http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/homomorphisms.pdf

This is called homomorphic function. F(x+y) = F(x) + F(y). Here is F(1-(x+y)) = F(1-x-y) = F(1)-F(x)-F(y).

http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/homomorphisms.pdf

@jay Thanks for this

@ jay is this similar to the orthogonal triple and double property gleason frame  function. It appears that it entails cauchy's equation on the unit trianle when you have any (1)given given two arbitary elements which sum to one,iff and only have function values which sum to 1 and(2) iffany given three which sum to one have function

(3) any given four which sum to one, the function values sum to one, and for any given three which sum to 2, the function values must sum to to 2

You've already got the answer to the question what's the name of the function of type (A). Of course, this is homomorphic function, because of each mapping: f: G --> G' with the property:      f(xoy) = f(x) o' f(y),  (where  (G,o) and (G',o') are two groups)

in algebra and mathematics generally called the homomorphism.

Dear Mirjana it appears to me that when you both functions of type (A) and type (B)

That is akin to the functions used by gleason he had odd-ness but his domain was different F(x)+F(-x)=0 I presume that is [-2,2] akin in someways to F(1-x)+F(x)=1 over [0,1] if one were to consider 0.5 the origin.  ( iam not sure which of his properties unit trace, self adjoint , any given orthogonal three which sum to one (that appears to be the homomorphic function)  corresponds to F(-x)+F(x)=0,

I presume that F(x)+F(-x)=0 specifies F(0)=0, ie F(0)=-F(-0)=-F(0) which entails 2F(0)=0 and thus F(0)=0 and given specifies F(2) given F(-2) on that domain and range, although given the homomorphism function it specifies F(1)=1 ie as 0+0+1 =F(0)+F(0)+F(1)=1 but as F(0)=0 presumably then F(1)=1

Arguably also with f(0)=0  one could get the second equation in the sense that if x+y+z=1, then x+y+0=1 so F(x)+F(y)+F(0)=1, yet F(0)=0

non-negativity and the any given three property (although I dont if it held everywhere)

 I call the homo-morphic  function

sum symmetry because I also noticed that if it made sense to use that as x+y=z+m\to F(x)=F(y)=F(M)+F(z) then where x=0 F(x)=0, that (x+y)+0\to z+m\ implies

(1) that x+y=z+m, and moreover that F(x+y)+F(0)=F(z)+F(m0, but as F(0)=0 this reduces to some form of cauchy equation F(x+y)=F(z)+(m), where x+y=z+m, ie F(z+m)=F(z)+F(M).  in effect giving you the symmetry equation as well any given two, what appears to be restricted form of cauchy equation.

It appears that the first two equations generalize to cauchy equation over the unit triangle due to the common terms x+y+z=1, x+ (Y+z)=1 using the second equation gives F(y+z)+F(x)=1, F(x)+F(Y)+F(Z)=1,using the first two equations

F(y+z)=1-F(X), F(Y)+F(Z)=1-F(x)= ie F(y+z)=F(z)+F(Y)

I thought that one (A)needed this to hold for infinitely many elements which sum to 1, that the function values do, but it appears that first two equations taken collectively appear to entail that at least for rationals,

I thought that  one needed an equation for at least two,distinct numbers, that one

So basically I am asking, is given one has F:[0,1]\to [0,1]

(A)(the homorphic functionx+y+z=1 \\\\leftrightarrowF(x)+F(y)+F(z)=1

(B)x+y=1\\\leftrightarrowF(x)+F(y)=1 ;(symmetry oddnes around 1/2)

(c) \sumn=1_x_ix+y+z+m=1 \\\leftrightarrowF(x_1)+F(x_2)+F(x_3+F(x_4)=1

(D) \sum^(n-3,4,5,6)_xi=2\leftrightarrow \sum^i=ni=iF(x)+F(y)+F(z)=2

(

(E)F strictly monotonic

are the further properties(C), , (D) redundantrequired for F under that domain and rangeF:[0,1] to [0,1]

Given F satisfies (A) the homomorphism equation and (B) symmetry,and(E) strictly monotone increasing, F:[0,1]\to [0,1] to give F(x)=x and continuity

I attached a copy of a paper where on page 25 they use both to get to cauchy equation although I am wondering if there it really constrains some of the irrational values whose sum >1

Mirjana Vukovic · and Jay

When you say a homo-rmorphic function are you referring to Cauchy's equation itself: F(x+y)=F(x)+F(y); or if (A) F(1-x-y)=1-F(x)-F(y) i presume you mean its a homomorphism of F(x+y)=F(x)+F(y);

Hence the name,

'the homomorphism', to distinguish it from other functional equations that are homomorphism of other 'lesser equations' given the hype/significance about cauchy's equation? (ie the 'one and only true' homomorphism etc)

. Or do you mean a homo-morphism of the identity function.As many functions are homomorphic if they preserve some certain structural properties.

I see that (A) is very close to being cauchy equation and given F(0)=0 gives cauch'ys equation,  and (C) as well as F(1)=1, likewise F(1)=1 or (C) likewise gives the same result. F(x+y)=F(x)+F(y) with F(0)=0 and F(1)=1, 

follow from  just one of   (1) F(1)=1, (2) F(0)=0  or (3) (C) F(1-x)=1-F(x). You dont need to presume them all, given one of the three it entails the rest given (A). I also see (C)  that it entails jensens equation but with a fixed pt as well.

If you mean its a hom-omophism of Cauchy's equation? as it very nearly directly implies. If you know of more detail or literature on (A) and where its referred to as 'the hom=omorphism 'that would be much appreciated/

with regard to (A) F(1-x-y)=1-F(x)-F(y)

and (C) F(1-x)=F(1-x) together it appears that these two just entail F(x+y)=F(x)+F(y) directly with F(1)=1 and thus gives F(x)=x over F[0,1]to [0,1]

where by this you mean that  the numerical representation is structure preserving with regard to addition.

an

The weaker F(1-x-y)=1- F(x)-F(y) where F: [0,1] to [0,1]

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

which appear to reduce to it , and to the identity, for on that restricted domain and range, given just one of F(0)=0 or F(1)=1 ,  ie F(1)=1 gives

F(1-x-y)=1- F(x)-F(y)

F(1)=F(1-0-0)=1-F(0)-F(0)=1-2F(0)

so if F(1)=1 then  1=1-2F(0) and -2F(0)=0 and thus

given F(0)=0 , and  letting y=0 in F(1-x-y)=1- F(x)-F(y) w

Then given F(0)=0 one cangets that F(1-x)= F(1-x-0)=1-F(x)-F(0)=1-F(x) ie

F(1-x)=1-F(x) and so we can use this to establish F(x+y)=F(x)+F(y)

by let x be x+y in F(1-x)=1-F(x) 

so F(1-(x+y)=1-F(x+y) or rather (1)1-F(x+y)=F(1-(x+y)

and  as original equation gives (2)F(1-x-y)=1- F(x)-F(y) 

we have that from (1) and (2)

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

-1*F(x+y)=-1*[F(x)+F(y) which is just F(x+y)=F(x)+F(y), in one sense stronger then cauchy equations as it gives F(1)=1 from F(0)=0 and vice versa, and in another way not quite as strong as it doesnt directly give F(0)=0 or entail cauchy equations

But it gives jensens equation really in a form ie x+y=z+m then 1-x-y=1-z-m and therefore

F(1-x-y)=F(1-z-m) and from the equation

x+y= z+m implies  1-x-y=1-z-m implies F(1-x-y)=F(1-z-m) implies 1-F(x)-F(y)=F(1-x-y)=F(1-z-m)=1-F(z)-F(m) implies 1-F(x)-F(y)=1-F(z)-F(m)

implies F(x)+F(y)=F(z)+F(m)

x+y=z+m implies F(x)+F(y)=F(z)+F(m). and now let z=1/2(x+y)= m=1/2(x+y).  F(z)=F(x)=F(1/2(x+y) a thus clearly z+m=1/2(x+y) +(1/2(x+y)=x+y 

so z+m=x+y,and thus F(z)+F(m)=F(x)+F(y) and thus

s t z=1/2(x+y)= m=1/2(x+y). F(z)=F(m)=F(1/2(x+y) 2F(1/2(x+y))=F(1/2(x+y)+F(1/2(x+y))=. F(z)+F(m)

so 2F(1/2(x+y))=F(z)+F(m)=F(x)+F(Yy) ie

2F(1/2(x+y) =F(x)+F(y) implies F(1/2(x+y)=F(x)/2+F(y)/2

. But in the presence of the singular fixed pt F(0)=0, or F(1)=1 it entails both the other fixed pt and cauchy's equations s (and in fact given F(0)=0, or F(1)=1, leads to the identity version of cauchys equation . It entails jensens equation in a sense (constant sum) x+y =z+m iff F(x)+F(y)=F(z)+F(m),  with a fixed pt F(1/3)=1/3 in addition, and is odd around that pt. ie F(1/3)=F(1-2/3)=F(1-1/3-1/3)=1-F(1/3)-F(1/3)

so F(1/3)=1-2F(1/3) and  gives 3F(1/3)=1 F(1/3)=1/3

2F(1/3)=2 times 1/3=2/3 and 1/3=F(1/3)=F(1- 2/3)=F(1-1/3-1/3)=F(1-[1/3+x]-(1/3-x)]= 1-F(1/3+x)-F(1/3-x) so 1/3=1-F(1/3+x)-F(1/3-x)  subtracting one from both sides

we have that  -2/3=-F(1/3+x)-F(1/3-x) multiplying through by -1 we have

2/3=F(1/3+x)+F(1/3-x)

which also gives 2F(1/3)=F(1/3+x)+F(1/3-x)  as F(1/3)=1/3 and that any two elements which sum to 2/3 in the domain have a function sum of 2/3

F(1/3-x)+F(1/3+x)=F(1-(   ie F(2/3)=F(1-1/3

I Presume you that  F(x+y)=F(x)+F(y) gives F(1+(-[x+y]))=F(1)+F(-[x+y]) cauchy additivity

=F(1)-F(x+y);   as cauchy equation is odd

=F(1)-F(x)+F(Y), by applying cauchy equation a second time.

F(0)=0 ie F(0)=F(0+0)=F(0)+F(0) thus F(0)=F(0)+F(0) so 0=F(0)-F(0)=F(0), ie F(0)=0

and, as F(0)=0, we can derive oddness 0=F(0)= F(-x+x)=F(-x)+F(x) and so F(-x)+F(x)=0 and thus, -F(x)=F(-x)

=F(1)-F(x)-F(y) a