Invariant means and iterates of mean-type mappings

A classical result states that for two continuous, strict means M,N:I2→I\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$M,\,N :I^2 \rightarrow I$$\end{document} (I is an interval) there exists a unique (M, N)-invariant mean K:I2→I\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K :I^2 \rightarrow I$$\end{document}, i.e. such a mean that K∘(M,N)=K\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K \circ (M,N)=K$$\end{document} and, moreover, the sequence of iterates ((M,N)n)n=1∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$((M,N)^n)_{n=1}^\infty $$\end{document} 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, without continuity assumption, (M,N)n→(K,K)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(M,N)^n \rightarrow (K,K)$$\end{document}.


Introduction
It is known that if M and N are continuous bivariate means in an interval I, and the mean-type mapping (M, N ) is diagonally contractive, that is, x ,y∈ I, x = y, 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 [6]). Moreover, in all these papers the uniqueness of the invariant mean was obtained under the condition that it is continuous. In [5,Theorem 4.5] it was shown that this condition can be relaxed. In Proposition 2.8 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 essentially weaker condition than the strictness of the means.
In Sect. 3 we give conditions under which the uniqueness of the invariant mean guarantee the relevant convergence of the sequence of iterates of the mean-type 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) In Sect. 4 we deliver certain conditions which guarantee uniqueness of invariant means.

Preliminaries
Let I ⊂ R be an interval. Recall that a function M : The mean M is called strict, if these inequalities are sharp unless x = y; For two means M, N : I 2 → I, a mean K : I 2 → I 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 (1) let us introduce the following notions: • left-strict in the first variable at the point x 0 , if for all t ∈ I, • 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, t < y 0 =⇒ M (t, y 0 ) < y 0 ; • 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 Example 2.5. The projective mean P 1 : R 2 (x, y) → x, is left and right-strict in the second variable, but it is neither left nor right strict at any point in the first variable. Similarly, the projective mean P 2 : R 2 (x, y) → y, 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 K : 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. Remark 2.7. If M : I 2 → R is a 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 Consequently, either Assume that x < y (which can be done without any loss of generality), we have In the first case, putting x 0 = x, t := y we have x 0 < t and M (x 0 , t) = x 0 so, in view of Definition 2.4, M is not right-strict in the first variable at the point x 0 . Moreover, putting t := x, y 0 := y we have t < y 0 and N (t, y 0 ) = y 0 , so N is not left-strict in the second variable at y 0 .
We omit similar considerations in the second case.

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 mean-type 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 inequality (2) 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 (2) 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 which implies (2) for the triple (x, y, 2) or However in the second subcase we get which implies that (2) is valid for no n ∈ N contradicting the assumption. As the converse implication is trivial, the proof is complete The only remaining case is that (M, N )(x, y) = (y, x) and (M, N )(y, x) = (x, y). Equivalently, However these equalities cannot be simultaneously valid as they are excluded by consecutive alternatives in (iii), respectively.

Uniqueness of invariant means and convergence of iterates of mean-type mappings
Invariant means and iterates of mean-type mappings 411 Proof. For x, y ∈ I. Let a be a sequence which is a shuffling of M n and N n , i.e a = (x, y, M 1 (x, y), N 1 (x, y), M 2 (x, y), N 2 (x, y), M 3 (x, y), N 3 (x, y), . . . ) .
By [6] 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 meantype 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 [6], we have that (without any assumptions for the means M and N ) 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 Finally, we can define K as a (common) pointwise limit of sequences (M n ) and (N n ), as they are equal.
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