Invariant means and iterates of mean-type mappings

TL;DR

This paper proves that for mean-type mappings with a unique invariant mean, the iterates always converge to that mean, even without assuming continuity, extending classical results.

Contribution

It generalizes the convergence of iterates of mean-type mappings to cases without continuity assumptions, given the uniqueness of the invariant mean.

Findings

Iterates of mean-type mappings converge to the invariant mean.

Uniqueness of the invariant mean guarantees convergence.

Convergence holds without continuity assumptions.

Abstract

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, $(M,N)^n \to (K,K)$.

Equations

$$\left|M\left(x,y\right)-N\left(x,y\right)\right|<\left|x-y\right|,x,y\in I,x\ne y,$$

$$\left|M_n\left(x,y\right)-N_n\left(x,y\right)\right|<\left|x-y\right|,$$

$$\min \left(x,y\right)\le M\left(x,y\right)\le \max \left(x,y\right)\text{, }x,y\in I\text{.}$$

$$\left|M\left(x,y\right)-N\left(x,y\right)\right|<\left|x-y\right|\text{, }x,y\in I\text{, }x\ne y\text{,}$$

$$t<x_0⟹M\left(x_0,t\right)<x_0;$$

$$x_0<t⟹x_0<M\left(x_0,t\right);$$

$$t<y_0⟹M\left(t,y_0\right)<y_0;$$

$$y_0<t⟹y_0<M\left(t,y_0\right).$$

$$\displaystyle M\text{ is left-strict if and only if }\>\>\big{(}x\neq y\Longrightarrow M(x,y)$$

$$\displaystyle M\text{ is right-strict if and only if }\>\>\big{(}x\neq y\Longrightarrow M(x,y)$$

$$\left(\min,\max\right)^{n}\left(x,y\right)=\left\{\begin{array}[]{ccc}\left(x,y\right)&if&x\leq y\\ \left(y,x\right)&if&x>y\end{array}\right.$$

$$|M(x,y)-N(x,y)|\ge |x-y|=|\max (x,y)-\min (x,y)|.$$

$$\min (x,y)\le M(x,y)\le \max (x,y)\text{ and }\min (x,y)\le N(x,y)\le \max (x,y)$$

$$\left|M_n\left(x,y\right)-N_n\left(x,y\right)\right|<\left|x-y\right|.$$

$$|M(x,y)-N(x,y)|\le |x-y|.$$

$$\text{either}|M(x,y)-N(x,y)|<|x-y|\text{ or }|M(x,y)-N(x,y)|=|x-y|.$$

$$|M_2(x,y)-N_2(x,y)|=|M(y,x)-N(y,x)|<|x-y|,$$

$$(M_2,N_2)(x,y)=(M,N)(y,x)=(x,y).$$

$$(M_n,N_n)(x,y)=\{\begin{array}{ll}(y,x)& \text{ if }n\text{ is odd,}\\ (x,y)& \text{ if }n\text{ is even,}\end{array}$$

$$\big{\{}(x,y),(y,x)\big{\}}\not\subset\big{\{}(M,N)^{n}(x,y)\colon n\in\mathbb{N}\big{\}}\>.$$

$$M(x,y)=y,N(x,y)=x,M(y,x)=x,\text{ and }N(x,y)=y.$$

$$\Big{\{}(x,\,y)\in I^{2}\colon\big{(}(M_{n},N_{n})(x,\,y)\big{)}_{n=1}^{\infty}\text{ converges to some point on the diagonal }\Big{\}}.$$

$$a=(x,y,M_1(x,y),N_1(x,y),M_2(x,y),N_2(x,y),M_3(x,y),N_3(x,y),\dots ).$$

$$L(x,y):=\underset{n\to \infty }{\mathrm{liminf}}{a}_n,U(x,y):=\underset{n\to \infty }{\mathrm{limsup}}{a}_n$$

$$\underset{n\to \infty }{\lim }{M}_n(x,y)=\underset{n\to \infty }{\lim }{N}_n(x,y),$$

$$a=(x,y,M_1(x,y),N_1(x,y),M_2(x,y),N_2(x,y),M_3(x,y),N_3(x,y),\dots ).$$

$$L(x,y):=\underset{n\to \infty }{\mathrm{liminf}}{a}_n,U(x,y):=\underset{n\to \infty }{\mathrm{limsup}}{a}_n$$

$$\left|M_n\left(x,y\right)-N_n\left(x,y\right)\right|<c\left|x-y\right|,$$

$$\min \left(x,y\right)\le M\left(x,y\right)\le \max \left(x,y\right)\text{, }\min \left(x,y\right)\le N\left(x,y\right)\le \max \left(x,y\right)\text{,}$$

$$\min \left(x,y\right)$$

$$\le M\left(M\left(x,y\right),N\left(x,y\right)\right)\le \max \left(M\left(x,y\right),N\left(x,y\right)\right)\le \max \left(x,y\right)\text{,}$$

Full text

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, $(M,N)^{n}\to(K,K)$.

1. Introduction

It is known that if $M$ and $N$ are continuous bivariable means in an interval $I,$ and the mean-type mapping $\left(M,N\right)$* is diagonally contractive, that is, *

[TABLE]

then there is a unique mean $K:I^{2}\rightarrow I$ that is $\left(M,N\right)$-invariant and, moreover the sequence of iterates $\left(\left(M,N\right)^{n}\right)_{n\in\mathbb{N}}$ of the mean-type mapping $\left(M,N\right)$ converges  to $\left(K,K\right)$ ([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 mean-type mapping (see Theorem 3.3). If $K,M,N:I^{2}\rightarrow I$ are arbitrary means and $K$ is a unique $\left(M,N\right)$-invariant mean, then the sequence of iterates $\left(\left(M,N\right)^{n}\right)_{n\in\mathbb{N}}$ of the mean-type mapping $\left(M,N\right)$ converges to $\left(K,K\right)$ 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 $\left(x,y\right)\in I^{2}\,,$ $x\neq y$, there is a positive integer $n=n\left(x,y\right)$ *such that *

[TABLE]

where $\left(M_{n},N_{n}\right):=\left(M,N\right)^{n}.$ Hereafter, whenever a pair $(M,N)$ is stated, the sequence $(M_{n},\,N_{n})$ defined in this way is also given.

2. Preliminaries

Let $I\subset\mathbb{R}$ be an interval. Recall that a function $M:I^{2}\rightarrow\mathbb{R}$ is called a mean in $I,$ if it is internal, i.e. if

[TABLE]

The mean $M$ is called: strict, if these inequalities are sharp unless $x=y$; symmetric if $M\left(x,y\right)=M\left(y,x\right)$ for all $x,y\in I$.

Given means $K,M,N:I^{2}\rightarrow I$ in an interval $I.$ The mean $K$ is called invariant with respect to the mean-type mapping $\left(M,N\right):I^{2}\rightarrow I^{2}$, briefly, $\left(M,N\right)$-invariant, if $K\circ\left(M,N\right)=K$ ([3]).

Remark 2.1**.**

If $M,N:I^{2}\rightarrow I$ are two means in an interval $I$ then, for all $n\in\mathbb{N}$, the mapping $\left(M_{n},N_{n}\right)$ is a mean-type mapping. Moreover every $(M,N)$-invariant mean is $(M_{n},N_{n})$-invariant for all $n\in\mathbb{N}$.

Remark 2.2**.**

([3]) If $K:I^{2}\rightarrow I$ is a continuous symmetric and strictly increasing mean, then for every mean $M:I^{2}\rightarrow I$ there exists a unique function $N:I^{2}\rightarrow I$ such that $K\circ\left(M,N\right)=K,$ and $N$ is a mean in $I$. Moreover, the mean $N$ is called complementary to $M$ with respect to $K$.

The following result is a consequence of bivariable version of Corollary 1 in [5] (see also [1], [2], [4]).

Theorem 2.3**.**

If $M,N:I^{2}\rightarrow I$ are continuous means such that the mean-type mapping $\left(M,N\right)$ is diagonally-contractive, i.e.

[TABLE]

then there is a unique mean $K:I^{2}\rightarrow I$ which is $\left(M,N\right)$-invariant and, moreover, the sequence of iterates $\left(\left(M,N\right)^{n}\right)_{n\in N}$ of the mean-type mapping $\left(M,N\right)$ converges (uniformly on compact subsets of $I^{2})$ to $\left(K,K\right)$.

To formulate some simple sufficient conditions guarantying (2) let us introduce the following notions:

Definition 2.4**.**

([6], [8]) Let $M:I^{2}\rightarrow\mathbb{R}$ be a mean in an interval $I\subset\mathbb{R}$ and let $x_{0},y_{0}\in I$. The mean $M:I^{2}\rightarrow\mathbb{R}$ is called:

left-strict in the first variable at the point $x_{0}$, if for all $t\in I$,

[TABLE]

right-strict in the first variable at the point $x_{0}$, if for all $t\in I$,

[TABLE]

left-strict in the second variable at the point $y_{0}$, if for all $t\in I$,

[TABLE]

right-strict in the second variable at the point $y_{0}$, if for all $t\in I$,

[TABLE]

The mean $M$ is called left-strict in the first variable in a set $A\subset I$, if it is left-strict in the first variable at every point $x_{0}\in A$; and $M$ is called left-strict in the first variable, if it is left-strict in the first variable at every point. (Analogously one can introduce the remaining notions).

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

[TABLE]

Example 2.5**.**

The projective mean $P_{1}\colon\mathbb{R}^{2}\ni(x,y)\mapsto x$, 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 $P_{2}\colon\mathbb{R}^{2}\ni(x,y)\mapsto y$, 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 $\left(P_{1},P_{2}\right)$ coincides with the identity of $\mathbb{R}^{2}$, and of course, the sequence of its iterates $\left(\left(P_{1},P_{2}\right)^{n}:n\in\mathbb{N}\right)$ converges to $\left(P_{1},P_{2}\right)$. Note that here $P_{1}\neq P_{2}$, moreover, every mean $M:\mathbb{R}^{2}\rightarrow\mathbb{R}$ is $\left(P_{1},P_{2}\right)$-invariant.

Example 2.6**.**

Consider the extreme mean $\min:\mathbb{R}^{2}\mathbb{\rightarrow R}\,$. Since, $t<x_{0}\Longrightarrow\min\left(x_{0},t\right)<x_{0}$ 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 $\max:\mathbb{R}^{2}\mathbb{\rightarrow R}$ 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 $n\in\mathbb{N}$, the $n$th iterate $\left(\min,\max\right)^{n}$ of mean-type mapping $\left(\min,\max\right)$ has the form

[TABLE]

and, of course, the sequence of iterates of $\left(\left(\min,\max\right)^{n}:n\in\mathbb{N}\right)$ converges.

On the other hand, one can check that a mean is invariant with respect to the mean-type mapping $\left(\min,\max\right)$ if and only if it is symmetric.

Remark 2.7**.**

If $M:I^{2}\rightarrow\mathbb{R}$ 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

Proposition 2.8**.**

Let $M,N:I^{2}\rightarrow I$ be two means in an interval $I$. If one of the following conditions is valid:

1. (i)

$M$* and $N$ are left-strict,* 2. (ii)

$M$* and $N$ are right-strict,*

then property (2) holds.

Proof.

Suppose to the contrary that there exists $x,\,y\in I$, $x\neq y$ such that

[TABLE]

Binding this inequality with

[TABLE]

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 $M,\ N\colon I^{2}\to I$, the mean-type mapping $(M,N)$ is weakly contractive if for every elements $x,\,y\in I$ with $x\neq y$ there is a positive integer $n=n\left(x,y\right)$ such that

[TABLE]

Next lemma provide a complete characterization of weak-contractivity in terms of $(M_{2},N_{2})$.

Lemma 2.9**.**

Let $M,\ N\colon I^{2}\to I$ be means. Mean-type mapping $(M,N)$ is weakly contractive if and only if mean-type mapping $(M_{2},N_{2})$ is contractive.

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

[TABLE]

Therefore

[TABLE]

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 $n\in\mathbb{N}$ 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

[TABLE]

which implies (3) for the triple $(x,y,2)$ or

[TABLE]

However in the second subcase we get

[TABLE]

which implies that (3) is valid for no $n\in\mathbb{N}$ contradicting the assumption.

As the converse implication is trivial, the proof is complete ∎

Proposition 2.10**.**

Let $M,N:I^{2}\rightarrow I$ be two means in an interval $I$. If the following conditions are valid:

1. (i)

$M$* is right-strict in the first variable or $N$ is left-strict in the second variable,* 2. (ii)

$M$* is left-strict in the second variable or $M$ is right-strict in the second variable or $N$ is left-strict in the first variable or $N$ is right-strict in the first variable,* 3. (iii)

$M$* is left-strict in the first variable or $N$ right-strict in the second variable,*

then the mapping $(M,N)$ is diagonally-contractive.

Proof.

Fix $x,\,y\in I$, $x\neq y$. It suffices to prove that

[TABLE]

First assume that $x<y$. Applying property (i) we obtain that $(M,N)(x,y)\neq(x,y)$. In view of property (iii) we get $(M,N)(y,x)\neq(y,x)$.

The only remaining case is that $(M,N)(x,y)=(y,x)$ and $(M,N)(y,x)=(x,y)$. Equivalently,

[TABLE]

However these equalities cannot be simultaneously valid as they are excluded by consecutive alternatives in (ii), respectively. The case $x>y$ is completely analogous. ∎

3. Uniqueness of invariant means and convergence of iterates of

mean-type mappings

Definition 3.1**.**

Let $M,N:I^{2}\rightarrow I$ be means in an interval $I$. Define the diagonal basin $\mathcal{A}_{M,N}\subset I^{2}$ by

[TABLE]

Theorem 3.2**.**

Let $M,N:I^{2}\rightarrow I$ be means in an interval $I$. Then $\mathcal{A}_{M,N}$ is a maximal subset of $I^{2}$ such that all $(M,N)$-invariant means are equal to each other.

Proof.

For $x,y\in I$. Let $a$ be a sequence which is a shuffling of $M_{n}$ and $N_{n}$, i.e

[TABLE]

By [7] we obtain that

[TABLE]

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 $(x,y)\in\mathcal{A}_{M,N}$.

If $L(x,y)=U(x,y)$ then we obtain that the sequence $a$ converges, whence $(M_{n}(x,y))_{n\in\mathbb{N}}$ and $(N_{n}(x,y))_{n\in\mathbb{N}}$ converge to a common limit, i.e. $(x,y)\in\mathcal{A}_{M,N}$.

Conversely, if $(x,y)\in\mathcal{A}_{M,N}$ then we get

[TABLE]

which implies that the sequence $a$ is convergent, and therefore $L(x,y)=U(x,y)$. As $L$ and $U$ is the smallest and the greatest $(M,N)$-invariant mean, we have that all $(M,N)$-invariant means have the same value at the point $(x,y)$. ∎

Theorem 3.3**.**

Let $K,M,N:I^{2}\rightarrow I$ be arbitrary means. $K$ is a unique $\left(M,N\right)$-invariant mean if and only if the sequence of iterates $\left(\left(M,N\right)^{n}\right)_{n\in N}$ of the mean-type mapping $\left(M,N\right)$ converges to $\left(K,K\right)$ pointwise in $I^{2}.$

Proof.

For $x,y\in I$. Let $a$ be a sequence which is a shuffling of $M_{n}$ and $N_{n}$, i.e

[TABLE]

By [7] we obtain that

[TABLE]

are the smallest and the biggest $(M,\,N)$-invariant means, respectively.

$(\Rightarrow)$ 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.

$(\Leftarrow)$ 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. ∎

4. Mean-type mappings with diagonally-contractive iterates and

invariant means

Theorem 4.1**.**

If $M,N:I^{2}\rightarrow I$ are continuous means such that $(M,N)$ is weakly contractive then there exists a unique mean $K:I^{2}\rightarrow I$ which is $\left(M,N\right)$-invariant and, moreover, the sequence of iterates $\left(\left(M,N\right)^{n}\right)_{n\in\mathbb{N}}$ of the mean-type mapping $\left(M,N\right)$ converges (uniformly on compact subsets of $I^{2})$ to $\left(K,K\right).$

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.

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 $K\colon I^{2}\to I$.

We can now apply Theorem 3.3 to obtain that $((M,N)^{n})_{n\in\mathbb{N}}$ converges to $(K,K)$ pointwise on $I^{2}$. Moreover, in view of [5, Corollary 4.4 (ii)], this convergence is uniform on every compact subset of $I^{2}$. ∎

Remark 4.2**.**

Let us emphasize that the continuity assumption in the previous theorem cannot be omitted; cf. [7, section 3.1] for details.

Theorem 4.3**.**

Let $M,N:I^{2}\rightarrow I$ be (not necessarily continuous) means and $c\in\left[0,1\right)$. If for every point $\left(x,y\right)\in I^{2}\,,$ $x\neq y$, there is a positive integer $n=n\left(x,y\right)$ such that

[TABLE]

where $\left(M_{n},N_{n}\right):=\left(M,N\right)^{n},$ then there is a unique mean $K:I^{2}\rightarrow I$ that is $\left(M,N\right)$-invariant and, moreover the sequence of iterates $\left(\left(M,N\right)^{n}\right)_{n\in N}$ of the mean-type mapping $\left(M,N\right)$ converges (uniformly on compact subsets of $I^{2})$ to $\left(K,K\right).$

Proof.

Take an arbitrary $\left(x,y\right)\in I^{2}$. Since $M$ and $N$ are means in $I,$ for every $n\in\mathbb{N}$, we have

[TABLE]

hence

[TABLE]

and

[TABLE]

whence

[TABLE]

and, by induction, for every $n\in\mathbb{N}$,

[TABLE]

Thus the sequence $\left(\min\left(M_{n}\left(x,y\right),N_{n}\left(x,y\right)\right):n\in\mathbb{N}\right)$ is increasing and the sequence $\left(\max\left(M_{n}\left(x,y\right),N_{n}\left(x,y\right)\right):n\in\mathbb{N}\right)$ is decreasing; and both are convergent. Clearly, if $x=y$ or $M_{k}\left(x,y\right)=N_{k}\left(x,y\right)$ for some positive integer $k$, then $M_{n}\left(x,y\right)=N_{n}\left(x,y\right)$ for all $n\geq k$, and consequently we have

[TABLE]

Assume that

[TABLE]

In particular, we have $M_{n}\left(x,y\right)\neq N_{n}\left(x,y\right)$ for every $n\in\mathbb{N}$. By the assumption, for every $n\in\mathbb{N}$ there is $k=k_{n}\in\mathbb{N}$ such that

[TABLE]

that is that

[TABLE]

which leads to a contradiction. Thus

[TABLE]

and consequently,

[TABLE]

Finally, let $K$ be a pointwise limit of $(M_{n})$ or $(N_{n})$, as they are equal. ∎