In all these examples, r and α are such that 0 < r < 1, α : = 1 / ( 1 − r ) > 1. Example 3.1 Let f ( x ) = x be the associated function of G.
In particular, every (symmetric) homogeneous monotone mean m is m 1 -decomposable for some homogeneous mean m 1. Proof Let f be the associated function of m and set α = 1 / ( 1 − r ) > 1.
Example 3.3 Let f ( x ) = ( x − 1 ) / ln x, x > 0, with f ( 1 ) = 1 be the associated function of L. A simple computation leads to f ( x α ) f ( x α − 1 ) = α − 1 α x α − 1 x α − 1 − 1, which is the associated function of E α − 1, α, that is, E α − 1, α π r = L.
Example 3.2 Let f ( x ) = ( x + 1 ) / 2 be the associated function of A. Clearly, we have f ( x α ) f ( x α − 1 ) = x α + 1 x α − 1 + 1, which is the associated function of G α − 1, α.
Example 3.5 Let f ( x ) = x − 1 2 a r c s i n x − 1 x + 1, x > 0, with f ( 1 ) = 1 be the associated function of P. Obviously, we have f ( x 2 ) f ( x ) = ( x + 1 ) arcsin x − 1 x + 1 arcsin x 2 − 1 x 2 + 1.
be the associated function of E p, q, with convenient forms for p = q ≠ 0 and p ≠ 0, q = 0. Using Proposition 3.1, we obtain f p, q π ( x 2 ) f p, q π ( x ) = ( x 2 q − 1 x 2 p − 1. x p − 1 x q − 1 ) 1 / ( q − p ), which after simplification remains f p, q π ( x 2 ) f p, q π ( x ) = ( x q + 1 x p + 1 ) 1 / ( q − p ).
