Invariant means and iterates of mean-type mappings

Classical result states that for two continuous and strict means $M,\,N \colon I^2 \to I$ ($I$ is an interval) there exists a unique $(M,N)$-invariant mean $K \colon I^2 \to I$, i.e. such a mean that $K \circ (M,N)=K$ and, moreover, the sequence of iterates $((M,N)^n)_{n=1}^\infty$ converge to $(K,K)$ pointwise. Recently it was proved that continuity assumption cannot be omitted in general. We show that if $K$ is a unique $(M,N)$-invariant mean then, under no continuity assumption, $(M,N)^n \to (K,K)$.


Introduction
It is known that if M and N are continuous bivariable means in an interval I, and the mean-type mapping (M, N ) is diagonally contractive, that is, (1) |M (x, y) − N (x, y)| < |x − y| , x, y ∈ I, x = y, then there is a unique mean K : I 2 → I that is (M, N )-invariant and, moreover the sequence of iterates ((M, N ) n ) n∈N of the mean-type mapping (M, N ) converges to (K, K) ( [5]).
For the results of this type, with more restrictive assumptions (see [1]). In particular, instead of (1) it was assumed that both means are strict; in [2] it was assumed that at most one mean is not strict, and condition (1) appeared first in [4] (see also [7]). Moreover, in all these papers the uniqueness of the invariant mean was obtained under the condition that it is continuous. In [5] it was shown that this condition is superfluous. In [6] it is shown that condition (1) holds if M is right-strict in both variables and N is left-strict in both variables (or vice versa), that is essentionally weaker condition than the strictness of the means.
In section 3 we give conditions under which the uniqueness of the invariant mean guarantee the relevant convergence of the sequence of iterates of the meantype mapping (see Theorem 3.3). If K, M, N : I 2 → I are arbitrary means and K is a unique (M, N )-invariant mean, then the sequence of iterates ((M, N ) n ) n∈N of the mean-type mapping (M, N ) converges to (K, K) pointwise in I 2 .) In section 4 we improve the above result, namely, we show that the result remains true on replacing the diagonal-contractivity of mean-type mapping, by demanding only that, for every point outside some iterate of the mean-type mapping is diagonally contractive, more precisely, that for every point (x, y) ∈ I 2 , x = y, there is a positive integer n = n (x, y) such that Hereafter, whenever a pair (M, N ) is stated, the sequence (M n , N n ) defined in this way is also given.

Preliminaries
Let I ⊂ R be an interval. Recall that a function M : x, y ∈ I. The mean M is called: strict, if these inequalities are sharp unless x = y; symmetric if M (x, y) = M (y, x) for all x, y ∈ I.
Given means K, M, N : I 2 → I in an interval I. The mean K is called invariant with respect to the mean-type mapping (M, N ) :  The following result is a consequence of bivariable version of Corollary 1 in [5] (see also [1], [2], [4]).  To formulate some simple sufficient conditions guarantying (2) let us introduce the following notions: right-strict in the first variable at the point x 0 , if for all t ∈ I, left-strict in the second variable at the point y 0 , if for all t ∈ I, right-strict in the second variable at the point y 0 , if for all t ∈ I, We adapt the convention that if a mean is left-strict (right-strict) in both variables then we call it simply left-strict (right-strict). Then

, is left and right-strict in the first variable, but it is neither left nor right strict at any point in the second variable.
The mean-type mapping (P 1 , P 2 ) coincides with the identity of R 2 , and of course, the sequence of its iterates ((P 1 , P 2 ) n : n ∈ N) converges to (P 1 , P 2 ). Note that here P 1 = P 2 , moreover, every mean M : Example 2.6. Consider the extreme mean min : R 2 → R . Since, t < x 0 =⇒ min (x 0 , t) < x 0 for every t; the mean min is left-strict in the first variable and in the second variable, but it is not right-strict in the first and the second variable. Moreover, the extreme mean max : R 2 → R is right-strict in the first and the second variable, but it is not left-strict in the first and the second variable.
For every n ∈ N, the nth iterate (min, max) n of mean-type mapping (min, max) has the form and, of course, the sequence of iterates of ((min, max) n : n ∈ N) converges. On the other hand, one can check that a mean is invariant with respect to the mean-type mapping (min, max) if and only if it is symmetric.

strict mean then it is left and right-strict in each of the two variables.
Using the definition of one-side strict means we get the following result Proof. Suppose to the contrary that there exists x, y ∈ I, x = y such that Binding this inequality with min(x, y) ≤ M (x, y) ≤ max(x, y) and min(x, y) ≤ N (x, y) ≤ max(x, y) we obtain that M (x, y) and N (x, y) are two different endpoints of the interval [min(x, y), max(x, y)]. Thus {M (x, y), N (x, y)} = {min(x, y), max(x, y)}.
On the other hand, if M and N are both left-strict then none of them can be equal to max(x, y). Similarly, if M and N are both right-strict then none of them can be equal to min(x, y). These contradictions end the proof.
2.1. Weakly contractive mean-type mappings. In this section we deal with some relaxation of the assumption (1). For two means M, N : I 2 → I, the meantype mapping (M, N ) is weakly contractive if for every elements x, y ∈ I with x = y there is a positive integer n = n (x, y) such that Next lemma provide a complete characterization of weak-contractivity in terms of (M 2 , N 2 ). Proof. We prove that if the inequality (3) is satisfied for some triple (x, y, n) then it is also valid for (x, y, 2). As M and N are means we get In the first case we obtain validity of (3) for the triple (x, y, 1) which, by mean property, implies its validity for (x, y, 2).
In the second case we get, by mean property, {M (x, y), N (x, y)} = {x, y}. If (M, N )(x, y) = (x, y) then (M, N ) n (x, y) = (x, y) for all n ∈ N contradicting the assumption. Thus we obtain M (x, y) = y and N (x, y) = x. Applying the same argumentation to the pair (y, x) we get that either However these equalities cannot be simultaneously valid as they are excluded by consecutive alternatives in (ii), respectively. The case x > y is completely analogous.  By [7] we obtain that L(x, y) := lim inf n→∞ a n , U (x, y) := lim sup n→∞ a n are the smallest and the biggest (M, N )-invariant means, respectively. It is necessarily and also sufficient to prove that L(x, y) = U (x, y) if and only if (x, y) ∈ A M,N .

Uniqueness of invariant means and convergence of iterates of mean-type mappings
If L(x, y) = U (x, y) then we obtain that the sequence a converges, whence (M n (x, y)) n∈N and (N n (x, y)) n∈N converge to a common limit, i.e. (x, y) ∈ A M,N .
By [7] we obtain that L(x, y) := lim inf n→∞ a n , U (x, y) := lim sup n→∞ a n are the smallest and the biggest (M, N )-invariant means, respectively. (⇒) As K is a unique (M, N )-invariant mean we get K = L = U . It implies that the sequence (a n ) is convergent to K(x, y). Therefore both M n and N n converge to K pointwise on I 2 . In particular its Cartesian product (M, N ) n = (M n , N n ) converges to (K, K) pointwise on the same set.
(⇐) We know that the sequence (a n ) is convergent to K(x, y), in particular K(x, y) = L(x, y) = U (x, y). As x, y were taken arbitrarily we have K = L = U . It implies that the smallest and the biggest (M, N )-invariant mean coincide providing the uniqueness. Proof. In view of Theorem 3.3 we get that (M 2 , N 2 ) is a contractive mean-type mapping, which is also continuous. Applying the result from [5] (see the beginning of this paper) we obtain that there exists a unique (M 2 , N 2 )-invariant mean. As every (M, N )-invariant mean is also (M 2 , N 2 )-invariant we get that there exists at most one (M, N )-invariant mean.

Mean-type mappings with diagonally-contractive iterates and invariant means
On the other hand, applying some general results from [7], we have that (under no assumptions for means) there exists at least one (M, N )-invariant mean. Therefore there exists a unique (M, N ) invariant mean; say K : I 2 → I.
Thus the sequence (min (M n (x, y) , N n (x, y)) : n ∈ N) is increasing and the sequence (max (M n (x, y) , N n (x, y)) : n ∈ N) is decreasing; and both are convergent. Clearly, if x = y or M k (x, y) = N k (x, y) for some positive integer k, then M n (x, y) = N n (x, y) for all n ≥ k, and consequently we have Assume that lim n→∞ min (M n (x, y) , N n (x, y)) < lim n→∞ max (M n (x, y) , N n (x, y)) .
In particular, we have M n (x, y) = N n (x, y) for every n ∈ N. By the assumption, for every n ∈ N there is k = k n ∈ N such that |M kn (M n (x, y) , N n (x, y)) − N kn (M n (x, y) , N n (x, y)) | < c |M n (x, y) − N n (x, y)| that is that |M n+kn (x, y) − N n+kn (x, y)| < c |M n (x, y) − N n (x, y)| , which leads to a contradiction. Thus Finally, let K be a pointwise limit of (M n ) or (N n ), as they are equal.