Does anyone know the general continuous functional form of these constraints. Given, Midpoint convexity, it appears to satisfy jensen equality; symmetry appears to give one the other half. Midpoint concavity. I am not sure what has to be satisfied in order to give the same results as I had to use F(0.5)=0.5 at the very least, and F(0)=0 (and presumably F(1)=1.
F:[0,1]\to[0,1]; both closed and bounded intervals
1. Strictly and monotonely increasing; F(x1)> F(x2) iff x1>x2, F(x1)=F(x2)\to x1=x2
2. At least uniformly continuous
1.F(1)=1, F(0.5)=0.5 F(0), F(1)=max, all x>0.5 F(x)>0.5, else if x
I have a somewhat more formal proof F(2x)=2F(x) as in 5 seems to be entailed and can be extended to F(4x)=F(x)+F(x)+F(x)+F(x)=4(F(x). Numerically, I can directly get all power of 3,5 and maybe all rationals to a degree (and otherwise bounded between the values of F(x) above and below. I am not sure what the precise form of jensen equation,is that must be satisifed can the presumption of the max point F(1)=1, F(0)=0 and F(0.5)=0. One can infer that if given the inequalitys;
for all pairs of points
F(X/2+Y/2)=F(1-X)/2+F(1-Y)/2 as well; and thus that they enforce equality. So F(x/2+y/2)=F(x)/2+F(y)/2 appears to holds or roughly so, although it appears to bit odd that it this would fall out just as a result of the symmetry
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