if F is strictly montone increasing function F:[0,1] such that F(0)=0 and F(1)=1 and satisfies jensens equation at x=0 and at x=1
ie the restricted form where F([1+x])=F(1)/2+F(x)/2
and F(x/2)=F(0)/2+F(x)/2
jensen's equality with y=0 using F(0)=0, F(1/2^nx)=1/2^nF(x) forall non-negative n, (doubling halving,quartering etc)
and jensen at one one, F(1)=1, F([1+x]/2=1/2+F(x)/2
I presume such functions will be effectively dyadic-ally linear, and given strict monotonic increasing will be linear (or will be if continuity has to be presumed)
I
and
presume it will if entails symmetry F(1-x)+F(x)=1, which given midpoint convexity at 1 and 0 collapses these into e qualities and gives this result for F strictly monotone increasing.
Moreover, it appears, that given jensen equation at 0 by itself , it also gives back jensens equality at one, and if the above does not hold true may be stronger as it does appears to entail F(x)=x, ie a function that symmetric odd around 1/2, \forall x in dom
(F)[0,1]F(1-x)+F(x)=1,s F strictly monotone increasingF:[0,1]\to[0,1], F(1)=1, (and thus F(0)=0, and F(1/2)=1/2) as result )and is a doubling function F(1/2^nx)=1/2^nF(x)
F(1)=1;; where
effectively entails jensen equality at 1 given jensen equality being satisfied at 0 by the equation below with F(0)=0 which gives F(1)=1 given symmetry and note that F(1/2)=1/2 is effectively buried in it).
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