiarithmetic mean, i.e., with M being of the form (2) M (x 0 , ..., x n ) = ϕ −1 1 n + 1 n i=0 ϕ(x i ) for all x 0 , . . . , x n ∈ I, where ϕ is a continuous bijection form I onto an interval J ⊂ R. Observe that it is enough to consider equation (1) with M being the arithmetic mean which follows from the following remark. It is easy to prove that any self-mapping satisfying (3) is injective (see, e.g. [2, Lemma 2.1]). Therefore, any continuous surjective solution to equation (3) is automatically continuous bijection; in particular it is strictly monotone.
The paper is organized as follows. In Section 2 we collect basic properties of polynomial-like iterative equations and preliminary information on linear homogeneous recurrence relations. In Section 3 we solve equation (3) in the cases where k = 0 or k = n. The case where 0 < k < n is treated in Section 4. The final section contains examples and remarks on the considered equation as well as some further problems.

Polynomial-like iterative equations and recurrence relations
Equation (3) is a very special case of an iterative equation belonging to an interesting and widely studied class of functional equations, called polynomial-like iterative equations, of the form where f k stands for the k-fold iterate of a self-mapping unknown function f (defined on an interval I ⊂ R) and a 0 , a 1 , . . . , a n are given real coefficients. It turns out that continuous solutions to (4) deeply depend on the roots of its characteristic equation which is obtained by assuming that f has the form f (x) = rx.
In general, it is very difficult to find all continuous functions satisfying equation (4). These difficulties follow from the non-linearity of the operator f → f n . Up to now the complete solution is known only in the case where n = 2 and I = R [11]. The problem still remains open even for n = 3 [7]. A partial solution in the case where n = 3 and I = R was obtained in [15] and some results in the case where n 3 and I = R can be found in [3,5,10,13,14,16].
It is known [9] that if a polynomial m i=0 b i r i divides a polynomial n i=0 a i r i and a function f satisfies m i=0 b i f i (x) = 0 then it satisfies also (4). One of methods for finding solutions to (4) is based on a reverse reasoning, i.e., if a continuous function f satisfies equation (4), then we want to find a divisor of the polynomial n i=0 a i r i such that f is a solution to the corresponding equation of lower 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 obtained in [1,2,12,17] and some of them will be crucial tools in this paper.
Equation (5) can be considered as the characteristic equation of the recurrence relation (6) n i=0 a i x m+i = 0 which might be obtained by choosing x 0 ∈ I arbitrarily and putting x m = f (x m−1 ) for every m ∈ N. It is easy to see that if a 0 = 0 and a function f satisfies (4), then it is injective (see, e.g., [2, Lemma 2.1.]). This observation implies that every continuous solution to (4) is strictly monotone. It means that the sequence (x m ) m∈N0 given by x 0 ∈ I and x m = f (x m−1 ) for m ∈ N is either monotone (in the case of increasing f ) or anti-monotone (in the case of decreasing f ). By antimonotone 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 f is bijective we can consider the dual equation Putting f −n (x) in place of x we see that f satisfies (4) if and only if f −1 satisfies (7). We can also extend the sequence (x m ) m∈N0 to the whole Z by setting x −m = f −1 (x −m+1 ) for m ∈ N. Then relation (6) is satisfied for m ∈ Z.
For the theory of linear recurrence relations we refer the reader, for instance, to [4, §3.2]. We shall 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 n r n + . . . + c 1 r + c 0 we denote by R(c n , . . . , c 0 ) the collection {(r 1 , k 1 ), . . . , (r p , k p )} of all pairs of its pairwise distinct (complex) 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.
Theorem 2.1. Assume that R(a n , . . . , a 0 ) = {(λ 1 , l 1 ), . . . , (λ p , l p ), Then a real-valued sequence (x j ) j∈N0 satisfies (6) 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 k , C k are polynomials whose degrees equal at most m k − 1, with φ k being an argument of µ k , for k = 1, . . . , q.

Cases k = 0 and k = n
In the case where k = 0 equation (3) takes the form Its characteristic equation is of the form which, after multiplication by r − 1, can be written as r n+1 − (n + 1)r + n = 0. Therefore, by the previous remarks, if a function f satisfies (8), then it also satisfies the equation f n+1 (x) − (n + 1)f (x) + nx = 0. Now, applying [2, Theorem 5.7], we obtain the following result.
Theorem 3.1. If f ∈ C(I) satisfies (8), then: (i) f (x) = x in the case where n is an odd number or in the case where n is an even number and I = R; where c is a constant and r 0 stands for the negative root of equation (9), in the case where n is an even number and I = R.
Now we shall consider the case where k = n. Then equation (3) takes the form and its characteristic equation is of the form We will need the following lemma which is an elaboration of [11,Theorem 9]; the crucial point is that an unknown function acts on a subinterval of the real line.
Proof. (i) In the case where f is increasing we can apply [2, Theorem 4.1 and Remark 4.4] to conclude that f (x) = x. Therefore, assume f is decreasing.
Fix x ∈ I and put x 0 = x and It is easy to see that the sequence (x m ) m∈N0 satisfies the relation By Theorem 2.1 we have x m = A + Bρ m for some constants A and B (depending on x). Consequently, there exists a finite limit lim m→∞ f m (x). Fix x, y ∈ I. Since f is decreasing, the sequence (f m (x) − f m (y)) m∈N0 is antimonotone. Moreover, it is convergent, so it must converge to zero. It means that the limit lim m→∞ f m (x) does not depend on x.
We can rewrite equation (12) as and, by a simple induction, we obtain Passing with m to ∞ gives Setting c = (1 − ρ) lim m→∞ f m (x) ends the proof of assertion (i).
(ii) It is enough to apply assertion (i) to the dual equation to (12).
3. If f ∈ C(I) satisfies (10), then: where c is a constant and r 0 stands for the negative root of equation (11), in the case where n is an even number.
In the case of odd n, g has the only extremum at 1, so it is the only real root of g. In the case of even n, g has extrema at 0 and 1, so it also has a negative root r 0 . Since g(0)g(−1) < 0, we have r 0 > −1 and since g ′ (r 0 ) > 0, it follows that r 0 is a single root.

Case 0 < k < n
In order to determine continuous solutions to equation (3) in the case where 0 < k < n we need information on roots of its characteristic equation which is of the form (14) (n + 1)r k = n i=0 r i .
We start with a general observation on complex roots of equation (14).
Lemma 4.1. All complex roots of equation (14) are in modulus less than 2n + 1.
This completes the proof.
Moreover, the set of all real roots of equation (14) coincides with the set of all solutions to the equation and G ′ (r) = (n + 1)r k−1 g(r). Note also that in the case where n − k is odd the function g is strictly convex with the global minimum at the point and, by (15), we have (17) g(r min ) < 0 if and only if r min = 1.
However, in the case where n − k is even the function g| (0,+∞) is strictly convex with the global minimum at the point r min and (17) holds, whereas the function g| (−∞,0) is strictly concave with the global maximum at the point r max = −r min and, by (15), we have Described properties of functions g and G will play a key role in the next lemma which will allow us to locate real roots of equation (14).
Moreover, in all the above cases r 1 , r 2 , r 3 , r 4 are single roots, except the case n = 2k in which r 1 is a double root.
(i) The assumption n = 2k implies r min = 1. It yields g(r) > g(r min ) = 0 for every r = r min . Consequently, the function G is strictly increasing, which jointly with (15) shows that r min is the unique real root of equation (14).
(ii) The assumption n < 2k implies r min > 1. We conclude that there exists exactly one t 0 ∈ R\{1} such that g(t 0 ) = 0 and, moreover, t 0 > r min . Consequently, the function G has exactly one local maximum at the point r 1 = 1 and exactly one local minimum at the point t 0 . This jointly with (15) shows that equation (14) has two real roots: r 1 = 1 and r 2 ∈ (t 0 , ∞). By Lemma 4.1 we have r 2 < 2n + 1.
(iii) The reasoning is similar as in (ii).
(v) The reasoning is similar as in (iv). Now observe that if k is even, then for every r ∈ (0, ∞) we have G ′ (r) > 0 ⇐⇒ g(r) > 0 and for every r ∈ (−∞, 0) we have The assumption n < 2k implies r min > 1. We conclude that there exists exactly one t 0 ∈ R\{1} such that g(t 0 ) = 0 and, moreover, t 0 > r min . Consequently, the function G has exactly one local maximum at the point r 1 = 1 and exactly two local minimums at points 0 and t 0 . This jointly with (15) shows that equation (14) has three real roots: r 1 = 1, r 2 ∈ (t 0 , ∞) and r 3 ∈ (−∞, 0). By Lemma 4.1 we have r 2 < 2n + 1 and sine in this case, we have r 3 > −1.
(vii) The reasoning is similar as in (vi).
(x) The reasoning is similar as in (ix). To prove the moreover part note that equality g ′ (r) = 0 can hold only for r ∈ {r min , r max }. Thus G ′′ (r i ) = (n + 1)r k−1 i g ′ (r i ) = 0 for every i ∈ {2, 3, 4} and for i = 1 in the case where n = 2k. If n = 2k, then G ′′ (r 1 ) = 0 and G ′′′ (r 1 ) = (2k + 1)(k + 1)k = 0. Now we want to obtain certain information on the location of non-real roots of equation (14). It will be more convenient for us to consider (16) instead of (14). Lemma 4.3. Assume that 0 < k < n and at least one of the numbers k and n is odd. Then for a non-real root z and a real root r 0 of equation (14) we have |z| = |r 0 |.
Proof. Let r 0 ∈ (0, ∞); in this case parity of k or n does not play any role. Suppose, towards a contradiction, |z| = r 0 . Then which is impossible. Now assume r 0 ∈ (−∞, 0) and as before, suppose that |z| = −r 0 . By Lemma 4.2 we need to consider only the case where n is odd.
The last lemma we need is basically Theorem 8 and Theorem 10(iii) from [11]; the only difference is that in our lemma the unknown function acts on a subinterval of the real line.
where a, b ∈ clI with a ≤ b and r 0 stands for the different from 1 positive root of equation (14), in the case where k is an odd number and n is an even number with n = 2k; (iii) f (x) = r 0 x + c, where r 0 stands for the negative root of equation (14), or f is of form (21), where a, b ∈ clI with a ≤ b and r 0 stands for the different from 1 positive root of equation (14), in the case where k is an even number and n is an odd number or both k and n are odd numbers.
Proof. The idea of the proof in each of the considered cases is the same and run in the following way. First we determine all real roots of equation (14)  Remark 4.6. In the case where I = R the surjectivity assumption in Theorem 4.5 is satisfied automatically. However, this assumption is not essential when considering the dual equation is not necessary.

Examples, remarks and problems
In this section we give some examples showing how our main results works. (ii) f (x) = x or f (x) = cx r0 , where c is a constant such that f (I) ⊂ I and r 0 stands for the negative root of equation (11), in the case where n is an even number.
Proof. It is enough to use Theorem 3.3 and Remark 1.1 with ϕ = exp.  Proof. It is enough to use Theorem 3.1 and Remark 1.1 with ϕ of the form ϕ(x) = x 1/p . The root r 0 has no influence for solutions to (22), because the interval {x 1/p : x ∈ I} cannot be equal to the whole real line. (ii) f (x) = x or f (x) = cx r0 , where c is a constant such that f (I) ⊂ I and r 0 stands for the negative root of equation (9), in the case where n is an even number and I = (0, ∞).
Proof. It is enough to use Theorem 3.1 and Remark 1.1 with ϕ = exp. Proof. Applying assertion (i) of Theorem 4.5 and Remark 1.1 with ϕ = log we conclude that (25) holds. Now we argue as in Corollary 5.6.
We finish this paper with two problems motivated by the main results.
Problem 5.8. Can we omit the sujectivity assumption on f in Theorem 4.5?
Problem 5.9. Determine all (bijections) f ∈ C(I) satisfying (3) in the case where both the numbers k and n are even.