Let's have a generalized f-mean: $$A(x,y)=f\left(\frac{f^{-1}(x)+f^{-1}(y)}{2} \right)$$ where $f$ is a strictly monotonic function. Since $A$ is a mean it follows that $$\min(x,y) \leq A(x,y) \leq \max(x,y)$$ Now because of the facts $$\lim_{n \to \space + \infty} \sqrt[n]{ \frac{x^n + y^n}{2} } = \max(x,y)$$ and $$\lim_{n \to \space - \infty} \sqrt[n]{ \frac{x^n + y^n}{2} } = \min(x,y)$$ it is clear that there exist an $N$ and $M$ such that $$\sqrt[N]{\frac{x^N + y^N}{2}} \leq A(x,y) \leq \sqrt[M]{\frac{x^M + y^M}{2}}$$ for every $(x,y)$ pair in a finite interval. From here on I can just assume (but not prove) that there exists a probability density function $p$ having the following property $$A(x,y) = \int^{+\infty}_{-\infty} p(t) \sqrt[t]{\frac{x^t + y^t}{2}} \, dt$$ This can be viewed as some sort of transorm of $f$, like $$p(t) = P\{f\}(t)$$ My question is given an $f$ or $A$ how can one find this function $p$ (or does my assumption above even hold)? If it's easy, can you provide an example for me for a given $A(x,y)$ like $\sqrt{xy}$?
Edit: Since for the arithmetic mean $p$ should be $\delta(t-1)$ we can allow $p$ to be a distribution, instead of a function. Other means would also require $p$ to be a generalized function.