Condition for weigthed quasi-arithmetic means A real mean is a symmetric function $M : \mathbb{R}\times\mathbb{R}\to\mathbb{R}$ with $M(x,x)=x$, which is monotonuos in each variable.
According to Wikipedia: https://en.wikipedia.org/wiki/Quasi-arithmetic_mean, and the german paper "Aufbau von Mittelwerten mehrerer Argumente. II." in which Georg Augmann proved in 1935 that for a mean that is an analytic and symmetric function, the balancing property
$$
M(M(x,M(x,y)),M(y,M(x,y)))=M(x,y)
$$
implies, that $M$ is a quasi-arithmetic mean, thus
$$
M(x,y)=f^{-1}(\frac{f(x)+f(y)}{2})
$$
for an monotonous, continous function $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$.
For a general weighted quasi-arithmetic mean:
$$
M(x,y)=f^{-1}(w\cdot f(x)+(1-w)\cdot f(y)),
$$
the balancing property is obviously not always true. Instead one can show rather easily, that some form of medialty property is true:
$$
M(M(a,b),M(c,d))= M(M(a,c),M(b,d)).
$$
If one assumes the medialty property (for $a<b<c<d$ or in general) are there any results like Augmann's? Is it possible to show that a mean which fulfills this condition and an analytic function is a weighted, quasi-artihmetic mean? Are there any related conditions impylying this form? 
$Edit:$ a quasi-arithmetic mean with $w\neq 0.5$ obviously doesn't satisfy symmetry any more and is thus no mean according to my previous unthoughtful definition. 

Theorem 1 on page 571  in this  paper seems to answer my question.
I might elaborate, once i figured out how to read their reported source and understand it.
