Is A, Strictly increasing , Symmetric Probability function with symmetric inverse probability,Midpoint Convex, iff it is midpoint concave?
Is this derivation (A)-rough proof (see below) valid?
And thus does F satisfy Jensen's equation if F is midpoint convex?. And what are the names of the properties in $(5.1)$ and $(5.2) $ ,a bi-conditional and inverse version of symmetry in nd $(6)$ the. $(5.1)$ and $(5.2) $ a form of dd-ness; leftright reflection symmetry
: 1.$$F:[0,1]\to[0,1]$$ 2. Strictly Monotone Increasing and in-jective. 3.$$F(0)=0, \,F(1)=1,\, F(0.5)=0.5$$
4.$$F\text{ is}\text{ Midpoint Convex},\forall x,y\in\text{dom}(F)=[0,1] F(\frac{x+y}{2})\leq(\frac{F(x)}{2}+\frac{F(y)}{2})$$ *5.Satisfies, the the bi-conditional symmetry properties 5.1.(1)$$\forall(x,(1-x))\in\text{Dom}(F)=[0,1]:F(x)+F(1-x)=1$$ : 5.2.(2)$$\forall(p,(1-p))\in\text{Im}(F)\subseteq [0,1]:F^{-1}(p)+F^{-1}(1-p)=1$$. $$(x,x_2)\in\text{dom}(F);\, [x_1+x_2=1][\leftrightarrow[F(x_1) + F(x_2)=F(1)=1]$$ $$(p_1,p_2)\in \text{codom}(F);\,[p_1+p_2=1]\leftrightarrow [F^{-1}(p_1) + F^{-1}(p_2)=F^{-1}(1)=1]$$ 6. $$∀(x_1,x_2)∈Dom(F)=[0,1]:[x_1+x_2]>1↔[F(x_1)+F(x_2)]>1$$ $$∀(x_1,x_2)∈Dom(F)=[0,1]:[x_1+x_2]<1↔[F(x_1)+F(x_2)]<1 $$∀ $$\forall(p_1,p_2)∈Im(F);[p_1+p_2]>1↔[F^{−1}(p_1)+F^{−1}(p_2)]>1$$ $$\forall(p_1,p_2)∈Im(F);[p_1+p_2]>1↔[F^{−1}(p_1)+F^{−1}(p_2)]
Does Midpoint Convexity(5), given symmetry conditions (5.1,5.2), andF injective (2), and real valued, (1), with boundary constraints (3) entail that F is Midpoint concavity and thus a Jensen function?
(A)Rough Proof $F:[0,1] \to [0,1]$; (A1)$F$ is in-jective and strictly monotonic increasing (A2) with $F(0)=0$, $F(0.5)=0.5$ $F(1)=1$
Premises $(1)$ Midpoint convexity $\forall(x,y)\in \text{dom}(F)=[0,1]:F(\frac{x+y}{2})\leq\frac{F(x)}{2}+\frac{F(y)}{2}$ $(1a)$ Midpoint convexity $F$ at all points and thus midpoint convex @$\frac{(1-y)+(1-x)}{2})$ this follows from $(1)$ $$ F(\frac{(1-y)+(1-x)}{2})\leq\frac{F(1-x)}{2}+\frac{F(1-y)}{2}$$ $(2)$ Symmetry:$\forall(x_1,x2)\in dom(F)=[0,1]: $(x_1+x_2)=1 \leftrightarrow F(x_1)+F(x_2)=1$
$(2a)$ Inverse symmetry: $\forall(p_1,p2) \in Im(F)=[0,1]: $(p_1+p_2)=1 \leftrightarrow F^{-1}(p_1)+F^{-1}(p_2)=1$
Notice that there is always one pair of domain values in the system for each mid point domain point, $m\in \text{dom}(F)$ , $m=\frac{x}{2} +\frac{y}{2}, \,\text{between}\, \frac{x}{2}\,{\&} \frac{y}{2},\text{is s.t}\, (1-m)\in\text{Dom}(F) \,\text{lies halfway between}\,\, \frac{1-x}{2}\in\text{dom}(F)\, {\&} \frac{1-y}{2}\in\text{dom}(F)$.
$$1-m=1-\frac{x+y}{2}= 1-(\frac{x}{2} + \frac{y}{2})$$
(1)$$=(1-\frac{x}{2}-\frac{y}{2})=\frac{1-x}{2}+\frac{1-y}{2}$$
(2) $$F(1-m)= F(1-\frac{x+y}{2})=F(\frac{2-(x+y)}{2})$$ $$=F(\frac{1 +1 -x-x}{2})=F(\frac{(1-x)+(1-y)}{2})$$
(2.1) From and (2) and (1) $$\frac{F(1-m)}{2} \leq \frac{F(1-x)}{2}+\frac{F(1-y)}{2}$$
By Midpoint Convexity and (2.1)
(3) $$F(\frac{(1-y)+(1-x)}{2})\leq\frac{F(1-x)}{2}+\frac{F(1-y)}{2}$$
Thus(2.1)
(3) $$F(1 - \frac{x+y}{2})\,\leq \, \frac{F(1-x)}{2}+\frac{F(1-y)}{2}$$
$$\text{And Midpoint convexity}$$ $$F(\frac{x+y}{2}) \leq \frac{F(x)}{2}+\frac{F(y)}{2}$$
(4)By the symmetry equations (1) an (2)**
(4.1)$$F(x)+F(1-x)=1$$ and (4.2)$$F^{-1}(p)+F^{-1}(1-p)=1$$
(4.3), $$F(t)+F(1-t)=1\, \Rightarrow\,F(1-t)=1-F(t)$$
From (4)$$\text{for any arbitrary t}\,\in \text{dom}(F)=[0,1];\,\text{there exists, one, and and only one}$$ $$ t_2=(1-t_1)\,t_2\in\, \text{dom(F)}=[0,1]\, [t_1+t_2=1]\quad\text{s.t;}$$
$$\forall (t_1, t_2)\in \text{dom(F)};\,[t_1+t_2=1]\, \Leftrightarrow\,[F(t_1)+F(t_2)=1]$$
(5)Thus letting $t=\frac{x+y}{2}$ in (4.3)
(5.1)$$F(t) +F(1-t)=1\rightarrow [(\frac{x+y}{2})+(1-\frac{x+y}{2})=1 \rightarrow F(\frac{x+y}{2})+ F(1-\frac{x+y}{2})=1$$
$$F(1-t)=F(1-\frac{x+y}{2})\,\text{which by (5) gives}$$
(5.2)$$F(1-t)=1-F(t)=1-F(\frac{x+y}{2})=F(1-\frac{x+y}{2})$$
So by (5.2)and by midpoint convexity** and (5.3)
(5.4) $$1-F(\frac{x+y}{2})= F(1-\frac{x+y}{2})= F(\frac{1-x}{2}+\frac{1-y}{2}) \leq F(\frac{1-x}{2}+F(\frac{1-y}{2}) $$
we derive (6):
(6) $$1-F(\frac{x+y}{2})\leq \frac{F(1-x)}{2}+\frac{F(1-y)}{2}$$
(7)By Symmetry again; however
$$1-F(x)=F(1-x)\,{\&}\,1-F(y)=F(1-y)$$
Thus from (6) and (7) we derive (8)
(8) $$1-F(\frac{x+y}{2})\leq \frac{F(1-x)}{2}+\frac{F(1-y)}{2}=\frac{1-F(x)}{2}+\frac{1-F(y)}{2}=1-[\frac{F(x)}{2}+\frac{F(y)}{2}]$$
and this entails (9)
(9) $$1-F(\frac{x+y}{2})\leq 1-[\frac{F(x)}{2}+\frac{F(y)}{2}]$$
Thus subtracting one from both sides we get (10)
(10)$$$-F(\frac{x+y}{2})\leq -[\frac{F(x)}{2}+\frac{F(y)}{2}]$$
but (10) implies (11)
(11)$$F(\frac{x+y}{2})\geq\frac{F(x)}{2}+\frac{F(y)}{2}$$
$$\text{but this is midpoint concavity}$$
$$\text{however, given F is midpoint convex}$$
$$\text{but by midpoint convexity}$$
(12)$$F(\frac{x+y}{2})\leq\frac{F(x)}{2}+\frac{F(y)}{2}$$
$$\text{then, at}\quad\frac{x+y}{2}\in\text{dom}(F)$$ $$\text{ given (11), and (12)}\,\text{F is midpoint convex}\,\text{and F is midpoint concave}$$
(13) $$\{[F(\frac{x+y}{2})\leq(\frac{F(x)}{2}+\frac{F(y)}{2})] \,\land\,[ F(\frac{x+y}{2})\geq(\frac{F(x)}{2}+\frac{F(y)}{2})] \}$$ $$(13)\Leftrightarrow F(\frac{x+y}{2})=(\frac{F(x)}{2}+\frac{F(y)}{2})$$
And from (13) see that, (C) Jensen's equation holds.
A similar argument is used to derive jensen's equation @ $1-m$ using $(1)$ $\in\text{Dom}(F)$.
(1)$$\text{ Use midpoint convexity at}\quad m=\frac{x+y}{2}\quad F(m)=F(\frac{x+y}{2})$$
$$F(\frac{x+y}{2})\leq(\frac{F(x)}{2}+\frac{F(y)}{2})\rightarrow 1-F(\frac{x+y}{2}) \geq 1-[\frac{F(x)}{2}+\frac{F(y)}{2}]$$
which by symmetry $F(x)+F(1-x)=1$ induces midpoint concavity @ $1-m=\frac{(1-x)+(1-y)}{2}\in \text{dom}(F)$, which by assumption is midpoint convex as well
as $F$ is midpoint convex at all points, and thus jensens equation
$$ \frac{(1-x)+(1-y)}{2}=1- \frac{x+y}{2} \rightarrow \frac{x+y}{2}+\frac{(1-x)+(1-y)}{2}=1 \rightarrow F(\frac{x+y}{2})+F(\frac{(1-x)+(1-y)}{2})=1 \rightarrow 1-F(\frac{x+y}{2}) = F(1- \frac{x+y}{2})=F( \frac{(1-x)+(1-y)}{2})$$
(2.1)$$F(\frac{(1-x)+(1-y)}{2})=1-F(\frac{x+y}{2}) \geq 1-[\frac {F(x)}{2}+\frac{F(y)}{2}]= \frac{1-F(x)}{2} + \frac{1-F(y)}{2}=\frac{F(1-x)}{2}+\frac{F(1-x)}{2}$$ using symmetry $1-F(y)=F(1-y)$ etc
but,$F$ is midpoint convex (everywhere;by assumption) therefore is midpoint convex at $1-m$ (as used in the previous derivation; ie by(4)) .
Which results in Jensen's equality (equation) at $1-m$ as well(3):
(3) $$F(\frac{(1-x)+(1-y)}{2})=F(\frac{1-x}{2})+F(\frac{1-x}{2})$$
See my post question for the additional properties that F has
satisfies.
References
1.Ger, Roman, Almost additive functions on semigroups and a functional equation, Publ. Math. 26, 219-228 (1979). ZBL0444.39004.
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5.[AczD] J. Acz´el, J. Dhombres, Functional equations in several variables.With applications to mathematics, information theory and to the natural and social sciences. Encyclopedia of Math. and its App., 31, CUP, 1989
One can derive continuity in multiple just by strict monotone continuity and the fact that midpoint so that its continuous at all points except @1 without symmetry due to the fact @0 one uses midpoint convexity and the fact that $F(0)=0$ to get that continuity @1: $$F(\frac{1}{2} x) \leq \frac{1}{2} F(x) \rightarrow F(\frac{1}{2^{n}}x) \leq \frac{1}{2^{n}} F(x) $$
by setting x=1 F(1)=1
$$ 0 \leq F(\frac{1}{2^{n}}) \leq \frac{1}{2^{n}} $$,
$$0 \leq F(\frac{1}{2^{n}} \times x) \leq [\frac{1}{2^{n}}\times F(x)] \rightarrow [0 \leq F(\frac{1}{2^{n}}\times 1)=F(\frac{1}{2^{n}})\leq [\frac{1}{2^{n}}\times F(1)] = \frac{1}{2^{n}} \times 1= \frac{1}{2^{n}}] $$ so that $n\to \infty$, or as $x\to 0$ from above, $F(x)\to 0$
And due to strict monotonicity (I presume one does not need to know much more about the limit from below at @0) and setting a particular fixed pt $F(1)=1, x=1$
, vanishes
Note however that $F(\frac{1}{2^{n}}) =\frac{1}{2^{n}})$ however due to symmetry
In any case to derive continuity @1 one uses the symmetry and that $$(A)[F(1-\frac{1}{2^{n}}x)+F(\frac{1}{2^{n}}x)=1 ]\rightarrow[F(1-\frac{1}{2^{n}}x)= 1- F(\frac{1}{2^{n}}x) $$
and that $$(B)F(\frac{1}{2^{n}}x) \leq \frac{1}{2^{n}}F(x) \rightarrow [1- F(\frac{1}{2^{n}}x)] \geq [1-\frac{1}{2^{n}}F(x)]$$
Using $(A)$ and $(B)$
$$ F(1-\frac{1}{2^{n}} x)=1-F(\frac{1}{2^{n}}x) \geq [1-\frac{1}{2^{n}}F(x)]$$ $$\leftrightarrow$$
Which reduces to: using $F(1)=1$ at the boundary and monotonicity
$$1 \geq F(1-\frac{1}{2^{n}} x) \geq 1-\frac{1}{2^{n}}F(x)$$
using $F(1)=1$;$x=1$
$$1 \geq F(1-\frac{1}{2^{n}}\times1)=1-F(\frac{1}{2^{n}}) \geq 1-\frac{1}{2^{n}} \times 1=1-\frac{1}{2^{n}}$$ so that $n\to \infty$, or as $x\to 1$ from below, $F(x)\to 1$
Although this is an equality as well. Technically In fact it appears that I can get all of the rationals, $F(r)=r$ (the dyadics and most non-dyadics) out just using midpoint convexity at $0$ and at $1$ as a result of symmetry and use an argument that agrees with F(x) on a dense set. This would limit the use of midpoint convexity, and make greater use of my symmetry equations I believe. It might require more use of my inverse symmetry.
$0$although it would be perhaps continuous at $1$ and $0$ and and strictly monotonic