Explicit forms of invariant means: complementary results to Gauss $\left( {\mathcal {A}},{\mathcal {G}}\right)$-theorem and some applications

Abstract

Explicit forms of invariant means for all mean-type mappings generated by the bivariable classical arithmetic, geometric and harmonic means (complementary to the Gauss $\mathcal {AGM}$-theorem) are given. A generalization for higher dimension mean-type mappings as well as two open problems and an application in determining implicit solutions of some functional equations, are presented.

1 Introduction

Every mean-type mapping, under some general conditions, has a unique invariant mean ([8], see also [7]), but it is a nontrivial problem to find its explicit form. Knowing the invariant mean of a mean-type mapping is important, as it facilitates, a significant in applications and a complete description of the dynamical system generated by the mapping (see, for instance, the Gauss method of calculating elliptic integrals, an explicit solution of some functional equations).

In this paper we are mainly interested in searching for the explicit forms of invariant means for the mean-type mappings involving the classical symmetric arithmetic, geometric and harmonic means and their natural generalizations.

Section 3 is devoted to bivariable means. To describe the obtained results, recall the basic notions in this case. 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. [6]). K is sometimes called the Gauss composition of the means M and N (cf. Daróczy and Páles [4]), and N is referred to as a complementary mean to M with respect to K (briefly, K-complementary mean to M) ([6], see also [12]). Under general conditions (Lemma 1), 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 $\left( K,K\right)$ ([8], also J.M. Borwein and P.B. Borwein [2, 7, 12]).

Denote by ${\mathcal {A}},{\mathcal {G}},{\mathcal {H}}$ the arithmetic, geometric and harmonic bivariable means in $\left( 0,\infty \right)$. The first (folklore) result of Sect. 3 says that the geometric mean ${\mathcal {G}}$ is invariant with respect to the mapping $\left( {\mathcal {A}},{\mathcal {H}}\right)$ (the invariance equality ${\mathcal {G}}\circ \left( {\mathcal {A}},{\mathcal {H}}\right) ={\mathcal {G}}$ is equivalent to the Pythagorean proportion); the second result, due to Gauss, says that $K_{{\mathcal {A}},{\mathcal {G}}}\circ \left( {\mathcal {A}},{\mathcal {G}}\right) =$ $K_{{\mathcal {A}},{\mathcal {G}}}$, where

$$\begin{aligned} K_{{\mathcal {A}},{\mathcal {G}}}\left( x,y\right) =\frac{\pi }{2}\left( \int _{0}^{ \frac{\pi }{2}}\frac{d\varphi }{\sqrt{x^{2}\cos ^{2}\varphi +y^{2}\sin ^{2}\varphi }}\right) ^{-1}, \end{aligned}$$

and according to the third result, $K_{{\mathcal {A}},{\mathcal {G}}}\circ \left( {\mathcal {A}},{\mathcal {G}}\right) =$ $K_{{\mathcal {H}},{\mathcal {G}}}$, where

$$\begin{aligned} K_{{\mathcal {H}},{\mathcal {G}}}\left( x,y\right) =\frac{2xy}{\pi }\int _{0}^{ \frac{\pi }{2}}\frac{d\varphi }{\sqrt{x^{2}\sin ^{2}\varphi +y^{2}\cos ^{2}\varphi }}. \end{aligned}$$

These results allow us to determine explicitly the invariant means for all mean-type mappings $\left( M,N\right)$ such that $M,N\in \left\{ {\mathcal {A}},{\mathcal {G}},{\mathcal {H}}\right\}$.

In Sect. 4, as corollaries, we formulate the quasiarithmetic counterparts of Theorems 1–3.

In [9] it was shown that the bivariable $\left( {\mathcal {A}}, {\mathcal {H}}\right)$-invariance of the geometric mean ${\mathcal {G}}$ (the Pythagorean harmony proportion) has a unique extension for means of k -variable, $k\ge 3.$ In case $k=3$ it has the form

$$\begin{aligned} \mathcal {G\circ }\left( {\mathcal {A}},{\mathcal {B}},{\mathcal {H}}\right) ={\mathcal {G}}\text {,} \end{aligned}$$

where

$$\begin{aligned} {\mathcal {A}}\left( x,y,z\right)= & {} \frac{x+y+z}{3},\quad {\mathcal {B}} \left( x,y,z\right) =\frac{yz+zx+xy}{x+y+z}, \\ {\mathcal {H}}\left( x,y,z\right)= & {} \frac{3xy}{yz+zx+xy},\quad \mathcal { G}\left( x,y,z\right) =\root 3 \of {xyz}. \end{aligned}$$

This fact allows us to get the higher dimension counterpart of Theorem 1 when $k=3$ (in particular, $\left( {\mathcal {G}},{\mathcal {G}},{\mathcal {G}} \right)$ is the limit of the sequence of iterates of the mean-type mapping $\left( {\mathcal {A}},{\mathcal {B}},{\mathcal {H}}\right)$). For each of the remaining three nontrivial mean-type mappings $\left( {\mathcal {A}},{\mathcal {B}},{\mathcal {G}}\right) ,$ $\left( {\mathcal {H}},{\mathcal {B}},{\mathcal {G}}\right)$, $\left( {\mathcal {A}},{\mathcal {G}},{\mathcal {H}}\right) ,$ there is a unique invariant mean. Referring to Theorems 2 and 3, we post open questions concerning their explicit forms. We end this section with a short discussion of the general case $k>3$.

In Sect. 5 we apply the results of Sect. 3 to determine the implicit forms of some functional equations.

2 Notations and auxiliary facts

Let $I\subset {\mathbb {R}}$ be an interval and $k\in {\mathbb {N}}$, $k\ge 2$ be fixed. A function $M:I^{k}\rightarrow {\mathbb {R}}$ is called a k-variable mean in I (see for instance [3]), if

$$\begin{aligned} \min \left( x_{1},\ldots ,x_{k}\right) \le M\left( x_{1},\ldots ,x_{k}\right) \le \max \left( x_{1},\ldots ,x_{k}\right) ,\quad x_{1},\ldots ,x_{k}\in I. \end{aligned}$$

The mean M is called

strict, if these inequalities are sharp for all nonconstant sequences $\left( x_{1},\ldots ,x_{k}\right) \in I^{k};$

symmetric, if $M\left( x_{\sigma \left( 1\right) },\ldots ,x_{\sigma \left( k\right) }\right) =M\left( x_{1},\ldots ,x_{k}\right)$ for each permutation $\sigma$ of the set $\left\{ 1,\ldots ,k\right\}$ and all $x_{1},\ldots ,x_{k}\in I.$

Assume that $I=\left( 0,\infty \right)$. A k-variable mean M in $\left( 0,\infty \right)$ is called homogeneous, if

$$\begin{aligned} M\left( tx_{1},\ldots ,tx_{k}\right) =tM\left( x_{1},\ldots ,x_{k}\right) ,\quad t,x_{1},\ldots ,x_{k}>0. \end{aligned}$$

Let us note some easy to verify facts.

Remark 1

If $M:I^{k}\rightarrow {\mathbb {R}}$ is a mean in I, then

• (i) M is reflexive, i.e. $M\left( x,\ldots ,x\right) =x$ for every $x\in I;$

• (ii) for every interval $J\subset I$, M is a mean in J and we have $M\left( J^{k}\right) =J;$ in particular, $M\left( I^{k}\right) =I.$

Remark 2

If a function $M:I^{k}\rightarrow {\mathbb {R}}$ is reflexive and (strictly) increasing in each variable, then it is a (strict) mean in I.

Let $M_{1},\ldots ,M_{k}:I^{k}\rightarrow I$ and ${\textbf{M}}:I^{k}\rightarrow I^{k}$ be the mean-type mapping ${\textbf{M}}:I^{k}\rightarrow I^{k},$ given by ${\textbf{M}}=\left( M_{1},\ldots ,M_{k}\right)$.

A mean $K:I^{k}\rightarrow I$ is called ${\textbf{M}}$-invariant (or invariant with respect to the mean-type mapping $\left( M_{1},\ldots ,M_{k}\right)$), if

$$\begin{aligned} K\circ \left( M_{1},\ldots ,M_{k}\right) =K. \end{aligned}$$

Clearly, if K is $\left( M_{1},\ldots ,M_{k}\right)$-invariant then

$$\begin{aligned} \min \left( M_{1},\ldots ,M_{k}\right) \le K\le \max \left( M_{1},\ldots ,M_{k}\right) \text {.} \end{aligned}$$

We need the following

Lemma 1

Let $M_{1},\ldots ,M_{k}:I^{k}\rightarrow I$ be continuous and strict means in I. Then

• (i) there is a unique $\left( M_{1},\ldots ,M_{k}\right)$-invariant mean $K:I^{k}\rightarrow I$;

• (ii) K is continuous and

  $$\begin{aligned} \min \left( M_{1},\ldots ,M_{k}\right) \le K\le \max \left( M_{1},\ldots ,M_{k}\right) ; \end{aligned}$$

• (iii) the sequence $\left( \left( M_{1},\ldots ,M_{k}\right) ^{n}:n\in {\mathbb {N}} \right)$ of iterates of the mean-type mapping $\left( M_{1},\ldots ,M_{k}\right)$ converges uniformly on compact subsets, and

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left( M_{1},\ldots ,M_{k}\right) ^{n}=\left( K,\ldots ,K\right) ; \end{aligned}$$

• (iv) if $M_{1},\ldots ,M_{k}$ are symmetric, then so is K;

• (v) if $\ I=\left( 0,\infty \right)$ and $M_{1},\ldots ,M_{k}$ are homogeneous, then so is K.

Results (i)–(iii) are a special case of the main result of [8]. Result (iv) is an obvious consequence of (iii).

In the monograph [2] (as well as in several later papers), even in case $k=2$, the uniqueness of the invariant mean is proved under additional assumption of the continuity of the invariant mean K. In [8] it is shown that this assumption is redundant. Recently in [10] it has been shown that the uniqueness of the $\left( M_{1},\ldots ,M_{k}\right)$-invariant mean implies result (iii).

Remark 3

For each permutation $\sigma$ of $\left\{ 1,\ldots ,k\right\}$ the dynamical systems $\mathbf {\xi }_{n+1}={\textbf{M}}\left( \mathbf {\xi }_{n}\right)$ and $\mathbf {\xi }_{n+1}^{\left[ \sigma \right] }={\textbf{M}}_{\sigma }\left( \mathbf {\xi }_{n}^{\left[ \sigma \right] }\right)$ ($n\in {\mathbb {N}}_{0}= \mathbb {N\cup }\left\{ 0\right\}$; $\mathbf {\xi }_{0}\in I^{k}$ ), generated respectively by the mean-type mappings $\mathbf {M:}=\left( M_{1},\ldots ,M_{k}\right)$ and ${\textbf{M}}_{\sigma }:=\left( M_{\sigma \left( 1\right) },\ldots ,M_{\sigma \left( k\right) }\right)$, are “isomorphic”, i.e. that $\mathbf {\xi }_{n}^{\left[ \sigma \right] }=\mathbf {\xi }_{\sigma \left( n\right) }$ ($n\in {\mathbb {N}}_{0}$), so it is enough to consider $\left( M_{\sigma \left( 1\right) },\ldots ,M_{\sigma \left( k\right) }\right)$ for one, arbitrarily chosen, $\sigma$.

Let ${\mathcal {A}}:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right)$, ${\mathcal {G}}:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right) ,$ ${\mathcal {H}}:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right)$ denote, respectively, the arithmetic, geometric and harmonic means, i.e.

$$\begin{aligned} {\mathcal {A}}\left( x.y\right) =\frac{x+y}{2},\quad {\mathcal {G}} \left( x,y\right) =\sqrt{xy},\quad {\mathcal {H}}\left( x,y\right) = \frac{2xy}{x+y}. \end{aligned}$$

Defining ${\mathcal {A}}:\mathop {\displaystyle \bigcup }\limits _{k=1}^{\infty }\left( 0,\infty \right) ^{k}\rightarrow \left( 0,\infty \right)$, ${\mathcal {G}}:\mathop {\displaystyle \bigcup }\limits _{k=1}^{\infty }\left( 0,\infty \right) ^{k}\rightarrow \left( 0,\infty \right) ,$ ${\mathcal {H}}:\mathop {\displaystyle \bigcup }\limits _{k=1}^{\infty }\left( 0,\infty \right) ^{k}\rightarrow \left( 0,\infty \right)$ by

$$\begin{aligned}{} & {} {\mathcal {A}}\left( x_{1},\ldots ,x_{k}\right) =\frac{x_{1}+\cdots +x_{k}}{k},\quad {\mathcal {G}}\left( x_{1},\ldots ,x_{k}\right) \\ {}{} & {} =\root k \of {x_{1}\cdot \cdots \cdot x_{k}},{\mathcal {H}}\left( x_{1},\ldots ,x_{k}\right) =\frac{k}{\frac{1 }{x_{1}}+\cdots +\frac{1}{x_{k}}},\quad k\in {\mathbb {N}}\text {,} \end{aligned}$$

respectively, allows us to use the symbols ${\mathcal {A}},{\mathcal {G}},{\mathcal {H}}$ for every k.

3 The classical bivariable means

First consider the mean-type mapping $\left( {\mathcal {A}},{\mathcal {H}}\right)$ (see, for instance, [11]).

Theorem 1

A mean $K:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right)$ is invariant with respect to the mean-type map $\left( {\mathcal {A}},{\mathcal {H}}\right)$, i.e.

$$\begin{aligned} K\circ \left( {\mathcal {A}},{\mathcal {H}}\right) =K, \end{aligned}$$

if, and only, if 

$$\begin{aligned} \ K={\mathcal {G}}. \end{aligned}$$

Moreover, the sequence $\left( \left( {\mathcal {A}},{\mathcal {H}}\right) ^{n}:n\in {\mathbb {N}}\right)$ of iterates of $\left( {\mathcal {A}},{\mathcal {H}} \right)$ converges uniformly on compact subsets of $\left( 0,\infty \right) ^{2}$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {A}},{\mathcal {H}}\right) ^{n}=\left( {\mathcal {G}},{\mathcal {G}}\right) . \end{aligned}$$

Proof

Note that for all $x,y>0$,

$$\begin{aligned} \mathcal {G\circ }\left( {\mathcal {A}},{\mathcal {H}}\right) \left( x,y\right) = \sqrt{\frac{x+y}{2}\frac{2xy}{x+y}}=\sqrt{xy}={\mathcal {G}}\left( x,y\right) , \end{aligned}$$

so the geometric mean ${\mathcal {G}}$ is $\left( {\mathcal {A}},{\mathcal {H}} \right)$-invariant.

Since the means ${\mathcal {A}}$ and ${\mathcal {H}}$ are continuous and strict, applying Lemma 1 with $k=2\,$, $M_{1}={\mathcal {A}}$ and $M_{2}={\mathcal {H}}$, we get the uniqueness of the invariant mean ${\mathcal {G}}$ and the convergence result. $\square$

Remark 4

Lemma 1.(iv) and the symmetry of the means ${\mathcal {A}},{\mathcal {H}}$ imply that the analogous result (with $K={\mathcal {G}}$) holds true for the mean-type mapping $\left( {\mathcal {H}},{\mathcal {G}}\right)$.

Remark 5

The identity

$$\begin{aligned} {\mathcal {A}}\left( x,y\right) -{\mathcal {H}}\left( x,y\right) =\frac{1}{2}\frac{ \left( x-y\right) ^{2}}{x+y},\quad x,y>0, \end{aligned}$$

and, the easy to verify inequalities

$$\begin{aligned} \mathcal {H\circ }\left( \left( {\mathcal {A}},{\mathcal {H}}\right) ^{n}\right) \le {\mathcal {G}}\le \mathcal {A\circ }\left( \left( {\mathcal {A}},{\mathcal {H}} \right) ^{n}\right) ,\quad n\in {\mathbb {N}}\text {,} \end{aligned}$$

imply, for every $x,y>0,$ the fast convergence of the sequence $\left( \left( {\mathcal {A}},{\mathcal {H}}\right) ^{n}\left( x,y\right) :n\in {\mathbb {N}} \right)$ to ${\mathcal {G}}\left( x,y\right) =\sqrt{xy}.$

Remark 6

The invariance equality $\mathcal {G\circ }\left( {\mathcal {A}},{\mathcal {H}} \right) ={\mathcal {G}}$ is equivalent to the Pythagorean harmony proportion

$$\begin{aligned} \frac{{\mathcal {A}}}{{\mathcal {G}}}=\frac{{\mathcal {G}}}{{\mathcal {H}}}. \end{aligned}$$

For the mean-type mapping $\left( {\mathcal {A}},{\mathcal {G}}\right)$ we have the following result of Gauss (see. for instance, [5]).

Theorem 2

A mean $K:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right)$ is invariant with respect to the mean-type mapping $\left( {\mathcal {A}}, {\mathcal {G}}\right)$, i.e.

$$\begin{aligned} K\circ \left( {\mathcal {A}},{\mathcal {G}}\right) =K, \end{aligned}$$

if, and only if,$\,\ K=K_{{\mathcal {A}},{\mathcal {G}}}$, where

$$\begin{aligned} \ K_{{\mathcal {A}},{\mathcal {G}}}\left( x,y\right) =\frac{\pi }{2}\left( \int _{0}^{\frac{\pi }{2}}\frac{d\varphi }{\sqrt{x^{2}\cos ^{2}\varphi +y^{2}\sin ^{2}\varphi }}\right) ^{-1},x,y>0. \end{aligned}$$

(1)

Moreover, the sequence $\left( \left( {\mathcal {A}},{\mathcal {G}}\right) ^{n}:n\in {\mathbb {N}}\right)$ of iterates of $\left( {\mathcal {A}},{\mathcal {G}} \right)$ converges uniformly on compact subsets of $\left( 0,\infty \right) ^{2},$

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {A}},{\mathcal {G}}\right) ^{n}=\left( K_{{\mathcal {A}},{\mathcal {G}}},K_{{\mathcal {A}},{\mathcal {G}}}\right) , \end{aligned}$$

and $K_{{\mathcal {A}},{\mathcal {G}}}$ is a strict, continuous and symmetric mean.

Proof

The function $K_{{\mathcal {A}},{\mathcal {G}}}$ given by (1) is a strict mean, as it is reflexive, i.e. $K_{{\mathcal {A}},{\mathcal {G}}}\left( x,x\right) =x$ for all $x>0$, and $K_{{\mathcal {A}},{\mathcal {G}}}$ is strictly increasing in each of the variables (see Remark 2). In view of the Gauss theorem (see [5]), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {A}},{\mathcal {G}}\right) ^{n}\left( x,y\right) =\left( K_{{\mathcal {A}},{\mathcal {G}}}\left( x,y\right) ,K_{{\mathcal {A}},{\mathcal {G}}}\left( x,y\right) \right) ,\quad x,y>0. \end{aligned}$$

It follows that, for all $x,y>0$,

$$\begin{aligned} \left( K_{{\mathcal {A}},{\mathcal {G}}}\left( x,y\right) ,K_{{\mathcal {A}},{\mathcal {G}}}\left( x,y\right) \right)= & {} \lim _{n\rightarrow \infty }\left( {\mathcal {A}},{\mathcal {G}}\right) ^{n+1}\left( x,y\right) \\= & {} \lim _{n\rightarrow \infty }\left( \left( {\mathcal {A}},{\mathcal {G}}\right) ^{n}\left( {\mathcal {A}}\left( x,y\right) ,{\mathcal {G}}\left( x,y\right) \right) \right) = \\= & {} \left( K_{{\mathcal {A}},{\mathcal {G}}}\left( {\mathcal {A}}\left( x,y\right) , {\mathcal {G}}\left( x,y\right) \right) ,K_{{\mathcal {A}},{\mathcal {G}}}\left( {\mathcal {A}}\left( x,y\right) ,{\mathcal {G}}\left( x,y\right) \right) \right) \\= & {} \left( K_{{\mathcal {A}},{\mathcal {G}}}\circ \left( {\mathcal {A}},{\mathcal {G}} \right) \left( x,y\right) ,K_{{\mathcal {A}},{\mathcal {G}}}\circ \left( \mathcal {A },{\mathcal {G}}\right) \left( x,y\right) \right) \text {, } \end{aligned}$$

whence $K_{{\mathcal {A}},{\mathcal {G}}}\circ \left( {\mathcal {A}},{\mathcal {G}} \right) =K_{{\mathcal {A}},{\mathcal {G}}},$ that is $K_{{\mathcal {A}},{\mathcal {G}}}$ is invariant with respect to the mapping $\left( {\mathcal {A}},{\mathcal {G}} \right)$.

The uniqueness of the $\left( {\mathcal {A}},{\mathcal {G}}\right)$-invariant mean and the result on convergence follow from Lemma 1 with $k=2,$ $M_{1}= {\mathcal {A}}$ and $M_{2}={\mathcal {G}}$. $\square$

Remark 7

Lemma 1.(iv) and the symmetry of the means ${\mathcal {A}},{\mathcal {G}}$ imply that the analogous result (with $K=K_{{\mathcal {G}},{\mathcal {A}}}=K_{{\mathcal {A}},{\mathcal {G}}}$) holds true for the mean-type mapping $\left( {\mathcal {G}}, {\mathcal {A}}\right) .$

Remark 8

The identity

$$\begin{aligned} {\mathcal {A}}\left( x,y\right) -{\mathcal {G}}\left( x,y\right) =\frac{1}{2} \left( \sqrt{x}-\sqrt{y}\right) ^{2},\quad x,y>0, \end{aligned}$$

and the obvious inequalities

$$\begin{aligned} G\mathcal {\circ }\left( \left( {\mathcal {A}},{\mathcal {G}}\right) ^{n}\right) \le K_{{\mathcal {A}},{\mathcal {G}}}\le \mathcal {A\circ }\left( \left( {\mathcal {A}},{\mathcal {G}}\right) ^{n}\right) ,\quad n\in {\mathbb {N}}\text {,} \end{aligned}$$

imply, for every $x,y>0,$ the fast convergence of the sequence$\left( \left( {\mathcal {A}},{\mathcal {G}}\right) ^{n}\left( x,y\right) :n\in {\mathbb {N}}\right)$ to $\left( K_{{\mathcal {A}},{\mathcal {G}}}\left( x,y\right) ,K_{{\mathcal {A}}, {\mathcal {G}}}\left( x,y\right) \right) .$ Gauss observed that this fact permits the calculation of the elliptic integrals. On May 30, 1799, in his Diary [5], denoting by ${\bar{\omega }}$ half of the length of Bernoulli’s lemniscate inscribed in the unit circle, he observed the following fascinating equality

$$\begin{aligned} K_{{\mathcal {A}},{\mathcal {G}}}\left( 1,2\right) =\frac{\pi }{{\bar{\omega }}}. \end{aligned}$$

For the mean-type mapping $\left( {\mathcal {G}},{\mathcal {H}}\right)$, applying Theorem 2, we prove the following

Theorem 3

A mean $K:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right)$ is invariant with respect to the mean-type mapping $\left( {\mathcal {H}}, {\mathcal {G}}\right)$, i.e.

$$\begin{aligned} K\circ \left( {\mathcal {H}},{\mathcal {G}}\right) =K \end{aligned}$$

if, and only if,  $K=K_{{\mathcal {H}},{\mathcal {G}}}$, where

$$\begin{aligned} K_{{\mathcal {H}},{\mathcal {G}}}\left( x,y\right) :=\frac{2xy}{\pi }\int _{0}^{ \frac{\pi }{2}}\frac{d\varphi }{\sqrt{x^{2}\sin ^{2}\varphi +y^{2}\cos ^{2}\varphi }},\quad x,y>0. \end{aligned}$$

(2)

Moreover, the sequence $\left( \left( {\mathcal {H}},{\mathcal {G}}\right) ^{n}:n\in {\mathbb {N}}\right)$ of iterates of $\left( {\mathcal {H}},{\mathcal {G}} \right)$ converges uniformly on compact subsets of $\left( 0,\infty \right) ^{2}$,

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {H}},{\mathcal {G}}\right) ^{n}=\left( K_{{\mathcal {H}},{\mathcal {G}}},K_{{\mathcal {H}},{\mathcal {G}}}\right) , \end{aligned}$$

and $K_{{\mathcal {H}},{\mathcal {G}}}$ is a strict, continuous and symmetric mean.

Proof

The function $K_{{\mathcal {H}},{\mathcal {G}}}$ is a strict mean, as $K_{{\mathcal {H}},{\mathcal {G}}}\left( x,x\right) =x$ for all $x>0$, and $K_{{\mathcal {H}}, {\mathcal {G}}}$ is strictly increasing in each of the variables (see Remark 2).

Note that

$$\begin{aligned} K_{{\mathcal {H}},{\mathcal {G}}}\left( x,y\right) =\frac{1}{K_{{\mathcal {A}}, {\mathcal {G}}}\left( \frac{1}{x},\frac{1}{y}\right) },\quad x,y>0 \text {,} \end{aligned}$$

(3)

where $K_{{\mathcal {A}},{\mathcal {G}}}$ is given by (1), which implies formula (2). Since, for all $x,y>0,$

$$\begin{aligned} {\mathcal {H}}\left( x,y\right) =\frac{2xy}{x+y}=\frac{1}{{\mathcal {A}}\left( \frac{1}{x},\frac{1}{y}\right) },\quad {\mathcal {G}}\left( x,y\right) =\frac{1}{{\mathcal {G}}\left( \frac{1}{x},\frac{1}{y}\right) }, \end{aligned}$$

making use of (3), the invariance equality $K_{{\mathcal {A}},{\mathcal {G}}}=$ $K_{{\mathcal {A}},{\mathcal {G}}}\circ \left( {\mathcal {A}},{\mathcal {G}}\right)$ guaranteed by Theorem 2, and again of (3), we get for all $x,y>0$,

$$\begin{aligned} \left( K_{{\mathcal {H}},{\mathcal {G}}}\circ \left( {\mathcal {H}},{\mathcal {G}} \right) \right) \left( x,y\right)= & {} \frac{1}{K_{{\mathcal {A}},{\mathcal {G}} }\left( \frac{1}{{\mathcal {H}}\left( x,y\right) },\frac{1}{{\mathcal {G}}\left( x,y\right) }\right) }=\frac{1}{K_{{\mathcal {A}},{\mathcal {G}}}\left( \frac{1}{ \frac{1}{{\mathcal {A}}\left( \frac{1}{x},\frac{1}{y}\right) }},\frac{1}{\frac{1 }{{\mathcal {G}}\left( \frac{1}{x},\frac{1}{y}\right) }}\right) } \\= & {} \frac{1}{K_{{\mathcal {A}},{\mathcal {G}}}\left( {\mathcal {A}}\left( \frac{1}{x}, \frac{1}{y}\right) ,{\mathcal {G}}\left( \frac{1}{x},\frac{1}{y}\right) \right) }=\frac{1}{K_{{\mathcal {A}},{\mathcal {G}}}\left( \frac{1}{x},\frac{1}{y}\right) } \\ {}= & {} K_{{\mathcal {H}},{\mathcal {G}}}\left( x,y\right) , \end{aligned}$$

which shows that $K_{{\mathcal {H}},{\mathcal {G}}}$ is invariant with respect the mean-type mapping $\left( {\mathcal {H}},{\mathcal {G}}\right)$. Now the result follows from Lemma 1. $\square$

Remark 9

Lemma 1.(iv) and the symmetry of the means ${\mathcal {H}},{\mathcal {G}}$ imply that the analogous result (with $K=K_{{\mathcal {G}},{\mathcal {H}}}=K_{{\mathcal {H}},{\mathcal {G}}}$) holds true for the mean-type mapping $\left( {\mathcal {G}}, {\mathcal {H}}\right)$.

Remark 10

The identity

$$\begin{aligned} {\mathcal {G}}\left( x,y\right) -{\mathcal {H}}\left( x,y\right) =\frac{\sqrt{xy}}{ x+y}\left( \sqrt{x}-\sqrt{y}\right) ^{2},\quad x,y>0, \end{aligned}$$

and, the easy to verify inequalities

$$\begin{aligned} \mathcal {H\circ }\left( \left( {\mathcal {H}},{\mathcal {G}}\right) ^{n}\right) \le K_{{\mathcal {H}},{\mathcal {G}}}\le G\mathcal {\circ }\left( \left( {\mathcal {H}},{\mathcal {G}}\right) ^{n}\right) ,\quad n\in {\mathbb {N}}\text {,} \end{aligned}$$

imply, for every $x,y>0,$ the fast convergence of the sequence $\left( \left( {\mathcal {H}},{\mathcal {G}}\right) ^{n}\left( x,y\right) :n\in {\mathbb {N}} \right)$ to $\left( K_{{\mathcal {H}},{\mathcal {G}}}\left( x,y\right) ,K_{ {\mathcal {H}},{\mathcal {G}}}\left( x,y\right) \right) .$

We conclude the considerations in this section with

Remark 11

For each mean-type mapping $\left( M,N\right)$ such that$\ M,N\in \left\{ {\mathcal {A}},{\mathcal {G}},{\mathcal {H}}\right\}$, the explicit form of the $\left( M,N\right)$-invariant mean is known. It is ${\mathcal {G}}$ for the mappings $\left( {\mathcal {A}},{\mathcal {H}}\right)$ and $\left( {\mathcal {H}}, {\mathcal {A}}\right)$ (Theorem 1); $K_{{\mathcal {A}},{\mathcal {G}}}$ for $\left( {\mathcal {A}},{\mathcal {G}}\right)$ and $\left( {\mathcal {G}},{\mathcal {A}}\right)$ (Theorem 2); $K_{{\mathcal {H}},{\mathcal {G}}}$ for $\left( {\mathcal {H}},\mathcal {G }\right)$ and $\left( {\mathcal {G}},{\mathcal {H}}\right)$ (Theorem 3); ${\mathcal {A}}$ for $\left( {\mathcal {A}},{\mathcal {A}}\right)$; ${\mathcal {H}}$ for $\left( {\mathcal {H}},{\mathcal {H}}\right)$; and ${\mathcal {G}}$ for $\left( {\mathcal {G}},{\mathcal {G}}\right)$.

4 Conclusions for bivariable mean-type mappings involving quasi-arithmetic means

Remark 12

Let $I,J\subset {\mathbb {R}}$ be intervals and $\varphi :I\rightarrow J$ be a continuous strictly monotonic function. If $M:J^{2}\rightarrow J$ is a mean in J then the function $M^{\left[ \varphi \right] }:I^{2}\rightarrow I$,

$$\begin{aligned} M^{\left[ \varphi \right] }\left( x,y\right) :=\varphi ^{-1}\left( M\left( \varphi \left( x\right) ,\varphi \left( y\right) \right) \right) ,\quad x,y\in I\text {,} \end{aligned}$$

is a mean in I, and it is called quasi-M-arithmetic.

Every mean of the form ${\mathcal {A}}^{\left[ \varphi \right] }$ is called quasiarithmetic.

The following is an easy consequence of Theorem 1.

Corollary 1

Let $I\subset {\mathbb {R}}$ be an interval and $\varphi :I\rightarrow \left( 0,\infty \right)$ be a continuous strictly monotonic function. Then a mean $K:I^{2}\rightarrow I$ is invariant with respect to the mean-type mapping $\left( {\mathcal {A}}^{\left[ \varphi \right] },{\mathcal {H}}^{\left[ \varphi \right] }\right)$, i.e. $K\circ \left( {\mathcal {A}}^{\left[ \varphi \right] },{\mathcal {H}}^{\left[ \varphi \right] }\right) =K$, if, and only if,

$$\begin{aligned} K={\mathcal {G}}^{\left[ \varphi \right] }. \end{aligned}$$

Moreover, the sequence $\left( \left( {\mathcal {A}}^{\left[ \varphi \right] }, {\mathcal {H}}^{\left[ \varphi \right] }\right) ^{n}:n\in {\mathbb {N}}\right)$ of iterates of $\left( {\mathcal {A}}^{\left[ \varphi \right] },{\mathcal {H}}^{ \left[ \varphi \right] }\right)$ converges uniformly on compact subsets of $I^{2}$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {A}}^{\left[ \varphi \right] }, {\mathcal {H}}^{\left[ \varphi \right] }\right) ^{n}=\left( {\mathcal {G}}^{\left[ \varphi \right] },{\mathcal {G}}^{\left[ \varphi \right] }\right) . \end{aligned}$$

Remark 13

The geometric mean ${\mathcal {G}}$, (as well as ${\mathcal {G}}^{\left[ \varphi \right] })$ is quasiarithmetic.

The suitable consequences of Theorems 2 and 3 read as follows.

Corollary 2

Let $I\subset {\mathbb {R}}$ be an interval and $\varphi :I\rightarrow \left( 0,\infty \right)$ be a continuous strictly monotonic function. Then a mean $K:I^{2}\rightarrow I$ is invariant with respect to the mean-type mapping $\left( {\mathcal {A}}^{\left[ \varphi \right] },{\mathcal {G}}^{\left[ \varphi \right] }\right)$, i.e. $K\circ \left( {\mathcal {A}}^{\left[ \varphi \right] },{\mathcal {G}}^{\left[ \varphi \right] }\right) =K$, if, and only if,

$$\begin{aligned} K=\left( K_{{\mathcal {A}},{\mathcal {G}}}\right) ^{\left[ \varphi \right] }. \end{aligned}$$

Moreover, the sequence $\left( \left( {\mathcal {A}}^{\left[ \varphi \right] }, {\mathcal {G}}^{\left[ \varphi \right] }\right) ^{n}:n\in {\mathbb {N}}\right)$ of iterates of $\left( {\mathcal {A}}^{\left[ \varphi \right] },{\mathcal {G}}^{ \left[ \varphi \right] }\right)$ converges uniformly on compact subsets of $I^{2}$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {A}}^{\left[ \varphi \right] }, {\mathcal {G}}^{\left[ \varphi \right] }\right) ^{n}=\left( \left( K_{{\mathcal {A}},{\mathcal {G}}}\right) ^{\left[ \varphi \right] },\left( K_{{\mathcal {A}}, {\mathcal {G}}}\right) ^{\left[ \varphi \right] }\right) . \end{aligned}$$

Remark 14

The arithmetic–geometric mean $K_{{\mathcal {A}},{\mathcal {G}}}$ is not quasiarithmetic, since

$$\begin{aligned} K_{{\mathcal {A}},{\mathcal {G}}}\left( K_{{\mathcal {A}},{\mathcal {G}}}\left( 1,200\right) ,K_{{\mathcal {A}},{\mathcal {G}}}\left( 10,100\right) \right) =44.722 \end{aligned}$$

and

$$\begin{aligned} K_{{\mathcal {A}},{\mathcal {G}}}\left( K_{{\mathcal {A}},{\mathcal {G}}}\left( 1,10\right) ,K_{{\mathcal {A}},{\mathcal {G}}}\left( 200,100\right) \right) =46.496, \end{aligned}$$

so $K_{{\mathcal {A}},{\mathcal {G}}}$ does not satisfy the bissymetry equation (cf. J. Aczél [1, p. 281]). It follows that for every continuous strictly monotonic $\varphi :I\rightarrow \left( 0,\infty \right) ,$ the mean $\left( K_{{\mathcal {A}},{\mathcal {G}}}\right) ^{\left[ \varphi \right] }$ is not quasiarithmetic.

Corollary 3

Let $I\subset {\mathbb {R}}$ be an interval and $\varphi :I\rightarrow \left( 0,\infty \right)$ be a continuous strictly monotonic function. Then a mean $K:I^{2}\rightarrow I$ is invariant with respect to the mean-type mapping $\left( {\mathcal {H}}^{\left[ \varphi \right] },{\mathcal {G}}^{\left[ \varphi \right] }\right)$, i.e. $K\circ \left( {\mathcal {H}}^{\left[ \varphi \right] },{\mathcal {G}}^{\left[ \varphi \right] }\right) =K$, if, and only if,

$$\begin{aligned} K=\left( K_{{\mathcal {H}},{\mathcal {G}}}\right) ^{\left[ \varphi \right] }. \end{aligned}$$

Moreover, the sequence $\left( \left( {\mathcal {H}}^{\left[ \varphi \right] }, {\mathcal {G}}^{\left[ \varphi \right] }\right) ^{n}:n\in {\mathbb {N}}\right)$ of iterates of $\left( {\mathcal {H}}^{\left[ \varphi \right] },{\mathcal {G}}^{ \left[ \varphi \right] }\right)$ converges uniformly on compact subsets of $I^{2}$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {H}}^{\left[ \varphi \right] }, {\mathcal {G}}^{\left[ \varphi \right] }\right) ^{n}=\left( \left( K_{{\mathcal {H}},{\mathcal {G}}}\right) ^{\left[ \varphi \right] },\left( K_{{\mathcal {H}}, {\mathcal {G}}}\right) ^{\left[ \varphi \right] }\right) . \end{aligned}$$

Remark 15

Similar calculations as in the previous remark show that neither $K_{ {\mathcal {H}},{\mathcal {G}}}$ nor $\left( K_{{\mathcal {H}},{\mathcal {G}}}\right) ^{ \left[ \varphi \right] }$ is a quasiarithmetic mean.

5 Higher dimensional case and some open questions

In the three dimensional case the arithmetic, geometric and harmonic means ${\mathcal {A}}:\left( 0,\infty \right) ^{3}\rightarrow \left( 0,\infty \right)$, ${\mathcal {G}}:\left( 0,\infty \right) ^{3}\rightarrow \left( 0,\infty \right)$ are of the form

$$\begin{aligned} {\mathcal {A}}\left( x,y,y\right) =\frac{x+y+z}{3}, \quad {\mathcal {G}} \left( x,y,z\right) =\root 3 \of {xyz}, \quad {\mathcal {H}}\left( x,y,z\right) =\frac{3xyz}{yz+zx+xy}, \end{aligned}$$

and the following counterpart of Theorem 1 holds true (see also [9] )

Theorem 4

A unique function ${\mathcal {B}}:\left( 0,\infty \right) ^{3}\rightarrow \left( 0,\infty \right)$ satisfying the invariance equality

$$\begin{aligned} \mathcal {G\circ }\left( {\mathcal {A}},{\mathcal {B}},{\mathcal {H}} \right) ={\mathcal {G}}\text {,} \end{aligned}$$

is of the form

$$\begin{aligned} {\mathcal {B}}\left( x,y,z\right) =\frac{xy+yz+zx}{x+y+z},\quad x,y,z>0, \end{aligned}$$

and it is a strict symmetric mean. Moreover, the sequence $\left( \left( {\mathcal {A}},{\mathcal {B}},{\mathcal {H}}\right) ^{n}:n\in {\mathbb {N}}\right)$ of iterates of the mean-type mapping $\left( {\mathcal {A}},{\mathcal {B}},{\mathcal {H}} \right)$ converges uniformly on compact subsets of $\left( 0,\infty \right) ^{3}$, and

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {A}},{\mathcal {B}},{\mathcal {H}}\right) ^{n}=\left( {\mathcal {G}},{\mathcal {G}},{\mathcal {G}}\right) . \end{aligned}$$

Proof

The first result is an easy calculation. The “moreover” result follows from Lemma 1. $\square$

Remark 16

Let $L,M,N\in \left\{ {\mathcal {A}},{\mathcal {B}},{\mathcal {H}},{\mathcal {G}} \right\}$ be such that $\left\{ L,M,N\right\} \ne \left\{ {\mathcal {A}}, {\mathcal {B}},{\mathcal {H}}\right\}$. In view of Lemma 1 there exists a unique $\left( L,M,N\right)$-invariant mean $K_{L,M,N}:\left( 0,\infty \right) ^{3}\rightarrow \left( 0,\infty \right)$ such that

$$\begin{aligned} K_{L,M,N}=K_{L,M,N}\circ \left( L,M,N\right) \text {;} \end{aligned}$$

moreover, the sequence $\left( \left( L,M,N\right) ^{n}:n\in {\mathbb {N}} \right)$ of iterates of the mean-type mapping $\left( L,M,N\right)$ converges uniformly on compact subsets of $\left( 0,\infty \right) ^{3}$, and

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( L,M,N\right) ^{n}=\left( K_{L,M,N},K_{L,M,N},K_{L,M,N}\right) . \end{aligned}$$

We know the explicit form of $K_{L,M,N}$ only in the following five cases:

• (i) (Theorem 4) $K_{{\mathcal {A}},{\mathcal {B}},{\mathcal {H}}}=$ ${\mathcal {G}}$;

• (ii) $K_{{\mathcal {A}},{\mathcal {A}},{\mathcal {A}}}=$ ${\mathcal {A}}$, $K_{ {\mathcal {B}},{\mathcal {B}},{\mathcal {B}}}=$ ${\mathcal {B}}$, $K_{{\mathcal {H}}, {\mathcal {H}},{\mathcal {H}}}=$ ${\mathcal {H}}$, $K_{{\mathcal {G}},{\mathcal {G}}, {\mathcal {G}}}=$ ${\mathcal {G}},$ (these formulas follow immediately from the reflexivity of a mean).

This remark and the Gauss Theorem motivate the following open problems:

Problem 1

Is it possible to determine the explicit formulas for the means $K_{{\mathcal {G}},{\mathcal {A}},{\mathcal {H}}}$, $K_{{\mathcal {G}},{\mathcal {A}},{\mathcal {B}}}$, $K_{{\mathcal {G}},{\mathcal {B}},{\mathcal {H}}}$?

Problem 2

Is it true that one of means $K_{{\mathcal {G}},{\mathcal {A}},{\mathcal {H}}}$, $K_{ {\mathcal {G}},{\mathcal {A}},{\mathcal {B}}}$, $K_{{\mathcal {G}},{\mathcal {B}},{\mathcal {H}}}$ has an explicit form like the mean

$$\begin{aligned}{} & {} K\left( x,y,z\right) \\ {}{} & {} \quad =\left( \frac{\pi }{2}\right) ^{2}\left( \int _{0}^{ \frac{\pi }{2}}\int _{0}^{\frac{\pi }{2}}\frac{d\varphi d\psi }{\sqrt{ x^{2}\cos ^{2}\varphi \cos ^{2}\psi +y^{2}\sin ^{2}\varphi \cos ^{2}\psi +z^{2}\sin ^{2}\psi }}\right) ^{-1}\text {?.} \end{aligned}$$

Note that Theorem 1 and Theorem 4 have the following extension to an arbitrary k-dimensional case (see [9])

Theorem 5

For a fixed positive integer $k\ge 2$ and $p\in \left\{ 1,\ldots ,k\right\} ,$ define ${\mathcal {M}}_{p}:\left( 0,\infty \right) ^{k}\rightarrow \left( 0,\infty \right)$ by

$$\begin{aligned} {\mathcal {M}}_{p}\left( x_{1},\ldots ,x_{k}\right) :=\frac{\mathop {\displaystyle \sum }\limits _{i=1}^{k}x_{i}\left( \mathop {\displaystyle \prod }\limits _{j=1}^{p-1}x_{i+j-1}\right) }{ \mathop {\displaystyle \sum }\limits _{i=1}^{k}\mathop {\displaystyle \prod }\limits _{j=1}^{p-1}x_{i+j-1}},\quad \text {where }x_{p+i}=x_{i}\,\ \text {for all }i\in \mathbb {N\,}\text {.} \end{aligned}$$

Then (i) for each $p\in \left\{ 1,\ldots ,k\right\}$, ${\mathcal {M}}_{p}$ is a strict symmetric mean;

(ii) ${\mathcal {M}}_{1}={\mathcal {A}}$ and ${\mathcal {M}}_{k}={\mathcal {H}}$, where

$$\begin{aligned} {\mathcal {A}}\left( x_{1},\ldots ,x_{k}\right) =\frac{x_{1}+\cdots +x_{k}}{k},\quad {\mathcal {H}}\left( x_{1},\ldots ,x_{k}\right) =\frac{k}{\frac{1}{x_{1}} +\cdots +\frac{1}{x_{k}}},\quad x_{1},\ldots ,x_{k}>0\text {;} \end{aligned}$$

(iii) the geometric mean ${\mathcal {G}}:\left( 0,\infty \right) ^{k}\rightarrow \left( 0,\infty \right) ,$

$$\begin{aligned} {\mathcal {G}}\left( x_{1},\ldots ,x_{k}\right) =\left( \mathop {\displaystyle \prod }\limits _{p=1}^{k}x_{p}\right) ^{1/k} \end{aligned}$$

is invariant with respect to the mean-type mapping $\left( {\mathcal {M}} _{1},\ldots ,{\mathcal {M}}_{k}\right)$, i.e.

$$\begin{aligned} \mathcal {G\circ }\left( {\mathcal {M}}_{1},\ldots ,{\mathcal {M}}_{k}\right) ={\mathcal {G}}; \end{aligned}$$

the sequence $\left( \left( {\mathcal {M}}_{1},\ldots ,{\mathcal {M}}_{k}\right) ^{n}:n\in {\mathbb {N}}\right)$ of iterates of the mean-type mapping $\left( {\mathcal {M}}_{1},\ldots ,{\mathcal {M}}_{k}\right)$ converges uniformly on compact subsets of $\left( 0,\infty \right) ^{k}$, and

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( {\mathcal {M}}_{1},\ldots ,{\mathcal {M}}_{k}\right) ^{n}=\left( {\mathcal {G}},\ldots ,{\mathcal {G}}\right) . \end{aligned}$$

Clearly, this result leads to similar open problems in the general case $k\ge 3$.

6 Applications in solving functional equations

Theorem 1 allows us to prove the following

Proposition 1

A function $F:\left( 0,\infty \right) ^{2}\rightarrow {\mathbb {R}}$, that is continuous at every point of the diagonal $\left\{ \left( x,x\right) :x>0\right\} ,$ satisfies the functional equation

$$\begin{aligned} F\left( x,y\right) =F\left( \frac{x+y}{2},\frac{2xy}{x+y}\right) ,\quad x,y>0, \end{aligned}$$

(4)

if, and only if, there is a single variable continuous function $f:\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} F\left( x,y\right) =f\left( \sqrt{xy}\right) \,,\quad x,y>0. \end{aligned}$$

Proof

Assume that F is continuous at every point of the diagonal. and satisfies (4), i.e. $F=F\circ \left( {\mathcal {A}},{\mathcal {H}}\right)$. Hence, by induction, $F=F\circ \left( \left( {\mathcal {A}},{\mathcal {H}}\right) ^{n}\right)$ for every $n\in {\mathbb {N}}$, where $\left( {\mathcal {A}}, {\mathcal {H}}\right) ^{n}$ is the nth iterate of $\left( {\mathcal {A}}, {\mathcal {H}}\right)$. In view of Theorem 1, making use of the assumed continuity of F on the diagonal, for all $x,y>0$, we obtain

$$\begin{aligned}{} & {} F\left( x,y\right) =\lim _{n\rightarrow \infty }F\circ \left( \left( {\mathcal {A}},{\mathcal {H}}\right) ^{n}\left( x,y\right) \right) =F\left( \left( {\mathcal {G}},{\mathcal {G}}\right) \left( x,y\right) \right) \\ {}{} & {} \quad =F\left( \sqrt{xy},\sqrt{xy} \right) =f\left( \sqrt{xy}\right) , \end{aligned}$$

where $f(t):=F\left( t,t\right)$ for $t>0.$ The converse implication is obvious. $\square$

Corollary 4

(i) A function $\phi :\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ is continuous at the point 1 and satisfies the functional equation

$$\begin{aligned} \phi \left( t\right) =\frac{2t}{t+1}\phi \left( \frac{\left( t+1\right) ^{2} }{4t}\right) ,\quad t>0, \end{aligned}$$

(5)

if, and only if,

$$\begin{aligned} \phi \left( t\right) =\phi \left( 1\right) \sqrt{t},\quad t>0, \end{aligned}$$

where $\phi \left( 1\right)$ is an arbitrary real number.

(ii) A function $\phi :\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ is continuous at the point 1 and satisfies the functional equation

$$\begin{aligned} \phi \left( t\right) =\frac{1+t}{2}\phi \left( \frac{4t}{\left( t+1\right) ^{2}}\right) ,\quad t>0, \end{aligned}$$

if, and only if,

$$\begin{aligned} \phi \left( t\right) =\phi \left( 1\right) \sqrt{t},\quad t>0, \end{aligned}$$

where $\phi \left( 1\right)$ is an arbitrary real number.

Proof

(i) Assume that $\phi :\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ satisfies equation (5). Define $F:\left( 0,\infty \right) ^{2}\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} F\left( x,y\right) :=y\phi \left( \frac{x}{y}\right) ,\quad x,y>0, \end{aligned}$$

and note that $\phi$ is continuous at the point 1 if and only if F is continuous at every point of the diagonal $\left\{ \left( x,x\right) :x>0\right\} .$ Replacing t in (5) by $\frac{x}{y},$ we have

$$\begin{aligned} \phi \left( \frac{x}{y}\right) =\frac{2\frac{x}{y}}{\frac{x}{y}+1}\phi \left( \frac{\left( \frac{x}{y}+1\right) ^{2}}{4\frac{x}{y}}\right) , \quad x,y>0, \end{aligned}$$

whence

$$\begin{aligned} y\phi \left( \frac{x}{y}\right) =\frac{2xy}{x+y}\phi \left( \frac{\frac{x+y}{ 2}}{\frac{2xy}{x+y}}\right) , \quad x,y>0, \end{aligned}$$

which shows that F satisfies the functional equation

$$\begin{aligned} F\left( x,y\right) =F\left( \frac{x+y}{2},\frac{2xy}{x+y}\right) ,\quad x,y>0. \end{aligned}$$

In view of Proposition 1, we have

$$\begin{aligned} y\phi \left( \frac{x}{y}\right) =f\left( \sqrt{xy}\right) , \quad x,y>0, \end{aligned}$$

and, taking $y=x$, we get $f\left( x\right) =\phi \left( 1\right) x$ for all $x>0$. It follows that

$$\begin{aligned} y\phi \left( \frac{x}{y}\right) =\phi \left( 1\right) \sqrt{xy},\quad x,y>0, \end{aligned}$$

whence

$$\begin{aligned} \phi \left( t\right) =\phi \left( 1\right) \sqrt{t},\quad t>0. \end{aligned}$$

Since the function $\phi \left( t\right) =\phi \left( 1\right) \sqrt{t}$ ($t>0$) satisfies (5), the converse implication holds true. This completes the proof of (i).

Setting $F:\left( 0,\infty \right) ^{2}\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} F\left( x,y\right) :=x\phi \left( \frac{y}{x}\right) ,\quad x,y>0, \end{aligned}$$

and arguing similarly as above, we can prove (ii). $\square$

Similarly, applying Theorem 2 and Theorem 3, we obtain the following results.

Proposition 2

A function $F:\left( 0,\infty \right) ^{2}\rightarrow {\mathbb {R}}$ that is continuous on the diagonal $\left\{ \left( x,x\right) :x>0\right\}$ satisfies the functional equation

$$\begin{aligned} F\left( x,y\right) =F\left( \frac{x+y}{2},\sqrt{xy}\right) ,\quad x,y>0, \end{aligned}$$

if, and only if, there is a single variable continuous function $f:\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} F\left( x,y\right) =f\left( K_{{\mathcal {A}},{\mathcal {G}}}\left( x,y\right) \right) \,,\quad x,y>0. \end{aligned}$$

Corollary 5

(i) A function $\phi :\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ is continuous at the point 1 and satisfies the functional equation

$$\begin{aligned} \phi \left( t\right) =\sqrt{t}\phi \left( \frac{t+1}{2\sqrt{t}}\right) ,\quad t>0, \end{aligned}$$

if, and only if,

$$\begin{aligned} \phi \left( t\right) =\phi \left( 1\right) \frac{\pi }{2}\left( \int _{0}^{ \frac{\pi }{2}}\frac{d\varphi }{\sqrt{t^{2}\cos ^{2}\varphi +\sin ^{2}\varphi }}\right) ^{-1},\quad t>0, \end{aligned}$$

where $\phi \left( 1\right)$ is an arbitrary real number.

(ii) A function $\phi :\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ is continuous at the point 1 and satisfies the functional equation

$$\begin{aligned} \phi \left( t\right) =\frac{1+t}{2}\phi \left( \frac{2\sqrt{t}}{t+1}\right) ,\quad t>0, \end{aligned}$$

if, and only if,

$$\begin{aligned} \phi \left( t\right) =\phi \left( 1\right) \frac{\pi }{2}\left( \int _{0}^{ \frac{\pi }{2}}\frac{d\varphi }{\sqrt{\cos ^{2}\varphi +t^{2}\sin ^{2}\varphi }}\right) ^{-1}, \quad t>0, \end{aligned}$$

where $\phi \left( 1\right)$ is an arbitrary real number.

Proposition 3

A function $F:\left( 0,\infty \right) ^{2}\rightarrow {\mathbb {R}}$ that is continuous on the diagonal $\left\{ \left( x,x\right) :x>0\right\}$ satisfies the functional equation

$$\begin{aligned} F\left( x,y\right) =F\left( \frac{2xy}{x+y},\sqrt{xy}\right) ,\quad x,y>0, \end{aligned}$$

if, and only if, there is a single variable continuous function $f:\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} F\left( x,y\right) =f\left( K_{{\mathcal {H}},{\mathcal {G}}}\left( x,y\right) \right) \,,\quad x,y>0. \end{aligned}$$

Similarly as Corollary

Corollary 6

(i) A function $\phi :\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ is continuous at the point 1 and satisfies the functional equation

$$\begin{aligned} \phi \left( t\right) =\sqrt{t}\phi \left( \frac{2\sqrt{t}}{t+1}\right) , \quad t>0, \end{aligned}$$

if, and only if,

$$\begin{aligned} \phi \left( t\right) =2\phi \left( 1\right) \frac{t}{\pi }\int _{0}^{\frac{ \pi }{2}}\frac{d\varphi }{\sqrt{t^{2}\sin ^{2}\varphi +\cos ^{2}\varphi }}, \quad t>0, \end{aligned}$$

where $\phi \left( 1\right)$ is an arbitrary real number.

(ii) A function $\phi :\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ is continuous at the point 1 and satisfies the functional equation

$$\begin{aligned} \phi \left( t\right) =\frac{2t}{1+t}\phi \left( \frac{1+t}{2\sqrt{t}}\right) , \quad t>0, \end{aligned}$$

if, and only if,

$$\begin{aligned} \phi \left( t\right) =2\phi \left( 1\right) \frac{t}{\pi }\int _{0}^{\frac{ \pi }{2}}\frac{d\varphi }{\sqrt{\sin ^{2}\varphi +t^{2}\cos ^{2}\varphi }}, \quad t>0, \end{aligned}$$

where $\phi \left( 1\right)$ is an arbitrary real number.

The above corollaries can be easily generalized as follows:

Remark 17

If $M,N:\left( 0,\infty \right) ^{2}\rightarrow \left( 0,\infty \right)$ are the continuous, homogeneous, symmetric means such that

$$\begin{aligned} \left| M\left( x,y\right) -N\left( x,y\right) \right| <\left| x-y\right| ,\quad x,y>0\text {, }x\ne y, \end{aligned}$$

then there exists K, their Gauss composition [8, 10]. Moreover, a continuous function $\varphi :\left( 0,\infty \right) \rightarrow {\mathbb {R}}$ that is continuous at 1,  satisfies the functional equation

$$\begin{aligned} \varphi \left( t\right) =N\left( t,1\right) \varphi \left( \frac{M\left( t,1\right) }{N\left( t,1\right) }\right) , \quad t>0, \end{aligned}$$

if and only if

$$\begin{aligned} \varphi \left( t\right) =\varphi \left( 1\right) K\left( t,1\right) ,\quad t>0. \end{aligned}$$