F(x)=x  ???

Set

(1) for all x in dom(F):F(2x)=2F(x),

(2) for all x in dom(F) F(1/2x)=1/2F(x).

Define A(x)=2x and B(x)= 1/2x; dom(F) =[0,1]=I. Then A maps I=[0,1] onto [0,2] and B maps I=[0,1] onto [1/2, \infty].

It seems that you need to make some corrections of equations (1) and (2).

Dear Olev, I apologize I meant that is not F(x)=x

Dear Professor Miodrag, I meant to say that the first equation only applies if x

essentially just a function that satisfy jensens equation for x_0=0 F(x+0=0)=0  ie \forall  y,in dom F(y/2)=F( 0/2 +y/2)=F(x_0/2+y/2)= F(x0=0)/2+F(y0)/2= F(y)/2

ie \forall N\ in non negative integers, for all y in dom(F)=[0,1]; F(y/[2^n])=F(y)/[2^n]

 How does (3) 1/2^n F(x)=1/2^n F(x)?

imply that the first derivative is constant/second derivative is 0 , (1)for all  F'(x/2^n) where n is non negative integer, and moreover that its therefore constant everywhere, 

(2) as 1/2^n is not even the set of all dyadics rationals (for all m/2^n, 1

If one can show that said  function, F, given it absolutely continuous and twice continously differentiable must be a midpoint convex or midpoint-concave function and thus convex or concave given continuity, then it would follow?. At present it cant say anything much about the function value of F for x>0.5 except that its a continous interval; the dyadics are of the form F(x)=x for and x in [0,1] cap{ h/2^n, \for all non-negative integers, \forall integers h, such that 00.5, except that its continuous interval and strict monotone, where where F(1)=1, that IM (F(x); x\in [0.5, 1]=[0.5,1] but that does not imply that F(x)=x for all x>0.5, as there are plenty of bijective and continuous self maps, that are not the identity, unless sujrectivity and the continuous first derivative leads toa contraction

then the fact that it has two end points and multitudinous fixed pts  in the interior for x0 and be absolutely continuous for a mere, doubling halving function with dom(F)=[0,1] and F(1)=1,  to have any chance of being uniquely linear/ identity. function.

I  Note F(0)=0 is redundant as its implicit in

(3); F(x/2)=1/2 F(x) gives F(0)=F(0/2)= 1/2 F(0) so F(0)=1/2F(0) implies F(0)-1/2F(0)=0 implies 1/2F(0)=0 implies F(0)=0/2=0 implies F(0)=0

linear function ; normally speaking it would have to be jensen one at least two pts,(1) h at 0.

The second derivative is zero/first derivative is constant over all reals?

So essentially all it is a    F:[0,1] to [0,1], where F(1)=1 with a F twice continuous differentiable, where the first derivative is positive. F'(x)>0 (this appears to implies all the rest) satisfying

for (3) all x in dom [0,1] , and for all non negative integers  n, F( ( 1/2)^n) x)=(1/2)^n F(x) .

-Note that F(0)=0 need not be specified as its derivable from (3) halving/doubling

- and that F sur-jective need not be specified given F is  continuous and F(1)=1 and F(0)=0 with, dom(F)=[0,1], codom(F)=[0,1] as the intermediate value theorem requires that F attains every values between 1,and 0 and IM(F)=codom(F) and F sur-jective

- note that strict monotony need not be specified as injective continuous functions over closed domains are  monotone  increasing, and as a result strictly monotone ( as in-jectivity and monotony imply strictly monotony that F(0)=0 and F(1) =1 tells you the direction (strictly increasing) ;

also neither codomain F=[0,1], absolute continuity, continuity, surjectivity, nor injectivity nor strict monotony  or monotony, need to be specified given positive continuous  first derivative;

which gives absolute continuity and strict montone increasing-ness . GIven that  F(1)=1,dom (F)=[0,1] and F halving,

as halving gives F(0)=0, as 0 in dom (F)=[0,1]

and from strict monotone increasingness, from positive first derivative, (which entails injectivity as well) entails that:

-as dom(F)=[0,1],  as  F(0)=0 and F(1)=1 and F strict monotone increasing codomain(F) =[0,1], 

-continuous first derivative entails absolute continuity entails continuity,  and given codom=[0,1] now and dom [0,1], and F(1)=1 and F(0)=0, and  that F is surjective as well from the intermediate value theorem. So I presume it would have been enough to say

F:[0,1] to R

(1), F(1)=1

for all x in dom (F)=[0,1] and for non-negative integers n F((1/2^n) x)=(1/2)^nF(x)

(entails F(0)=0)

with a continuous positive first derivative. F'(x)>0 (as this entails injective and strictly monotone as well as F absolute continuous, as its continuous differentiable, - it also entails then that as F from strict monotone increasing  that as F(0)=0 (from above), F(1)=1,and dom=[0,1] that:codom(F)=[0,1] and

(2)that as continuous first derivative implies continuous differentiable implies absolute continuity implies continuity, and domF=[0,1], F(0)=0, F(1)=1 and now that codom(F)=[0,1]] that F is surjective given the intermediate value theorem, the end points, the domain and co-domain and continuity.

and thus continuity need not be specified

- alternatively I could have just said:

F  F:[0,1] to R] and F(1)=1 and (3) halving,F(1/2x)=1/2F(x)

where F has a defined continuous first derivative that is positive F'(x)>0, over its entire domain.

 - (1)As a continuous first derivative implies absolute continuity, -

-(2)that said first continous first derivative is positive that F is injective and strictly increasing,

-(3)halving F(1/2x)=1/2F(x) gives F(0)=0,

-(4)Given (A) continuity and inferred from  absolute continuity, inferred  from continuous first derivative, and (B) F(1)=1 ,F(0)=0, (F(0)=0, from halving) (D), and that dom (F)=[0,1], the intemediate value theorem gives that [0,1] subset im (F); and

- (5)strict monotony (from positive first continuous derivative) and dom (F)=[0,1], F(1)=1 and F(0)=0, gives codom(F)=[0,1]

and thus IM (F)subset codomain=[0,1] (by definition of codomain)

(5) and (6) then specifiy the codomain =[0,1], and that codomain=[0,1]=im(F)=[0,1] as  co-domain=[0,1] , and [0,1] subset im(F) by the intemediate value theorem, give co-domain =[0,1] subset im (F), gives

codomain subset IM (F)

and from the definition of co domain;  Im (F)subset codom,

so IM(F)=codom(F)  as Im(F)subset codomain, codomain subset IM(F),  and we know that is codomain is [0,1[

(A) continuity, (B)F(1)=1, (C), F(0)=0 (D) dom (F)=[0,1] gives that F attains every value between 0 and 1, from the intermediate value theorem so [0,1] subset im(F).

F is surjective and thus bijective over the domain [0,1],

-as strict monotonic increasing-ness f from positive first derivative implies that codom(F)=[0,1] , Given F(1)=1 and F(0)=0. THen as, for all x in dom(F)=[0,1], then for all x in dom(F);  x_0=0

See  Cauchy's functional equation is the functional equation

f ( x + y ) = f ( x ) + f ( y ) . 

Solutions to this are called additive functions. Over the rational numbers, it can be shown using elementary algebra that there is a single family of solutions, namely f : x ↦ c x  for any rational constant c . Over the real numbers, f : x ↦ c x , now with c  an arbitrary real constant, is likewise a family of solutions; however there can exist other solutions that are extremely complicated. However, any of a number of regularity conditions, some of them quite weak, will preclude the existence of these pathological solutions. For example, an additive function f : R → R  is linear if it is continuous.

Dear Professor Mateljevic; A lot of the time there is a great distinction made between linearity and additivity, I think that most of the time they mean (additivity in the weaker sense of kolomogorov, of disjoint pairwise unions of events). Cauchy additivity is a unique-ness kind of condition really (that is much stronger). It appears that for all extents and purposes, at least in my context that continuity is almost automatic (for example (non negativity with F(1)=1 over the domain [0,1], even where cauchy equation is confined to the unit triangle is sufficient;

ie F;[0,1] to [0,1]

F(1)=1 , where

[for all x,y in [0,1]=dom(F) s.t 01 , I presume however, that this does not apply to rational x, as they will trivial be F(x)=x,

and so the function sums of pairs  rational x in dom (F)=[0,1] x+y=z+m>1 will sum to the right rational function sum value, equal to the domain value x+y=1.5 iff F(x)+F(y)=1.5 for example; nonetheless. I presume its due F(0)=0, which due to doubling gives the larger values. I presume that cauchy additivity over the unit triangle is closest to Jensen equality ove

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

F(0)=0, F(1)=1  jensen equality is restricted to

forall x,y in [0,1] dom(F)  0

However i was interested only in forall x in domF=[0,1] for a continuous and strictly monotonic increasing and bi-jective Function F [0,1] to [0,1]

where F(1)=1 and F F(0)=0 that satisfies merely

\forall x in [0,1)=dom(F);F((1/2)*x)=1/2*F(x) 

this on first glance cant get anything more then F(x)=x for all 1/ 2^n (not dyadic fractions in [0,1] or in [0,0.5] but only 1/dyadic powers

F((1/2)^n)=(1/2)^n for non negative integers n>=0,  all it  really say anything about F(x) for x.>0.5 is  F(3/4)=1/2F(3/8) and F(1)=1 etc,

F(1)=1 F(0.5)=0.5 F(1/4)=1/4 F(1/8)=1/8 F(1/6)=1/16.. F(1/32)=1/32 etc ... F(0)=0; it does not generalize to anything more

F(1)=1 F(0.5)=0.5 F(1/4)=1/4 F(1/8)=1/8 F(1/6)=1/16... and F(0)=0 and that about it. Unless surjectivity and continuity and strict increasing ensure that conditional on there being 3/4 in IM(F) and 3/8 being in IM(F) and IM(F)=dom(F)=codom(F)=[0,1] that the identy is the only consistent solution due to these weak conditional connections F(3/8)=1/2 F(3/4)

on first glance directly, not about F(3/8) ie between F(1/4)=1/4 F(1/2)=1/2

@William Balthes

I am agree  that  there is a great distinction made between linearity and additivity.

See, for example,

  @Miodrag Mateljević, @M. Knezevic, @M.  Svetlik, Distance in the absolute plane and cauchy functional equations,Filomat 31(11):3585-3592 · January 2017

DOI: 10.2298/FIL1711585M.

@miodrag.   Thanks for this Professor Mateljevic. I am beginning to think that over certain domains a great deal of functional equations are over-determined (what one do with Cauchy equations over the unit interval to the unit interval) one can essentially do with a midpoint convexity, or midpoint concavity, or even  restricted versions theirof (mid-convex at top and bottom 0 and 1, F:[0, 1 ] to [0,1] F(0)=0, F(1/2) =1/2 F(1)=1, F monotonic increasing  or strictly monotone increasing and (continuity?)

. or at least with forall x in dom (F)=[0,1]: F(1-x)+F(x)=1 F(1-x)+F(x)=1and F(0)=0 and F (or striclty) monotonnic increasing  with F midpoint convex at 1 and 0 as,

F(0)=0 and F(1-x)+F(x)=1  gives F(1)=1,

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

and monotonic increasing F , dom (F)=[0,1] and F(0)=0, F(1)=1,  marks out the codomain to be [0,1],

and midpoint convexity at 1  with F(1) =1  F(1/2+x/2)0 otherwise and strictly monotonic as a function of pi in [0,1] independent of which probability vectors

every equivalent class p1=p2=p1 v2=p2v3=p1v3=0.4 must be assign dthe same function value iff  F( p1)=F(p2)=F(p1) v2=F(p2v3=0.4)=F(0.4)p1v3=F(0.4)=0.4 (for example  p1=p2=p1 v2=p2v3=p1v3=0.4 ....0 1>F(p)>0

\forall x in dom F(1/2x)= 1/2 F(x)

forall x in dom  F(1/3x)=F(1/3) F(x)

forall x in dom F(1/5x)=1/5F(x)

forall x in dom F(1/7x)=1/7F(x) up to 19 times or more,,at least appear to generalizes as far I know

\forall (p in domain) F(1-x-y)+F(x)+F(y)=1

x+y+z1 x+y+z>1 iff F(x)+F(z)+F(y)>1

F(x)+F(y)+F(z)=1 iff x+y+z =1

\forall  x,y, z,m [0,1] where x+y=z=m in [0,2] x+y=z+m iff F(x)+F(y)=F(m)+F(z)

[0,2] x+y>|=p>=0 1 p1>p2>p3>0 iff 1>F(p1)>F(p2)>F(p3)>

0F(0)=0 F(1)=1   F(p1)+F(p2)+F(p3)=1  where these domain is over the entire set of all convex combinations and equivalences class and context free, particular for any event in the structure  p{1,2,3{, p{1,2,3j in [0,1]forall vectors  vi k , vj whether vk =vj or nregardless of whether its p1 on vector =0.6=p3 on vector on some other vector ,  where the domain is the canonical simplex :

That if:              \forall p1, p2, p3) forall (vectors v1, v2) whether v2=v1 or not, whether  p v=p vj     iff   F(pv)=F(pvk)

for all, vectors in the simplex and for all events p{1,2,3{ and all events where p1, p2, p3 in said probability vector where a

                                   \forall (p1, p2 , p3 ) foall vectors (v, v2 whether v2=vk or not in the simplex domain ) p{1,2,3,}>p{1,2,3}v iff    \forall (v , v2) where F(p{1,2,3{v)>F(p2v).

Then if the domain is separable and closed under the all convex combinations, of p1 +p2 +p3 that sum to one  and in [0,1] each set a probability vectors and literally ranked numerically every which way and between (then so long as the dimension are at least three it appears to give rise to something akin to cauchy additivity, when the both entities have unions as well, and its closed under convex combinations, and these modally ranked by =, these will be mapped onto atomic events on other vectors, but I note that even without these; it appears that quantum like modally ranked systems all have a similar issue when there are infinitely many probability vectors, and there another probability function and one can only presumed that the the respects the ranked order globally, and additivity each space locally (when it comes to 3 , one quickly finds that the sums are immediately constrained due to the global equalities in the rank, and common terms on other bases).

   p1>pv iff     F(p1v)< F(p2v)

F(0) iff pi=0, F(1) =1 otherwise, 1>F(p)>0i

ff pi=1 F(1/3)=1/3 everywhere  F(1/2) =1/2 everywhere for instance,

and F(p1)+F(p2)+F(p3) is forced to sum to one as at all horizontal vectors.

Even iff they are on distinct bases.

That is, a probability function : \forall elements in all vectors in the probability simplex -> [0,1].

for all vectors in probsimplex: x+y+z=1 and F(x)+F(y)+F(z)=1

F(1)=1 and F(0)=0, whenever x=0 etc F(x=0)=0, whenever y=0 F(y=0)=0 etc.

That is we treat it as a single variable function on the simplex, over the domain prescribed equivalence classes (any element of a vector, with the same domain value/magnitude is assigned the same function value, as any other element of any vector (whether the vector/cartesian point is common/the same or not, whose coordinate magnitude (domain value) is equal, whether its both are "x's" or one is an "x" and the other is a "z" or a "y". It is also injective it is only assigned the same function value value as the members of its unique equivalence class. That is, we let F be strictly monotone over the equivalence classes=[0,1]=dom(F)

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

x^>x1^ if and only if F(x^)> F(x1^) (the entire 0.6 class is assigned a common/the same number (function value) in [0,1], the entire 0.7 class is assigned a common/the same function value (number) in [0,1], and the number assigned to said 0.7 class (or more accurately to every member of said class) is greater then the number that is assigned to every member of the 0.6 class.

x=0.6, for example denotes the equivalence class of all elements of all vectors (cartesian points) in the simplex whose value is 0.6 (ie all occurences of an x=0.6 , and all occurrence of a y=0.6, and all occurences of a z=0.6 in any vector, are treated as just x^=0.6, the equivalence class of all 0.6 valued events, and all are assigned the same function value, where strict monotone increasing meant to mean, that every element of the same equivalence class are assigned the same value, such that for any such equivalence classes, if one corresponds to a greater domain value such as 0.7 then every element of the 0.7 class (is not only assigned the same value as each other and distinct from any element that is an element of a distinct class) but greater then any element that is a member of a lower class (ie the function value assigned to the 0.6 class, F(x^=0.6) and thus to all of its members is smaller then the value assigned by F to the 0.7 class, F(x^=0.7) domain values

forall (x^)\in (0,1): F(x^)>0

ranked numerically over a separable probabilistic domain (the canonical 2 simplex, that is closed under all convex combinations of n non-negative reals that sum 1, where the function, F, is strictly ranked by elements in each vector (cartesian point), the equivalence class each corresponding to a domain element in [0,1] if the dimensions are only two, will not reproduce the probability simplex, at least not necessarily.

if dim>=3. Then without building union events for each world or cartesian point (including them in the boolean algebra of each point, where the x,y ,z denote the domain probabilities for the three mutually exclusive and exhaustive atomic events of that point/world/space). Then given the normalization constraint on each vector and the equivalence class notion (same domain probability same range/image probability, arbitrarily everywhere, for any event in the structure). Then one only need to get one extra fixed point (ie using the edges) (x=1/2, y=1/2, z=0) to essentially get all of them (at least every rational).

ie the 2, domain prob=1/2 events x, y above will be assigned the same function value (member of the same equivalence class) and the z point will be assigned the function value 0

And they have a function sum of 1.

F(1/2)=1/2 in this point but due to the equivalence class notion holding jeverywhere this will hold everywhere. One already has F(1/3)=0 and the top and bottom element F(0)=0 and F(1)=1, and so one can immediately solve for F(1/6) @ (1/2, 1/3, 1/6) using normalization of the function values and the 2 derived fixed points at 1/2, and 1/3 giving F(1/6)=1/6. One gets doubling and the function sum of 1 constraint (over the equivalence classes) for any given 2 elements whose domain sum is 1, and for any arbitrary three elements whose domain sum is 1 (equivalence classes), whether these points lie on the same vector or not. This is a result of the equivalence class notion the sum of 1 along a vector, and closure under convex combinations. For any three points whose domain sum is 1 and they do not lie on a common point, one can find a vector which has the same values in it, and by the equivalence class notion must have the same function values as the function values of the other three points of interest. They of the 2 sets of 3 points must have the same function sum : and the second set is just 3 points (atomic events) lying on the same vector (ie their function sum must be 1). Hence the other three points (despite lying on distinct vectors from another) are forced to have a function sum of 1, also.