Consider the following 2 bijective function
F:[0,1] to [0,1]
where (1) F strictly increasing, and absolutely continuous function on [0,1] and
and satisfies (2)), (3) and (4)
(2) F(0)=0, with F(1)=1 and F(1/2)=1/2
ie (3) \forall x in dom(F)=[0,1] F([1 +x]/2)
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notes
Note, also that star convex functions at 0, with F(0)=0 , ie F is super additive over the positive reals. see below and as the domain is positive and the co-domain non-negative theF is non-negative which given super-addivity implies and F monotone increasing,see below
as it F is a bijection, it is injective , and thus injective and given monotone increasing, it is strictly monotone increasing, see below (by definition I think)
and as its a bijective and thus surjective, it is a strictly monotone increasing surjection on a closed domain with the endpoint defined, which are generally continuous.
Note also that apart from all this F(x)=x for all reals >0.5 due to , \forall x\in dom (F):[0,1], \forall real t\in [0,1]: F(tx)=0.5; F(x)>=x but as we already have that
\forall x in dom (F)=[0,1], F(x)=x for x>0.5 x in dom(F) =[0,1] implies x=F(y)+0=F(y)
x>y, implies (y+(x-y))=x>y,
y+(x-y)=x implies F(x)=F(y+(y-x)>=F(y)+F(y-x). by super-additivity
so x
where as x>y and domain is positive =[0,1], 1>(x-y)>0; so (x-y) in dom (F)=[0,1]; so F(x-y) is defined and element of codom(F)=[0,1] , ie F(x-y)>=0 and all elements of codomain either =0 implies >=0 or in (0,1) and thus >0implies >=0 ie F>=0
so , (y+(x-y))=x implies F(x)=F(y+(y-x)>=F(y)+F(y-x) super additive and F is a function
implies that as : x>y ,x=y+(x-y)=>y,&
x> y implies that (x-y) in dom(F)=[0,1],implies F(x-y)>=0
x> y, implies,[ F(x-y)>=0 & F(x)=F(y +(x-y)) >=F(y)+F(y-x)].>=F(y)+0=F(y) implies
x>y i,mplies F(x)>=F(y) monotone, increasing,
(
F strictly monotone increasing as F b-ijective and thus injective and as 'monotone increasing and injective,' it is strictly monotone increasing; which appear to be true by defintiion.
as x>y also implies x !=y, so F(x)!= F(y) by injection
and as x>y implies x>=y implies F(x)>=F(y) by monotone increasing above
F[(x)>=yand F(x)!=F(y)]implies F(x)>F(y) hence strictly monotone increasing.
as x>y, implies x!=y, [F(x)!=F(y)injection
and x> y implies x>=y implies F(x)>=F(y)monotony
as x>y, implies[ x!=y, & x>=y] implies [F(x)!=F(y)& F(x)>=F(y) ]implies F(x)>F(y)so
x>y implies F(x)>F(y)
Thus F strictly monotone increasing.
That F is continuous follows from a theorem concerning strictly monotone surjective/bijective increasing function on closed intervals with range an interval. I suppose as its bijection F;[0,1] to [0,1] its IM(F)=codomain =[0,1] and thus its range is an interval and as strictly monotone it is continuous, and I think this generally implies that onclosed and bounded domains that F is uniformly continuous as well.
note that F(0)=0,gives
and F(1)=1, gives entails \forall real x in dom (F): F(x)
Dear @William Balthes,
F(x)=x is answer.· S
Miodrag Mateljevic, Miloljub Albijanic, Lagrange’s Theorem, Convex Functions and Gauss Map, Filomat 31:2 (2017), 321-334 DOI 10.2298/FIL1702321Mee
Thanks note that midpoint convexity is restricted. I presume that strict monotone increasing and absolute continuity is required. The answer would not follow from strict monotone increasing-ness alone without continuity being presumed. Would it fall out of just continuity and strict monotony.
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