Do there exist two functions $f$ and $g$ continuous and strictly increasing $[0,1] \to \mathbf{R}$ such that
$$ f^{-1}\left(\frac{1}{3} f(x) + \frac{2}{3} f(y)\right)<g^{-1}\left(\frac{1}{3} g(x) + \frac{2}{3} g(y)\right)$$
and
$$ f^{-1}\left(\frac{2}{3} f(x) + \frac{1}{3} f(y)\right)>g^{-1}\left(\frac{2}{3} g(x) + \frac{1}{3} g(y)\right)$$
for all $0\le x < y \le 1$?




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This question, related to generalized means (characterized by Kolmogorov-Nagumo and de Finetti), arises from the fact that such system has a [infinitely many] solution in the case that the sign of the inequalities are the same, e.g. $(f(t),g(t))=(t^\alpha,t^\beta)$ for some $\alpha \neq \beta$. What happens if the signs are reversed?



Symmetric inequality on generalized means
No. Denote $f(x)=a$, $f(y)=b$, then $g(f^{-1}(t))=h(t)$. Applying $g$ to both parts of first inequality we get $h(\frac13 a+\frac23 b)<\frac13 h(a)+\frac23h(b)$. It means that $h$ is convex (proof below), analogously the second inequality means that $h$ is concave.


Lemma. If $h$ is continuous on $[0,1]$ and $h(tx+(1-t)y)<th(x)+(1-t)h(y)$ for fixed $t\in (0,1)$ and all $x<y$, then $h$ is convex.


Proof. Consider the set $S$ of $s\in [0,1]$ such that $h(sx+(1-s)y)\leq s\cdot h(x)+(1-s)\cdot h(y)$ for $x\leq y$. It is clearly closed since $h$ is continuous. Assume that $S\ne [0,1]$, then the complement $[0,1]\setminus S$ is open, take any (maximal by inclusion) interval $(\alpha,\beta)\subset [0,1]\setminus S$, $\alpha\in S,\beta\in S$. We easily get $t\alpha+(1-t)\beta\in S$, a contradiction.


