Hello,

 please state clearly the question.

The attached paper contains it seems details enough to solve your question.

With F(0)=0 it looks like that Jensen and Cauchy are equivalent.

Best Regards,

 Gabriel

Yes one can derive cauchy's equation from jensens equation with  F(0)=0.

Although its unclear whether it whether it generalizes completely to rational or finite sums of irrational numbers; (in the non-continuous case) I believe it generalizes to the former and to the latter but its more round about about. Some suggest that its only only d-yadically rationally homogeneous (on maths stack exchange) which appears false;  that is, before one applies continuity?  I believe that it does, and it can be proven, but to do it rigorously, one cant just say that one can derive cauchys equation of form F(x+y)=F(x)+F(y) for all pairs and then uses cauchy's equation from that point onward to show rational homogeneity, P(sigma x)=sigma Pr(x) for all real x and rational sigma.

Ie That is, to forget that F(x+y)=F(x)+F(y)  on the basis of jensens equation for all pairs and a  doubling (jensen equation) and then forget about that fact from that point onward. And just use, substitution on F(x+y)=F(x)+F(y), as if it were a cauchy function to begin with  (using the standard techniques for cauchy's equation to get from F(x+y)=F(x)+F(y) for all reals to F(sigmax) =sigma F(x) for all real x and rational sigma (which is quite easy). iy, and F(2x)=2F(x) gives you subadditivity with midpoint concavity and sub-additivity with midpoint concavity and jensens equation is just the fusion of the two); ironic  becausehe  convexity is generally associated with super-additivity with F(0)=0 For instance midpoint convex structure aren't quiite rationally super-additive even with F(0)=0, lest sublinear models would be both super-additive and sub-additive (they are continuous and convex with F(0)=0 and F(sigmax)=sigma F(x) por the positive integers and some tiems all positive reals,; they are also sub-additive and midpoint vonvex. And presumably both super-additive and sub-additive when continuous

if one wants to be rigorous (ie let x=y+t) to get that F(x+y+t)=F(y)+F(t)+F(x); one needs to back to Jensen equation and derive however, that from it, however, indirect (it can be done of course. But its in the affine class, which is slightly distinct from the linear class, despite the fact that at the end of the day, it will have the same result as cauchy's equation. 

I fi

Moreover, I am wondering whether the conditions for continuity are slightly harder to satisfy.  With cauchy's equations is generally automatic in the context of a probability function ( i find it hard to believe to be honest) but one apparently  does not even require monotonicity, or strict monotonicity; but merely that

2.F is a function s.t F:[0,1] to [0,1],

2.F satisfies Cauchy equation over entire domain [0,1}

3.and F(1)=1;

From F being a positive domain to a positive co-domain, one can get to non-negativity, and F being bounded monontonic and is apparently automatically continuous.(apparently according to a theorem of azcel see 1989 Domhres and Azcel collorary 9) ; confirmed when  question i posed on maths stack exchange; ie the about constraints entail F(x)=x) One does not even have to directly specify that its injective or monotonic, bounded-ness let alone continuous. I still refuse to believe it. Even it i understand the proof. In any case is the same true for jensens equation but with F(0)=0 in addiition,

Dear William:

It is possible to give a complete overview of the solutions of Cauchy's equation, namely

F(x+y) = F(x) + F(y)

without any continuity assumption. That equation is indeed equivalent to Jensen's. That is helpful.

The basic idea is that of a Hamel basis: we have the remarkable theorem that every vector space has an (algebraic) basis. That is, if V is an arbitrary vector space (no finite dimensionality required) then there exists a set e_alpha of vectors in V such that

1) the set is linearly independent, that is, no finite linear combination of e_alpha can be zero, unless all coefficients vanish.

2) the set generates the whole vector space, that is, any element x of V can be expressed as a finite linear combination of e_alpha's

Given that theorem (quite non-trivial, it requires the axiom of choice) you immediately see that any linear map from V to V, say, can be uniquely specified by giving its values on the e_alpha, the extension being afterwards unique.

Now apply this to the reals, being a vector space V over the rationals. Then there exists a set e_alpha with the required properties: any x can be expressed as a finite rational combination of e_alpha's in an unique way. It is therefore enough to specify the values of f(x) on all e_alpha's and you can uniquely extend it to all reals.

Now any linear map with respect to the rationals satisfies Cauchy, and viceversa. It is then easy to show, for example, that any noncontinuous Cauchy map must be unbounded on every interval no matter how small. Any non-continous Cauchy function inherits from the Hamel basis a substantial amount of awfulness.

Thanks for this, by this by e alphas, I presume you mean the maximum element such $F(1)$ or all rationals (which is a consequence of it). What is the purpose then of people attempting show, after they have established their function, satisfies cauchy's equation, that their function is continuous. I presume its when they do not know the value of the upper- bound.

Or that they can use the field automorphism to show that its either F(x)=x or F(x)=0, and so by other means (I suppose by showing that there exists some such x neq  0 such that F(x)=0 without having to directly specify the value of F(1)=1

I presume in most cases, they have not specified some upper bounded, or whether its domain is non-negative or some such that. Or are funnily enough trying to use continuity before-hand to ensure that their function satisfies cauchy's equation

I presume, that the bounded-ness assumptions for cauchys equation are much esier to satisfy, essentially setting the upper bound fixes. It unlike in say mid-point convex functions where one has to also establish that its closed unter limits points, and not has a lim sup etc, and not merely an maximum upper bound

I suppose the hard yard, is then getting to cauchy's equation really. Once that is established the unique-ness of the representation is established. So give you know more about Physics and perhaps Gleasons theorem then I (it might be different as  I am not sure if he presumed F(1)=1) but that would it half correct to say, that the master task that gleason established was the linearity of the frame function, as a function of the amplitude moduli squared (which is really its homogeneity).

And that continuity was a consequence of that, and not conversely.  So that perhaps when people ask, how did Gleason establish that function must be continuous; I guess the better question or what is often really meant, is , the homogeniety, even if rational, or rather how did he establish that it must satisfy cauchy's equation .

Likewise I cannot see much of the purpose saying that linearity is the superposition principle or cauchy's equation plus 1 pt homogeneity for all reals; cauchy's equation is just a means to that end (real valued 1 pt homogeineity). As one gets to to real valued homogeneity for all real delta, and all real x through cauchy;s equation  (and if what is said above is correct) essentially immediately once one gets to cauchy's equations

Moreover, once e has merely 1pt homogeniety for all real delta and all real x, one can get F(x)=x  from F(0,1) = domain of F(1)=1 and one can derive cauchy's equation directly from it, 1=pt homogeneity, when it applies to all real (x,y) and all real (t) F(tx)=tF(x) ie F(x)= F( (x/[x+y]) times (x+y))=x/(x+y)\timesF(x+y)

do the same for F(y) add them together

F(x)+F(y)= x/(x+y)F(x+y)+y/(x+y)F(x+y)= (x+y)/(x+y)F(x+y)=F(x+y)

as well as the the so called sup-position principle. So 1pt homogeneity  its not as its a property for which which something needs to be added to it appears, (Except maybe F(1)=1 to get the linear function one wants), that is one does not need to add cauch'y equation to it,  if  1 pt homogeneity applies for all real delta, and all real x,. And the converse, almost appears to be the casefor cauchy's equatioon); no real need for automorphism equations and few other constraints

if its really true F:[0,1] to [0,1]

F(1)=1

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

entails immediate continuity, F(x)=x for all reals in [0,1] even if its somewhat restricted as one say a great deal about the function sums of irrationals,x in [0,1], yin [0,1] but x+y>1 not in [0,1] and so cauchy equation does not directly apply (it still directly applies for rationals where x+y >1) of course because F(x)=x for all rationals so x+y =F(x)+F(y) which is stronger. So I guess I thought one generally has to add something (why is there all this talk about continuity at a pt, continuity, strict or non strict monotonicity, or bounded on a set of positive measure) when in most applications a domain like this would taken for granted and I suppose must already meet those conditions? IF this Function must be continuous without even specifity monotonicity

and not in the domain of the function ( I suppose they will be indirectly connected to other events)

and derive cauchy equation from it, for x,y rational or irrational