Numbers x and 1-x   are symmetric wrt 1/2. Take  an arbitrary  function $f$   defined  for x \leq 1/2, and    f(1/2)=1/2 and extend  to   x \geq 1/2  by   f(x)= 1-f(1-x).

Thank for this;@miodrag

in my case, does this cover both the equation (the inverse form), ie they are bi-conditionals, and the fact that F(0)=0 and F(1)=1 F:[0,1]to [0,1] where,  F and F-1 strictly monotonic increasing and continuous

. Do you know what their official names, are -- and if i can find this in a textbook somewhere? If you have  reference I would much appreciate it (i know other functions with property in papers, but does azcel mention it anywhere in his books for example)

I think its a kind of -odd-ness (if you were to consider 1/2) the origin for example.

Some call the first equation, 'it left right reflection symmetry' balance, others call it central symmetry if F(0)=0 and F:[0,1] to [0,1] where F(1)=1 and F(0)=0. Not sure what the inverse function version is, which makes it a bi-conditional.

I am not sure what the inverse version is called,. Its only really symmetric once F(0)=0 otherwise one side below 0.5 might be out wack.

by which I mean if x1, x2

\forall (x1, x2)\in dom (F)s.t( x1+x2)neq=1, then F(x)+F(x1)\neq 1

\forall (p1, p2)\in codom(F)s.t (p1+p2)neq=1, then F-1(p)+F(p1)\neq 1

So its biconditional: I presume that any in-jective function:  F:[0,1] to [0,1] with F(0)=0, F(0.5)=0.5, F(1)=1,  F continuous which satisfies (1)the first identity

(1)\forall (x1, x2)\in dom (F)s.t( x1+x2)=1, then F(x)+F(x1)= 1, would satisfy

both of (A) and (B) as bi-conditionals , where (A) is the bi-conditional from of the first identity, where (B) is the bi-conditional form of the second identity.

And that if in addition F is  strictly monotone increasing and not merely, injective, would satisfy all of these below. So I presume that Minkowskis equation mark function would satisfy all of these conditions, wherever the inverse is defined? Except that it not endomorphic on the unit interval

(A)\forall (x1, x2)\in dom (F)s.t( x1+x2)=1, \leftrightarrow F(x)+F(x1)= 1

(B)\forall (p1, p2)\in codom(F)s.t (p1+p2)1, leftrightarrow then F-1(p)+F-1(p1)=1

\forall (x1, x2)\in dom(F)s.t (x1+x2)1

\forall (p1, p2)\in codom(F)s.t (p1+p2)1

. I presume its property that conway'ss function has inverse function property. Although I presume that any injective and strictly monotonic function satisfying F(x)+F(1-x)=1 would satisfy the (2) F-1(x)+F-1(1-x)=1 ;

Dear  William Balthes,

I do not know  the name for the identities (2) and (3) in the functional analysis literature.

In connection with your question we can

consider continuous solutions of equation

(1) F(1-x)+F(x)=1 for all x in [0,1].

For example $F(x)=x$ and $F(x)= sin2 (\pi x/2) $ are solutions.

At a first glance one can guess that $F(x)=x$ is only convex solutions of (1).

Dear William Balthes,

Suppose that  we consider consider continuous solutions of equation

(1) F(1-x)+F(x)=1 for all x in [0,1].

Numbers x and 1-x   are symmetric wrt 1/2.  Let   L be the graph  of    restrictation  F

for x \leq 1/2, and    F(1/2)=1/2. By symmetry wrt  to x=1/2 and then wrt  y=1/2 we can get the  complete graph of F.

It seems that using this we can prove  that any injective and strictly monotonic function satisfying F(x)+F(1-x)=1 would satisfy the (2) F-1(x)+F-1(1-x)=1.

Use the graphs of  F  and  F-1  are symmetric wrt y=x.

Dear Professor Miodrag Mateljeviv I thought that was the case, I just wanted to make sure I inferred myself that any injective F should satisfy (2) and if strictly increasing would satisfy the inequalities (3) ie

As (2)F-1(x)+F-1(1-x)=1, when attached to (1) F(x)+F=1,

can be read as x1+x2=1 if and only F(x1)+F(x2)=1 then we know that as if F, is injective, that if  F-1(p2)+F-1(1-p1 )neq 1,where p2=1-p1, and (p1,p2\in im(F) then sum p2+ p1=1 but their corresponding domain elements(x1,x2)\in domF(); x1=F-1(p) x2=F-1(1-p1 ) do not sum to one; and as x1+x2\ not equal 1 entails  x2 not equal to 1-x1.

Then as dom (F) =[0,1] there always exists some x3\in domF(x), st, x3=1-x1  F(x3)=F(1-x1) so  where F(x1)=p as x1=F-1(p).

So by (1)   F(x3)+F(x1)=1 , So F(x3)+p=1and and thus F(x3)=1-F(x1)=1-p,  so F(x3)=1-p, 

Yet  as  x2neq 1-x1, and  x3=1-x1 , so x2neq x3,. yet x2=F-1(1-p), which entails F(x2)=1-p, violating inject-ivity as x2 not equal x3  entails F(x2) not equal F(x3) but  ; F(x2)=F(x3)=1-p,

As for (3)x1+x21 ;

p1+p21 ;

We need only prove the 'if then' differection (3.1) and (3.3), the same arguments gives the 'if then', direction for (3.2) and (3.4), and 'if then of each these four, entailss' the only; the 'if then claims 'of F-1 (3.3) and (3.4) claims being /entailing the 'only if F claims in (3.1 and 3.2) and conversely the if then claims of (3.1 and 3.2) giving the only if claims for (3.3 and 3.4)

for the' if then 'of (3.1)

x1+x2x2, implies x4>x2, which entails x21-F(x6)=1-p1, so p2>1-p1 which contradicts the fact that p1+p2

@Miroslav Thats what I thought: That one could in effect consider, x= 1/2 the origin; as in the context of probability function F, F:[0,1] to [0,1], F(0)=0, F(1)=1 F(1/2)=1/2 and F strictly monotonic increasing and continuous, where the domain and range are [0,1],  1/2 is the midpoint/origin of that interval, in the same sense that 0 is the midpoint/origin of the real line.

Its often called a symmetry condition, or a symmetric probability function; central symmetric and left right reflection symmetry. I considered, that if one were to view 0.5; that this would be a general form of an odd -ness in the context of probability functions F:[0,1] to [0,1] given that it does not make a great deal of sense. in that context to speak of negative values. So just as the origin is  0 and actually the midpoint in some sense on a standard graph, which spans the positive and negative numbers:

Then in the context of probability F(0.5)=0.5 is the midpoint/origin, as it were, on [0,1], and with F(0)=0, F:[0,1] to [0,1] with F stricttly monotonic and continuous, F(1)=1, and F(0.5) (which are consequences) of this,

I thought that in two dimensions rotation by 180 degree symmetry (odd-ness) would be just left -right reflection symmetrey

So if one were to change the domain and range :to F1: [1/2, -1/2] to [-1/2 to 1/2];

F(0.5)=0.5 goes \to F1(0)=0,

F(0)=0 goes to F1(-1/2)=-1/2 ;

F(1)=1 goes to F1(1/2)=1/2

\forall (x)\in dom(F)[0,1];F(x) +F(1-x)=1,

goes to \forall (x) in  dom (F1) [1/2, -1/2] ],F1(x)+F1(-x)=0

\forall (x) in Im(F)\subset[0, 1][:(F)^(-1)(x)+(F)^(-1) (1-x)=1] goes to 

   \forall (x) in Im(F1)\subset[-1/2, 1/2]:(F1)^(-1)(-x)+(F1)^(-1)(x)=0

or rather ] \forall (x,y) in dom F1:([1/2, -1/2 ] x+y =0 \leftrightarrow (iff) F1(x) +F1(y)=0 and forall (p1, p2)\in IM(F1) =[-1/2,1/2] p1+p2=0 leftirghtarrow (iff) F1^(-1)(p1)+F^(-1)(p2)=0; as these used to be biconditioals as below

\forall (x,y)\in dom(F)[0,1]: [x+y=1 leftirghtarrow (iff) F(x)+F(y)=1]

\forall (p1,p2 )\in( Im(F)\subset[0,1]) :[ p1+p2=1 leftirghtarrow (iff) F-1(p1)+F-1(p2)=1

I presume the strict monotonicity and inequalities:

\forall (x,y) in dom(F)=[0,1]: [x+y>1 iff F(x)+F(y)>1]

\forall (x,y) in dom(F)=[0,1]: [x+y1]

\forall (p1,p2) in Im(F)[0,1]: p1+p20]

\forall (p,p2) in IM(F1)=[-1/2,1/2]: [p1+p2>0 iff [F1^(-1)(x)+F1^(-1)(y)]>]?

\f

orall (p1, p2)\in IM(F) p1+p2=1 leftirghtarrow (iff) F-1(p1)+F-1(p2)=1 goes to 

F(1-x) +F(x)=1

F-1(p)+F-1(p)=0

F1(-x)+F1(x)=0=F1(0) whi

I presume one would subtract 0.5 from the domain in the translation,  and for the at least for the symmetry eqiations and fixed pts and range values fixed pts.

F(0.75) becomes F1(0.75-0.25)=F1(0.25)=F-0.5 where F1 is the new function

F(x)+F(1-x)=1, as already entails  F(x+0.5)+F(0.5-x)=1= F(1)=2F(0.5); I presume one subtracts 0.5 from within the F itself  rather than as change of variable. Although one cannot just do this for the function values; I presume one would have to preserve ratios, in some sense so that F1(0.3)=0.1 F(x+-0.5)= 0.1+0.5

ie if for F:[0,1] to [0,1]  F(0.8)=0.6, where it must be bigger than 1/2 and its ranked due to strict monotonicity would this translate to: F1(0.3)=0.1  where its thus just F(x)=M corresponds to F1(x+[ -0.5]) =M-0.5 and conversely F1(x+ 0.5)=M1 +0.5 I suppose that would preserve ratios of lengths as the domain interval range lengths remain 1. I am not sure given this constraints

c

I presume the inequalities may not always easily translate,,

So that given (1) F(x)+F(1-x)=1 which entails already on F:[0,1] to [0,1] (by change of variable, so long x+0.5 in dom (F)=F: F(x+0.5)+F(0.5-x)=1= F(1)=2F(0.5)=2\times 0.5     , I presume one subtracts 0.5 from each F within in the inequality, within F, and from each function value in that inequality y

(1) goes to (2)F1(x)+F1(-x)= F1([x+0.5] - 0.5)+ F1((0.5-x) -0.5)=

F([x+0.5] - 0.5+0.5)-0.5+ F(0.5-x -0.5+0.5)-0.5=F(x+0.5)-0.5 +F(0.5-x)-0.5=F(x +0.5)+F(0.5-x)-0.5-0.5=[F(x +0.5)+F(0.5-x)]-1, where [F(0.5-x)+F(0.5+x)]=1 for the original F as its equivalent to F(1-x)+F(x)=1,  so [F(0.5-x)+F(0.5+x)]--1= F(1-x)+F(x)-1=1-1=0;

or one consider the ratio of the sum, when considering inequalities and sums in general, of course they do not sum to the unit event any more ie F1(-x)+F1(x)=0=F1(0) which is the translation of F(0.5)=0.5 not F1(-x)+F1(x)=0.5=F1(0.5) which is the translation of F(1)=1 ; but I presume that the first is more correct?

Not F1(-x)+F1(x)= F1([x+0.5] - 0.5)+ F1((0.5-x) -0.5)=F1(1-0.5)  =F1(0.5)=0.5\neq 2F1(0.5-0.5)=2\times 0=0

Not

F1(x-0.5) +F1(1-x-0.5)) =1-(0.5+0.5)=0 Fl(x-0.5)+F(0.5-x)=0

@Miodrag Mateljević; Indeed given Strict monotonic F,

F: [0,1] to [0,1]

F(0)=0;F(1)=1, F(0.5)=0.5

F(1-x)+F(x)=1

F strictly montonic

F:[0,1] to [0,1] which gives three fixed pts  one can show that jensens equality holds,  and continuity follows: ,

from midpoint convexity and F(1-x)+F(x)=1and F strictly monotonic (in fact only monotonicty is required for continuity although it appears it may not be required at all if jensens equality follows as I claim to have derived). I note that given monotonicity or strict monotonicity one can get F(x)=x just with midpoint convexity at 0 and 1 (mid-star convex at two pts) and from which F(x)=x, (note that only F(0)=0 needs to be stipulated, the other two follows from symmetry). midpoint convexity gives one inequality symmetry gives the other ie F(0.75)

This works for even star convexity or midpoint convexity restrict for 0 and 1

My worry is that any convex function is strictly monotonic increasing and continuous from the unit interval to the unit interva,l with the three fixed pts F(0)=0 , F(1)=1 and F(0.5)=0.5 will collapse into F(x)=x I think, rendering my symmetry requirements almost redundant

. So I want to weaken midpoint convexity as much as possible; it works even if F is midpoint convex at two pts, 1 and 0 , given symmetry 

) F(x)=x for all rationals and F is strictly monotonic F(1)=1 and F(0)=0 as it still entails doubling F(2x)=2x. I dont think mid-star convexity just at 0 or at a single pt will do it

nor would it work I think if F is strictly monotonic and continuous, where symmetry is not presumed, yet the three fixed pts above still hold and midpoint convex at those two pts ( i hope not) then again my symmetry would be redundant. I note that midpoint convex functions are rationally convexity (corresponding even without symmetry) could not one show that for all x>0.5F(x)=x (before continuity is applied) and even without symmetry but just from F:[0,1] to [0,1] F(1)=1 F(0)=0 F(0.5) and F perhaps strictly monotonic or injective

ie (1)use midpoint convextiy at 1 and 0 with F(1)=1 and F(0)=0 and the fact that it entails dyadic convexity or otherwise, use F(0.5)=0.5 and F(1)=1 to get this result F(3/4)leq 3/4 but also

(2)using rational convexity , F(0)=0,( sigma =2/3,x=0, y=0.75) and F(0.5)=0.5, which entails 0.5=F(0.5)=F(2/3\times 3/4)\leq 2/3 F(3/4) which entails 3/2\times0.5\leq F(3/4) which entails3/4 \leq  F(3/4) which is F(3/4)\geq 3/4 I hope this is false, as that follows otherwise without symmetry, and entails F(3/4)=3/4 and presumably for all ratiional (and this is upwards toward 1, which is where midpoint convex functions have trouble with continuity- it may not now, if its literally linear in that segment, especially given strict monotoniicity or just monotone increasing)

It works with symmetry thankfully but I hope only because of it (i note that star convex functions will be linear for all x>0.5 (or the fixed pt on that domain and range, with three fixed pts, two being at the end pt). I do not think mid-star convexity at a single pt will be enough to get my result even with symmetry- does it generalize to dyadics.

F(1)=1, follows from F(0)=0 ; I have already derived this, I want to use somethign weaker, One only needs monotonicity for another proof but it might be the case that none of this is required given it satisfies jensens equation and maybe its inverse form (ie often like cauchy equation with F(1) and F(0)=0, if the domain and range are non -negative, then jensen equality functions are generally continuous automatically apparently , However, despite this I want to keep strict monotonically and both symmetry equations my only worry is that with three fixed points a

if u introduce new variable s=t-0.5 and new function q(s)=f(s)-0.5 then the equation is rewritten as q(s)=q(-s), i.e. it is just a condition of antisymmetry.

Dear Professor Mukanova, by anti-symmetry do  do you mean-'oddness';

q(s)= -q(-s) rather than q(s)=q(-s)

with regard to F(t)+F(1-t)=1 equation , but around 0.5 (rather then 0)

Is anti symmetry a generalized around a pt odd-ness

However, I note that https://en.wikipedia.org/wiki/Parity_(mathematics), that odd-ness, sometimes is called anti-symmetry. Is anti symmetry the general name for odd-ness

F(1-x)+F(x)=1 expresses a kind of symmetry insofar as it reflects, a kind of equispacing or symmetry between function points around the the centre; For a function where F;[0,1] to [0,1] and F(0)=0 .

e F(0.5+s)+F(0.5-s)=1=2 *(0.5) =2F(0.5),

as F(1/2)=1/2; because F(1-0.5)+F(0.5)=1 so as F(1-0.5)=F(0.5) then  2F(0.5)=1 so  F(1/2)=1/2

  where its odd (in a sense around 0.5), ie centrally symmetric, left-right symmetric (its has the second minkowski question mark function property; https://en.wikipedia.org/wiki/Minkowski%27s_question_mark_function)

?(1-x) =1-?(x) where ? is the function.)

https://en.wikipedia.org/wiki/Minkowski%27s_question_mark_function

so          F(1/2)=F([1/2+x]/2 +[1/2-x]/2

and F(1/2-x)+F(1/2+x)=2F(1/2) so F(1/2-x)/2+F(1/2+x)/2=2F(1/2)/2=F(1/2)

so F(1/2-x)/2+F(1/2+x)/2=F(1/2)=F([1/2+x]/2 +[1/2-x]/2)

ie F([1/2+x]/2 +[1/2-x]/2)= F(1/2-x)/2+F(1/2+x)/2

I

(A) forall x,y in dom(F)=[0,1]  it says x+y=1 then F(x)+F(y)=1

I presume that for a strictly monotone increasing and surjective  function that satisfies the stronger (A2) where its iff, rather then if, then as in (A),

would be a function with an odd inverse function where its defined.

(A2)  \forall x,y in dom(F)=[0,1] x+y=1 iff F(x)+F(y)=1

i

where F:[0,1] to [0,1]; particularly when F(0)=0, its often called  a symmetric transformation probability function in certain cases,

I am also wondering about the bi-conditional form (C) for a strictly monotone increasing and continuous F:[0,1] to [0,1] where F(0)=0, and

and satisfies the stronger  (C) and the inequality variants

(C) 1 .for all x,y in dom(F)=[0,1] x+y=1 iff F(x)+F(y)=1;and

     (C1) for all (x,y) in Dom(F) =[0,1] x+y >1 iff F(x)+F(y)>1

(C1for all (x,y) in Dom(F) =[0,1] x+y 1

for all (x,y) in Dom(F) =[0,1] x+y

If it helps, The finite function satisfyiing f(x) + f(x-1) + f(x+1) ... = 1 is called atomic function

https://ru.m.wikipedia.org/wiki/Атомарная_функция

discovered in 1970th by Academician Rvachev.

Let me know if more info required.

Hi William,

to add to my last response, I believe the function you are looking for is called 'atomic function' shown in attached file and described in the link:

https://solitonscientific.github.io/AtomicString/AtomicString1.html

It is known for 50 years, with hundreds of papers/books published, mainly in Russian.

Recently I created some generalizations of AF called Atomic Strings which allows to represent AF via

combination of AStrings as shown on picture attached. AStrings and Atomic Functions are featured in my recent paper Article Atomic Strings and Fabric of Spacetime

I am also in AU, you can contact me if needed, and join my ResearchGate projects where AStrings are to be used in many branches of science.

Dear Sergio and Miroslav,

I believe that the first half of (A) :

\forall x in [0,1]:(A)F(1-x)+F(x)=1 .or \forall (x, y in [0,1]); x+y=1 then F(x)+F(y)=1

not the inverse /inequality partspart entailed by strict monotony which is the second half(

(A_1)\ foral(l (m,t)\inIM(F)\subseteq [0,1])F^-1(1-m)+F^-1(t)=1, or in otherowrds, which I presume to be equivalent:

\forall (m,t)\inIM(F)\subseteq [0,1];m=F^-1(x), t=F^-1(y)) :F(x)+F(y)=1 \to x+y=1

where (A) and (A_1) to give symmetry biconditional with (Ab), the bi-conditional : x+y =1 iff F(x)+F(y)=1

@ miroslav, I presume that injection by itself would be sufficient for the bi-conditional from (A),and the inverse function or the biconditional version of (A) whenver its its defined,

(one does not strict monotony) unless one requires the inequalities . The for the inverse function which are redundant, as defined through image are redunant (not codomain) I think .

inequalites

(AI1)\forall (x,y in [0,1]:x+y > 1 \iff F(x)+F(y)>1

(AI2)\forall (x,y in [0,1]):x+y

Dear, Sergio ; thatnks for that,A

are you refrring to (A) F(1-x)+F(x)=1or the stronger (B) F(1-(x+y)=1-F(x)-F(y)? (frame function really I call cauchy normalization).

(A) all x in [0,1]:(A)F(1-x)+F(x)=1 . (A)is sometimes called symmetry or oddness of the unit interval or at 1/2. Its a property of the minkowski question mark function :

https://en.wikipedia.org/wiki/Minkowski%27s_question-mark_function..

see http://www.linas.org/math/chap-minkowski.pdf. I recommend her 2008 article on arvix. by Lina Vepsas

Article Brunn-Minkowski inequalities in product metric measure spaces

https://en.wikipedia.org/wiki/Minkowski_functional

Sometimes called left- right reflection symmetry, central symmetry. Its been knnown about in utility antheory and Fechner distributions analysis in psychopshysics for some time

see the segal link See the segal link above in the question rank order and indifference, half through:

https://www.bc.edu/content/dam/files/schools/cas_sites/.../1993_jme_order.pdf

or half fthrough the warticle near the end , where (A) is refered as 'symmetry' in symmetric probability function but als i quantum mechanics:

Ssometimes (A) is icombined with (B) as a2-outcome frame function in povm s . These at must operate like 1-norm porbbailiites, its used in stead of oddness F(-x)=-F(x), which is considered weaker for some reason, I know gleason used oddness and (B).

I presume its derived through complementation and rank, or union + rank) . That how I did. in my system for info on thisthis see, condition in born rule type derivations for povms dim=2. on that: Although not information is given on how they arrived thtes equations. its ewasy to derive cauchy equation over the unit triangle from (A) and (B). F:[[0,1]\to [0,1]

\forall(x,y in [0,1] x+y1 \iff F-1(x)>G^-1(x)

\forall (x in IM(F))\forall y in IM (G) x+y>1 \iff F^-1(x)1 \iff F(x)>G(x)

\forall (x in[0,1]=dom(F))\forall y in)dom(F)(G) x+y>1 \iff F(x)

Hi William, I am referring to f(1-x) + f(x) = 1 with symmetry at x=1/2. It is featured in links

I sent you or my recent paper

https://www.researchgate.net/publication/320457739_Atomic_Strings_and_Fabric_of_Spacetime

Where it is used to build flat and curved spacetime quantum using exactly that property on lattice And explain gravity.

Hi William,

The link for downloadable paper regarding atomic string Astring(x) function is Article Atomic Strings and Fabric of Spacetime

Hi William,

The link for downloadable paper regarding atomic string Astring function is Article Atomic Strings and Fabric of Spacetime

Hi William,

The link for downloadable paper regarding atomic string Astring function is

Article Atomic Strings and Fabric of Spacetime

. Unlike others, the AString is infinitely smooth, with derivatives expressed via itself and the derivative of it is a well-known solitary atomic function.

Hi Sergio,

I believe that its often called oddness (around 1/2) or symmetry. I was just wondering if it has any special name apart from this. That its referred to as a kind of atomic function, is also interesting. I will look into that. You appear to have accidentally posted your answer above about five times.

In any case, thanks for the links. I will look at the above paper when I get time.

William

Hi Sergio,

I believe that its often called oddness (around 1/2) or symmetry. I was just wondering if it has any special name apart from this. That its referred to as a kind of atomic function, is also interesting. I will look into that. You appear to have accidentally posted your answer above about five times.

In any case, thanks for the links. I will look at the above paper when I get time.

William

Hi Sergio,

I believe that its often called oddness (around 1/2) or symmetry.

I was just wondering if it has any special name apart from this. That its referred to as a kind of atomic function, is also interesting. I will look into that. You appear to have accidentally posted your answer above about five times.

In any case, thanks for the links. I will look at the above paper you mention, when I get time.

This if probability function.

Dear Jamshed Nashir, sort of. Its describing a relationship between 2 probability functions.

hough in this, case its a numerical probabilistic like function function, whose domain=[0,1], (is real valued, not a set of events or sets). F's domain (and not just its image as in a standard probability function), is real valued. F's domain, is the image, of another "probability function", PR, over the same set of events, rather than that , " set of events", themselves.

I did not mention the second function above (its just the domain, x in [0,1]

above: Unlike a probability function, Dom(F) \subseteq [0,1], rather than, a set of sets, such as Dom(F)\subseteq P(\sigma), where P denotes the powerset, and sigma: the set of mutually exclusive and exhaustive atomic events

So here, F:[0,1]\to [0,1] rather than from : P(\sigma)\to ]0,1]

. Rather than a function, that takes in sets (rather than numbers), and whose image, is in [0,1].

In my case, the 'symmetry functional equation' is a property of a function from [0,1] to [0,1], rather than a function, whose domain is is some boolean algebra / or set of sets, and maps these to values \in [0,1] .

So, in my context, the symmetry equation, is saying that when the domain probability function ascribes 2 probability values to 2 distinct events, (A and ~A, for example), where PR(A)=x, and PR(~A)=1-x, which sum to 1, then, the second probabilistic function: F. must also ascribe, 2 values \in [0,1], to these events, which sum to 1.

In this case, the values assigned to the 2 events by the second function, could have have been described as, F(A) and F(~A), instead of F(1-x) and F(x), where these sum to 1 whenever the first function, PR, ascribes : ie as: "(1) PR(A)+PR(~A)=1, then F(A)+F(~A)=1".

Numerically, I described this as (2): F(x)+F(1-x)=1, instead for ease. Letting, x, 1-x stand for the values of the first function, and F(x) and F(1-x) stand for the values of the second function)

That is, t was easier to model the second function, F, as a normal numerical function (numerical domain and range, both = [0,1]). Where F's whose domain simply is, the set of all probability values assigned by the first function PR, [0,1]. In contrast, to having its domain being= domain of the first function, PR, as a set of sets, or events themselves. That is, rather than, a second probability function (over the same, a non-numerical domain: the set of events/the same algebra), its modelled as a function fo the probability values assigned to those events (rather than the events themselves).

So in this case: the domain values in [0,1], : "x, 1-x" , are probability values assigned to the events, A and ~A respectively, by the first probability function, PR. Not, the events, A and ~A, (as these would not be which would not be numbers, in [0,1] much less the probability values, PR(A)=x, PR(~A)=1-x), but rather the events: A, ~A, themselves).

F as per the the symmetry equation, F(x)+F(1-x)=1, models this property of the second function, F .

I only described one function (the second function, F) in the question. as the antecedent condition: "PR(A)+PR(~A)=1, in (1), as well as the consequent condition: F(A)+F(~A)=1 " in (1), is implicit when numerically described as in (2),

The antecedent condition, x+y=1, ("whenever the sum of the domain values=1):

Is described by the domain constraints, (x, 1-x), in (2);

And the consequent condition, (the sum of the function values of F), must be 1:

F(x)+F(y)=1, is, also, described by the same equation (2), F(x)+F(1-x)=1, as well : ie "if x+y=1, then F(x)+F(y)=1 "(the consequent: F(x)+F(y)=1, holds) whenever the antecedent, x+y=1, holds, is described, by the symmetry equation: (2) F(1-x)+F(x)=1. ,"ie where y=1-x, in the symmetry equation ( 2), not just the consequent condition in (1), above (F(A)+F(~A)=1), in F(1-x) +F(x)=1.

Another difference, which is not quite relevant for the question, but more about how I got to it, or why I am interested, is that in my case, there may be an entire equivalence class of events, that are, assigned the same domain value x, by the first function, PR, (say x=0.6 , denote the set of events with an objective probability of 0.6 by the first function,) and 0.4 describes the entire set of events assigned the objective probability 1-x=0.4, by the first function. So the events underlying the values, 0.4, and 0.6, may not be mutually exclusive or exhaustive events within the same boolean algebra, in the traditional sense. (of the form of A and ~A, respectively), but could be 2 hypothetical versions of the same event described by A, but where the probability values of the same event "described by A" differs between the 2 different probability spaces that are part of the same modal structures (an infinite spectrum, of connected finite probability spaces/worlds, with condiitonal logic connections between their truth values/hypothetical truth values, if A occurs in probability space 1: < A, 0.4 , ~A 0.6 >, then in some hypothetical world 2:, ~A obtains (~A2, w hypothetical ~A) in (A2=0.6, ~A2= 0.4), and A (denoted A_2)) fails to occur in world 2 where PR(A)=0.4+PR(A2)=0.6=1 rather then, PR(A)+PR(~A)=1 or Pr(A)+PR(~A2)=1, (in this latter case, the probability of A plus the probability of its hypothetical negation in some other world does not equal 1, but rather they have the same pr value 0.4, so PR(A)+PR(~A2)=0.4+0.4\neq 0.8

In Segal (1993, p. 392) (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwjv9-rgteLcAhUpBcAKHbpbDx8QFjABegQICBAC&url=https%3A%2F%2Fwww.researchgate.net%2Fprofile%2FWilliam_Balthes%2Fpost%2FF1-x_Fx1_what_is_the_name_of_this_functional_equation_and_in_biconditional_form%2Fattachment%2F59d626756cda7b8083a23273%2FAS%3A520622069514240%401501137382731%2Fdownload%2FOrder%2Bindifference%2Band%2Brank-dependent%2Bsymemtric%2Bsegal.pdf&usg=AOvVaw2dVsStwNm7DL-5igrOBsAb), it is a "symmetric probability function"

Cheers,

Jean

Dear Jean,

I know this, I was wondering if it has other names within the functional equation literature. This was where I first discovered a name in the context of utility representation theorems and decision theory, and which is why I use it in my context of a subjective probability function, as a numerical function, of another probability function (objective probability). The subjective probabilities, image values, sum to 1 when the their correspnding pre-images, objective probability values (the domain values) sum to one. Its often

I guess , as I and others have noted before, that its just oddness around 1/2.its just a translated version of oddness F(x)=-F(-x). Oddness around 1/2 ie letting x=1/2+x in the equation. F(1/2+x)+F(1/2-x)=2F(1/2)=2 *1/2=1, etc. The equation occurs in other contexts, like the minskowski question mark function for example/

We may also say that F is a function which commutes with the function

n: x -> 1-x (often called the standard fuzzy negation),

i.e., if we perform n and then F, or F and then n, we get the same result.

This is a property studied in detail in algebra.

I think, is the characteristic function. you can learn more in probability theory.