Complementary means with respect to a nonsymmetric invariant mean

It is known that if a bivariate mean K is symmetric, continuous and strictly increasing in each variable, then for every mean M there is a unique mean N\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$N\,$$\end{document} such that K is invariant with respect to the mean-type mapping M,N,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\left( M,N\right) ,$$\end{document}which means 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}$$K\circ \left( M,N\right) =K$$\end{document} and N is called a K-complementary mean for M (Matkowski in Aequ Math 57(1):87–107, 1999). This paper extends this result for a large class of nonsymmetric means. As an application, the limits of the sequences of iterates of the related mean-type mappings are determined, which allows us to find all continuous solutions of some functional equations.


Introduction
Let K, M and N be bivariate means in an interval I. The mean K is called invariant with respect to the mean-type mapping (M, N ) : I 2 → I 2 , briefly (M, N )-invariant, if K • (M, N ) = K (cf. [3]). This K is sometimes called the Gauss composition of the means M and N (cf. [2]), and N is referred to as a complementary mean to M with respect to K (briefly, K-complementary mean to M ) ( [3], see also [6]).
It is known that, under general conditions, the invariance relation guarantees the convergence of the sequence ((M, N ) n : n ∈ N) of iterates of the mean-type mapping (M, N ) to the mean-type (K, K). This fact has important applications in effectively solving some functional equations.
In [3] it was shown that if the mean K is symmetric, continuous and strictly increasing in each variable, then for every mean M there is a unique K-complementary mean.
The main results of this paper, Theorems 1 and 2, omitting the symmetry condition, give essential generalizations of this result. In particular, Theorem 1, where the symmetry of K is replaced by the implication "x < y =⇒ K (x, y) ≥ K (y, x) for all x, y ∈ I", says that for every bivariate mean M in I there is a unique mean M [K] such that K is M, M [K] -invariant. Moreover, M [K] is strict or continuous if so is M , and both M and M [K] are symmetric if so is K. Theorem 2 is a dual counterpart of Theorem 1. Here the symmetry of K is replaced by the implication "x < y =⇒ K (x, y) ≤ K (y, x) for all x, y ∈ I". For every mean M it provides a unique mean M [K] such that K is M [K] , M -invariant, and has similar additional properties as M [K] .
A simple example shows that the assumed implications in Theorems 1 and 2 are essential. Moreover it is shown (Remark 3 and Example 2) that in these results the assumption that K is strictly increasing cannot be replaced by the strictness of K.
Applying Theorems 1 and 2 we give conditions under which the sequences of mean-type mappings M, M [K] and M [K] , M converge to (K, K) (Theorem 3 and Theorem 4). This permits, in particular, to determine effectively all continuous functions F : I 2 → R satisfying each of the functional equations x,y ∈ I.

Complementary means
In the sequel I ⊂ R stands for an interval.  [3]). In the case when K is unique, it is sometimes called the Gauss composition of the means M and N (cf. [2]). If K is a unique (M, N )-invariant mean we say that N is a complementary mean to M with respect to K (briefly, N is K-complementary to M ).

Vol. 96 (2022)
Complementary means with respect to a nonsymmetric 45 Theorem 1. Let a continuous mean K : I 2 → I be increasing in the first variable and strictly increasing in the second one. Suppose that K satisfies the following condition: Then (i) for every mean M : Proof. Take an arbitrary mean M : I 2 → I and x, y ∈ I. We have Since, by the reflexivity of every mean, K (x, x) = x, the strict monotonicity of K in the second variable implies that M [K] (x, y) := z = x is the only solution of this equation.
Assume that x < y.
In this case we have Of course, ϕ is continuous. For t = x, making use first of the inequality x ≤ M (x, y) and implication (1), and then from the inequality M (x, y) ≤ y and the monotonicity of K in the second variable, we have For t = y, the inequality x ≤ M (x, y) and the monotonicity of K in the first variable give By the strict monotonicity of K in the second variable, such a z is unique.
Since in the case y < x we can argue similarly, the proof of (i) is complete.
(ii) This is a trivial consequence of (2). (iii) Assume, on the contrary, that M [K] is not strict, so there are x 0 , y 0 ∈ I, In the first case, from (2), we would have and, from (1), . But this is a contradiction as K is strictly increasing in the first variable and x 0 < M (x 0 , y 0 ) < y 0 . Since in the case when M [K] (x 0 , y 0 ) = y 0 we can argue similarly, this completes the proof of (iii).
(iv) We omit the simple argument of this part. To show that assumption (1) is essential, consider the following Example 1. Let I = R and K : R 2 → R be a weighted arithmetic mean, i.e. K (x, y) = ax + (1 − a) y for some fixed a ∈ (0, 1), and let M : R 2 → R be a mean. A simple calculation shows that a function N : R 2 → R satisfies (2) if and only if Vol. 96 (2022) Complementary means with respect to a nonsymmetric 47 Assume that for every weighted arithmetic mean M (x, y) = bx + (1 − b) y, b ∈ [0, 1], the function N is a mean in R, that is that is a mean in R. Clearly, N is a mean for every b ∈ [0, 1] if, and only if, We can also argue as follows. The function N given by (3) is a mean for every mean M if and only if or, equivalently, if and only if, for every mean M , and for all x, y ∈ R, Taking here first M = min and then M = max we obtain, respectively, the inequalities ax + (1 − a) y ≤ a min (x, y) + (1 − a) max (x, y) , x,y ∈ R, It is easy to see that each of these inequalities holds, if and only if a ≤ 1 2 . Note that with K (x, y) = ax + (1 − a) y, the inequality a ≤ 1 2 is equivalent to implication (1).
If K satisfies the "opposite" condition to (1), interchanging the roles of variables in K and applying Theorem 1, we obtain the following Theorem 2. Let a continuous mean K : I 2 → I be strictly increasing in the first variable and increasing in the second one. Suppose that K satisfies the following condition:

Then
(i) for every mean M : I 2 → I there is a unique mean M [K] : x,y ∈ I;

Some applications
Applying Theorem 1 and the main result of [5] (see also [4]) we obtain the following Using this result we prove the following if and only if there is a single variable continuous function ϕ : y)) , x,y ∈ I.
(ii) Suppose that K satisfies the condition  y)) , x,y ∈ I.
Proof. Assume first that F : I 2 → R is continuous on the diagonal Δ I 2 and satisfies equation (4), that is Hence, by induction, Since (K (x, y) , K (x, y)) belongs to the diagonal Δ I 2 for every (x, y) ∈ I 2 , by (5) and the continuity of F on Δ I 2 implies that for every (x, y) ∈ I 2 , Setting ϕ (t) := F (t, t) , t ∈ I , we conclude that F (x, y) = ϕ (K (x, y)) for all (x, y) ∈ I 2 .
To prove the converse implication, take an arbitrary function ϕ : I → R and put F := ϕ • K. Then, for all x, y ∈ I, making use of the K-invariance with respect to M, M [K] , we have which completes the proof of (i).
We omit the similar proof of (ii).
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