1 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 \(\left( M,N\right) :I^{2}\rightarrow I^{2},\) briefly \(\left( M,N\right) \)- invariant, if \(K\circ \left( M,N\right) =K\) (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 \(\left( \left( M,N\right) ^{n}:n\in {\mathbb {N}}\right) \) of iterates of the mean-type mapping \(\left( M,N\right) \) to the mean-type \(\left( K,K\right) \). 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 “\(x<y\Longrightarrow K\left( x,y\right) \ge K\left( y,x\right) \) for all \(x,y\in I\)”, says that for every bivariate mean M in I there is a unique mean \(M_{\left[ K\right] }\) such that K is \(\left( M,M_{\left[ K \right] }\right) \)-invariant. Moreover, \(M_{\left[ K\right] }\) is strict or continuous if so is M, and both M and \(M_{\left[ K\right] }\) 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 “\(x<y\Longrightarrow K\left( x,y\right) \le K\left( y,x\right) \) for all \(x,y\in I\)”. For every mean M it provides a unique mean \(M^{\left[ K\right] }\) such that K is \(\left( M^{\left[ K\right] },M\right) \)-invariant, and has similar additional properties as \(M_{\left[ K\right] }\).

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 \(\left( M,M_{\left[ K\right] }\right) \) and \(\left( M^{ \left[ K\right] },M\right) \) converge to \(\left( K,K\right) \) (Theorem 3 and Theorem 4). This permits, in particular, to determine effectively all continuous functions \(F:I^{2}\rightarrow {\mathbb {R}}\) satisfying each of the functional equations

$$\begin{aligned}&F\left( M\left( x,y\right) ,M_{\left[ K\right] }\left( x,y\right) \right) =F\left( x,y\right) \quad x,y\in I,\\&F\left( M^{\left[ K\right] }\left( x,y\right) ,M\left( x,y\right) \right) =F\left( x,y\right) ,\ \ \ \ x,y\in I. \end{aligned}$$

2 Complementary means

In the sequel \(I\subset {\mathbb {R}}\) stands for an interval.

A function \(M:I\times I\rightarrow I\) is called a mean in I,  if

$$\begin{aligned} \min \left( x,y\right) \le M\left( x,y\right) \le \max \left( x,y\right) ,\quad x,y\in I. \end{aligned}$$

A mean M is called strict, if for \(x\ne y\), both these inequalities are sharp; and symmetric, if \(M\left( x,y\right) =M\left( y,x\right) \) for all \( x,y\in I\,.\)

Every mean is reflexive, that is \(M\left( x,x\right) =x\) for all \( x\in I\).

Remark 1

If a function \(M:I\times I\rightarrow {\mathbb {R}}\) is reflexive and (strictly) increasing in each of the variables, then it is a (strict) mean in I.

Let \(K,M,N:I^{2}\rightarrow I\) be means. 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 \(K\circ \left( M,N\right) =K\) (cf. [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 \(\left( M,N\right) \)-invariant mean we say that N is a complementary mean to M with respect to K (briefly, N is K-complementary to M).

Recall that if M, N are continuous and strict means, then there exists a unique \(\left( M,N\right) \)-invariant mean (cf. [1], also [2, 4, 5]).

Theorem 1

Let a continuous mean \(K:I^{2}\rightarrow I\) be increasing in the first variable and strictly increasing in the second one. Suppose that K satisfies the following condition:

$$\begin{aligned} x<y\Longrightarrow K\left( x,y\right) \ge K\left( y,x\right) , \quad x,y\in I. \end{aligned}$$
(1)

Then

  1. (i)

    for every mean \(M:I^{2}\rightarrow I\) there is a unique mean \(M_{\left[ K\right] }:I^{2}\rightarrow I\) such that K is \(\left( M,M_{\left[ K\right] }\right) \)-invariant, i.e.

    $$\begin{aligned} K\left( M\left( x,y\right) ,M_{\left[ K\right] }\left( x,y\right) \right) =K\left( x,y\right) , \quad x,y\in I; \end{aligned}$$
    (2)
  2. (ii)

    if for a symmetric mean M, the mean \(M_{\left[ K\right] }\) is symmetric, then K is symmetric;

  3. (iii)

    if M is strict then so is \(M_{\left[ K\right] }\);

  4. (iv)

    if M is continuous then so is \(M_{\left[ K\right] }\).

Proof

Take an arbitrary mean \(M:I^{2}\rightarrow I\) and \(x,y\in I\). We have

$$\begin{aligned} \min \left( x,y\right) \le M\left( x,y\right) \le \max \left( x,y\right) . \end{aligned}$$

By setting \(z=M_{\left[ K\right] }\left( x,y\right) ,\) the equation (2) becomes

$$\begin{aligned} K\left( M\left( x,y\right) ,z\right) =K\left( x,y\right) . \end{aligned}$$

If \(x=y\) then \(M\left( x,y\right) =K\left( x,y\right) =x\) so this equation simplifies to

$$\begin{aligned} K\left( x,z\right) =x. \end{aligned}$$

Since, by the reflexivity of every mean, \(K\left( x,x\right) =x,\) the strict monotonicity of K in the second variable implies that \(M_{\left[ K\right] }\left( x,y\right) :=z=x\) is the only solution of this equation.

Assume that \(x<y.\)

In this case we have

$$\begin{aligned} x\le M\left( x,y\right) \le y. \end{aligned}$$

Define the function \(\varphi :\left[ x,y\right] \rightarrow {\mathbb {R}}\) by

$$\begin{aligned} \varphi \left( t\right) :=K\left( M\left( x,y\right) ,t\right) -K\left( x,y\right) , \quad t\in \left[ x,y\right] . \end{aligned}$$

Of course, \(\varphi \) is continuous.

For \(t=x\), making use first of the inequality \(x\le M\left( x,y\right) \) and implication (1), and then from the inequality \(M\left( x,y\right) \le y\) and the monotonicity of K in the second variable, we have

$$\begin{aligned} \varphi \left( x\right) =K\left( M\left( x,y\right) ,x\right) -K\left( x,y\right) \le K\left( x,M\left( x,y\right) \right) -K\left( x,y\right) <0 . \end{aligned}$$

For \(t=y\), the inequality \(x\le M\left( x,y\right) \) and the monotonicity of K in the first variable give

$$\begin{aligned} \varphi \left( y\right) =K\left( M\left( x,y\right) ,y\right) -K\left( x,y\right) \ge K\left( x,y\right) -K\left( x,y\right) =0. \end{aligned}$$

The Darboux property of \(\varphi \) implies that there is \(z\in \left[ x,y \right] \) such that \(\varphi \left( z\right) =0\) , i.e. such that

$$\begin{aligned} K\left( M\left( x,y\right) ,z\right) =K\left( x,y\right) . \end{aligned}$$

By the strict monotonicity of K in the second variable, such a z is unique. Setting here \(M_{\left[ K\right] }\left( x,y\right) :=z\), we get

$$\begin{aligned} K\left( M\left( x,y\right) ,M_{\left[ K\right] }\left( x,y\right) \right) =K\left( x,y\right) . \end{aligned}$$

Since in the case \(y<x\) we can argue similarly, the proof of (i) is complete.

  1. (ii)

    This is a trivial consequence of (2).

  2. (iii)

    Assume, on the contrary, that \(M_{\left[ K\right] }\) is not strict, so there are \(x_{0},y_{0}\in I\), \(x_{0}<y_{0},\) such that \(M_{\left[ K\right] }\left( x_{0},y_{0}\right) =x_{0}\) or \(M_{\left[ K\right] }\left( x_{0},y_{0}\right) =y_{0}\).

In the first case, from (2), we would have

$$\begin{aligned} K\left( M\left( x_{0},y_{0}\right) ,x_{0}\right) =K\left( M\left( x_{0},y_{0}\right) ,M_{\left[ K\right] }\left( x_{0},y_{0}\right) \right) =K\left( x_{0},y_{0}\right) , \end{aligned}$$

and, from (1),

$$\begin{aligned} K\left( x_{0},y_{0}\right) \ge K\left( y_{0},x_{0}\right) , \end{aligned}$$

whence \(K\left( M\left( x_{0},y_{0}\right) ,x_{0}\right) \ge K\left( y_{0},x_{0}\right) \). But this is a contradiction as K is strictly increasing in the first variable and \(x_{0}<M\left( x_{0},y_{0}\right) <y_{0} \).

Since in the case when \(M_{\left[ K\right] }\left( x_{0},y_{0}\right) =y_{0}\) we can argue similarly, this completes the proof of (iii).

  1. (iv)

    We omit the simple argument of this part. \(\square \)

Definition 1

Under the conditions of Theorem 1, the mean \(M_{\left[ K\right] }\) is called K-complementary (or K-right complementary) to the mean M.

Under the conditions of this theorem, part (i) can be strengthened as follows:

Remark 2

For every mean \(M:I^{2}\rightarrow I\) there is a unique function \( N:I^{2}\rightarrow I\) such that

$$\begin{aligned} K\left( M\left( x,y\right) ,N\left( x,y\right) \right) =K\left( x,y\right) ,\quad x,y\in I, \end{aligned}$$

holds and, moreover, N is a mean.

To show that assumption (1) is essential, consider the following

Example 1

Let \(I={\mathbb {R}}\) and \(K:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}\) be a weighted arithmetic mean, i.e. \(K\left( x,y\right) =ax+\left( 1-a\right) y\) for some fixed \(a\in \left( 0,1\right) \), and let \(M:{\mathbb {R}} ^{2}\rightarrow {\mathbb {R}}\) be a mean. A simple calculation shows that a function \(N:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}\) satisfies (2) if and only if

$$\begin{aligned} N\left( x,y\right) =\frac{a\left( x-M\left( x,y\right) \right) }{1-a}+y,\quad x,y\in {\mathbb {R}}. \end{aligned}$$
(3)

Assume that for every weighted arithmetic mean \(M\left( x,y\right) =bx+\left( 1-b\right) y\), \(b\in \left[ 0,1\right] \), the function N is a mean in \({\mathbb {R}}\), that is that

$$\begin{aligned} N\left( x,y\right) =\frac{a\left( 1-b\right) }{1-a}x+\left( 1-\frac{a\left( 1-b\right) }{1-a}\right) y, \quad x,y\in {\mathbb {R}}, \end{aligned}$$

is a mean in \({\mathbb {R}}\). Clearly, N is a mean for every \(b\in \left[ 0,1 \right] \) if, and only if,

$$\begin{aligned} \frac{a\left( 1-b\right) }{1-a}\le 1,\quad b\in \left[ 0,1\right] . \end{aligned}$$

Since the function \(\left[ 0,1\right] \ni b\longmapsto \frac{a\left( 1-b\right) }{1-a}\) is decreasing, this inequality holds true iff \(\ \frac{ a\left( 1-0\right) }{1-a}\le 1\), that is iff

$$\begin{aligned} a\le \frac{1}{2}. \end{aligned}$$

We can also argue as follows. The function N given by (3) is a mean for every mean M if and only if

$$\begin{aligned} \min \left( x,y\right) \le \frac{a\left( x-M\left( x,y\right) \right) }{1-a} +y\le \max \left( x,y\right) ,\quad x,y\in {\mathbb {R}}, \end{aligned}$$

or, equivalently, if and only if, for every mean M, and for all \(x,y\in {\mathbb {R}}\),

$$\begin{aligned}&ax+\left( 1-a\right) y-\left( 1-a\right) \max \left( x,y\right) \\&\quad \le aM\left( x,y\right) \le ax+\left( 1-a\right) y-\left( 1-a\right) \min \left( x,y\right) . \end{aligned}$$

Taking here first \(M=\min \) and then\(\ M=\max \) we obtain, respectively, the inequalities

$$\begin{aligned}&ax+\left( 1-a\right) y\le a\min \left( x,y\right) +\left( 1-a\right) \max \left( x,y\right) ,\quad x,y\in {\mathbb {R}},\\&\left( 1-a\right) \min \left( x,y\right) +a\max \left( x,y\right) \le ax+\left( 1-a\right) y,\quad \, x,y\in {\mathbb {R}}. \end{aligned}$$

It is easy to see that each of these inequalities holds, if and only if \( a\le \frac{1}{2}\).

Note that with \(K\left( x,y\right) =ax+\left( 1-a\right) y\), the inequality \( a\le \frac{1}{2}\) 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 \(K:I^{2}\rightarrow I\) be strictly increasing in the first variable and increasing in the second one. Suppose that K satisfies the following condition:

$$\begin{aligned} x<y\Longrightarrow K\left( x,y\right) \le K\left( y,x\right) , \quad x,y\in I. \end{aligned}$$

Then

  1. (i)

    for every mean \(M:I^{2}\rightarrow I\) there is a unique mean \(M^{\left[ K \right] }:I^{2}\rightarrow I\) such that K is \(\left( M,M^{\left[ K\right] }\right) \)-invariant, i.e.

    $$\begin{aligned} K\left( M^{\left[ K\right] }\left( x,y\right) ,M\left( x,y\right) \right) =K\left( x,y\right) ,\ \ \ \ x,y\in I; \end{aligned}$$
  2. (ii)

    if for some symmetric mean M, the mean \(M^{\left[ K\right] }\) is symmetric, then K is symmetric;

  3. (iii)

    if M is strict then so is \(M^{\left[ K\right] }\);

  4. (iv)

    if M is continuous then so is \(M^{\left[ K\right] }\).

Definition 2

Under the conditions of Theorem 2, the mean \(M^{\left[ K\right] }\) can be referred to as K-complementary (or K-left complementary) to the mean M.

From Theorems 1 and 2 we obtain the following improvement of Remark 1 in [3]:

Corollary 1

If \(K:I^{2}\rightarrow I\) is a continuous strictly increasing in each variable and symmetric mean, then for every mean \(M:I^{2}\rightarrow I\) there is a unique function \(N:I^{2}\rightarrow I\) such that \(K\circ \left( M,N\right) =K\), moreover N is a mean and

$$\begin{aligned} M_{\left[ K\right] }=N=M^{\left[ K\right] }. \end{aligned}$$

We end this section with

Remark 3

In Theorems 1 and 2 (as well as in [3], Remark 1), the strict increasing monotonicity in each variable of the invariant mean \( K:I^{2}\rightarrow I,\) cannot be weaken by the assumption that K is a strict mean.

To show it consider the following

Example 2

It is known that the contra-harmonic mean \(K:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right) \),

$$\begin{aligned} K\left( x,y\right) =\frac{x^{2}+y^{2}}{x+y}, \end{aligned}$$

is not strictly increasing. The symmetry of K implies that condition (1) of Theorem 1 is satisfied. Taking for M the arithmetic mean \(A\left( x,y\right) =\frac{x+y}{2}\) for \(x,y>0\), by a simple calculation, we get

$$\begin{aligned} K_{\left[ A\right] }\left( x,y\right) =\frac{x^{2}+y^{2}+\sqrt{2}\sqrt{ x^{4}+y^{4}}}{2\left( x+y\right) },\quad x,y>0. \end{aligned}$$

Of course, \(K_{\left[ A\right] }:\) \(\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right) \) and \(K_{\left[ A\right] }\) is reflexive, i.e. \(K_{ \left[ A\right] }\left( x,x\right) =x;\) so \(K_{\left[ A\right] }\) is a bivariate pre-mean in \(\left( 0,\infty \right) .\) But, as

$$\begin{aligned} K_{\left[ A\right] }\left( 1,10\right) >11, \end{aligned}$$

\(K_{\left[ A\right] }\) is not a mean.

(It can be verified similarly, that \(K_{\left[ G\right] }\), \(K_{\left[ H \right] }\) where G and H stand, respectively, for the geometric and harmonic mean, are pre-means, but not means.)

3 Some applications

Applying Theorem 1 and the main result of [5] (see also [4]) we obtain the following

Theorem 3

Suppose that a mean \(K:I^{2}\rightarrow I\) is continuous and strictly increasing in each variable.

  1. (i)

    If K satisfies the condition

    $$\begin{aligned} x<y\Longrightarrow K\left( x,y\right) \ge K\left( y,x\right) , x,y\in I, \end{aligned}$$

    then for every continuous and strict mean \(M:I^{2}\rightarrow I,\) the sequence \(\left( \left( M,M_{\left[ K\right] }\right) ^{n}:n\in {\mathbb {N}} \right) \) of iterates of the mean-type mapping \(\left( M,M_{\left[ K\right] }\right) :I^{2}\rightarrow I^{2}\) converges uniformly on compact sets to the mean-type map \(\left( K,K\right) \).

  2. (ii)

    If K satisfies the condition

    $$\begin{aligned} x<y\Longrightarrow K\left( x,y\right) \le K\left( y,x\right) , \quad x,y\in I, \end{aligned}$$

    then for every continuous and strict mean \(M:I^{2}\rightarrow I,\) the sequence \(\left( \left( M^{\left[ k\right] },M\right) ^{n}:n\in {\mathbb {N}} \right) \) of iterates of the mean-type mapping \(\left( M^{\left[ k\right] },M\right) :I^{2}\rightarrow I^{2}\) converges uniformly on compact sets to the mean-type map \(\left( K,K\right) \).

Using this result we prove the following

Theorem 4

Let a mean \(K:I^{2}\rightarrow I\) be continuous and strictly increasing in each variable and \(M:I^{2}\rightarrow I\) be an arbitrary strict and continuous mean.

  1. (i)

    Suppose that K satisfies the condition

    $$\begin{aligned} x<y\Longrightarrow K\left( x,y\right) \ge K\left( y,x\right) , x,y\in I. \end{aligned}$$

    Then a function \(F:I^{2}\rightarrow {\mathbb {R}}\) continuous at every point of the diagonal \(\Delta \left( I^{2}\right) :=\left\{ \left( x,x\right) :x\in I\right\} \) satisfies the functional equation

    $$\begin{aligned} F\left( M\left( x,y\right) ,M_{\left[ K\right] }\left( x,y\right) \right) =F\left( x,y\right) , \quad x,y\in I, \end{aligned}$$
    (4)

    if and only if there is a single variable continuous function \(\varphi :I\rightarrow {\mathbb {R}}\) such that

    $$\begin{aligned} F\left( x,y\right) =\varphi \left( K\left( x,y\right) \right) , \quad x,y\in I. \end{aligned}$$
  2. (ii)

    Suppose that K satisfies the condition

    $$\begin{aligned} x<y\Longrightarrow K\left( x,y\right) \le K\left( y,x\right) , x,y\in I. \end{aligned}$$

    Then a function \(F:I^{2}\rightarrow {\mathbb {R}}\) continuous at every point of the diagonal \(\Delta \left( I^{2}\right) \) satisfies the functional equation

    $$\begin{aligned}&F\left( M^{\left[ K\right] }\left( x,y\right) ,M\left( x,y\right) \right) \\&\quad =F\left( x,y\right) ,\quad x,y\in I, \end{aligned}$$

    if and only if there is a single variable continuous function \(\varphi :I\rightarrow {\mathbb {R}}\) such that

    $$\begin{aligned} F\left( x,y\right) =\varphi \left( K\left( x,y\right) \right) ,\quad x,y\in I. \end{aligned}$$

Proof

Assume first that \(F:I^{2}\rightarrow {\mathbb {R}}\) is continuous on the diagonal \(\Delta \left( I^{2}\right) \) and satisfies equation (4), that is

$$\begin{aligned} F\circ \left( M,M_{\left[ K\right] }\right) =F. \end{aligned}$$

Hence, by induction,

$$\begin{aligned} F=F\circ \left( M,M_{\left[ K\right] }\right) ^{n},\quad n\in {\mathbb {N}}, \end{aligned}$$
(5)

where \(\left( M,M_{\left[ K\right] }\right) ^{n}\) is the nth iterate of \( \left( M,M_{\left[ K\right] }\right) \). By Theorem 3 the sequence of mean-type mappings \(\left( M,M_{\left[ K\right] }\right) ^{n}\) converges to the mean-type mapping \(\left( K,K\right) :I^{2}\rightarrow I^{2}\); that is

$$\begin{aligned}&\lim _{n\rightarrow \infty }\left( M,M_{\left[ K\right] }\right) ^{n}\left( x,y\right) =\left( K\left( x,y\right) ,K\left( x,y\right) \right) ,\quad \left( x,y\right) \in I^{2}. \end{aligned}$$

Since \(\left( K\left( x,y\right) ,K\left( x,y\right) \right) \) belongs to the diagonal \(\Delta \left( I^{2}\right) \) for every \(\left( x,y\right) \in I^{2}\), by (5) and the continuity of F on \(\Delta \left( I^{2}\right) \) implies that for every \(\left( x,y\right) \in I^{2}\),

$$\begin{aligned} F\left( x,y\right)= & {} \lim _{n\rightarrow \infty }F\left( \left( M,M_{\left[ K \right] }\right) ^{n}\left( x,y\right) \right) =F\left( \lim _{n\rightarrow \infty }\left( M,M_{\left[ K\right] }\right) ^{n}\left( x,y\right) \right) \\= & {} F\left( K\left( x,y\right) ,K\left( x,y\right) \right) . \end{aligned}$$

Setting

$$\begin{aligned} \varphi \left( t\right) :=F\left( t,t\right) ,\quad t\in I \, , \end{aligned}$$

we conclude that \(F\left( x,y\right) =\varphi \left( K\left( x,y\right) \right) \) for all \(\left( x,y\right) \in I^{2}\).

To prove the converse implication, take an arbitrary function \(\varphi :I\rightarrow {\mathbb {R}}\) and put \(F:=\varphi \circ K\). Then, for all \( x,y\in I,\) making use of the K-invariance with respect to \(M,M_{\left[ K \right] }\), we have

$$\begin{aligned} F\left( M\left( x,y\right) ,M_{\left[ K\right] }\left( x,y\right) \right)= & {} \left( \varphi \circ K\right) \left( M\left( x,y\right) ,M_{\left[ K \right] }\left( x,y\right) \right) \\= & {} \varphi \left( K\left( M\left( x,y\right) ,M_{\left[ K\right] }\left( x,y\right) \right) \right) =\varphi \left( K\left( x,y\right) \right) \\= & {} F\left( x,y\right) , \end{aligned}$$

which completes the proof of (i).

We omit the similar proof of (ii). \(\square \)