ie for example, of (2) let

sqrt (2) stand for an arbitrary irrational (even though its rational

[x_1+y_1+z_1,=,x_1+,y_1+,z_1+K_1=sqrt(2)] leftrightarrowF(x)+F(y)+F(z)=F(x_1)+(y_1)+F(z_1)+F(k_1)=x+y+z ...=x_1+y_1+z_1+k_1=sqrt(2),,

The weaker case of (2a) is where [x_1+y_1+z_1,=,x_1+,y_1+,z_1+K_1= rational, but some of the (x,y) are (x_i,y_i) are irrational, and where one must show that

(x)+F(y)+F(z)=F(x_1)+(y_1)+F(z_1)+F(k_1)=x+y+z ...=x_1+y_1+z_1+k_1

Or that(2b) which is probably more difficult [x_1+y_1+z_1,=,x_1+,y_1+,z_1+K_1= irrational, but some of the (x,y) are (x_i,y_i) are rational

both being stronger than (3), weaker then (2) which only demands that

[x_1+y_1+z_1,=,x_1+,y_1+,z_1+K_1=sqrt(2)] leftrightarrowF(x)+F(y)+F(z)=F(x_1)+(y_1)+F(z_1)+F(k_1) and perhaps with the requirement

closest approximating rational from below1  (outside the domain) then if, some of the   the x,y,z ...,x_1,y_1,z_1  are irrrational valued  but the where the sums,, x+y+z ...=x_1+y_1+z_1 =rational , one can show the stronger (2),

and perhaps also when , when x+y+z >1 ,but  some of the x_1 are rational, but the sums x+y+z ...=x_1+y_1+z_1 =irrational one can perhaps show the stronger (2), in

And one can show (2) when x+y+z >1  when its completely rational; sum and summands

as one one can clearly show the stronger F(x)=x for all rational x in the domain, in any case

,and  one can show  that cauchy's eq holds, , whether the sum, or summands or both, are irrational/ rational or NOT irrational/ rational,so long as x+y+z is in the domain (so that it even makes sense to do so) and otherwis

e, (2) or (3) where (2) is the stronger it seems.

so perhaps that is why continuity is automatic, cauchys eq is stronger when extemd then rational homogeniety,,

where F(0)=0 and F(1)=1 , and thus F(x)=x for all rationals, with F:[0,1]\to[0,1] and F strictly monotonic increasing as it puts stronger bounds on the structure. I presume that strict monotonicity will at least ensure that the irrational fall within the right bounds (not merely that they are positive valued) which does not seem sufficient to either entail monotonicity, or continuity and render continuity automatic, perhaps I was forgetting the bigger picture; that is that cauchy's equation actually bounds the values even more-so, that is of the irrationals, particularly, if x+y+z need not be in the domain.

Yes, such functions are continuous and the reason is that the range is bounded. It is enough to note that any non-void interval contains a Hamel basis H, i.e. a basis of R over the rationals. Therefore, the restriction of F to H can be uniquely extended to an additive function, say A. This aditive function is bounded on [0,1] and thus continuous. Because F(1) = A(1)= 1 we have A(x) = x. Last thing to observe is that A is equal to F on [0,1]. This is because A is unique, F satisfies the conditional Cauchy equation and they agree on a Hamel basis.

An alternative argument is to extend F to an additive function assuming rational homogeneity, but this will require a bit more work. You will get an additive extension of F which is bounded on [0,1] and thus is continuous.

I presume that rational homogeineity will hold for all F(\sigma x)=\sigma F(x), where x is rational,sigma rational, so long that sigma x is in the domain.

So F(x)=x over [0,1] without even 'mononoticity being presumed'? Would this be the case if F(0)=0 and jensen's equality were to hold instead.

I presume you are considering it a 'conditional cauchy equation', because x+y>1, may not be in the domain, of thus,  cauchy equation does not apply\for all (x,y)\in: as x+y may not be in the domain. Though I presume it will be continuous and identical to F(x)=x over [0,1] in any case, I presume it will entail that that for rationals F(x)+F(y)=x+y even when x+y >1, as F(x)=x for all rationals,

s

If instead jensens equation were to hold for all x,y in the domain instead, or 'merely rational homogeineity' and not cauchy equations, would monotone increasing be required to say that because F(x)=x over all rationals, and agrees with a continuous function such G over a dense set, where G(x)=x for all reals in [0,1] that by strict montoone increasing the functions  are identical for all x in (0,1).

So I presume that cauchy's equation, even in this restricted form, has an, additional property, which does not require monotonicity/strict monotone increasing  to be presumed; it entails F(x)=x over all rationals, but I suppose that also has some implications for the irrational values

.I presume that this restricted form 'conditional cauchy equation' is akin to jensens equation as x/2+y/2 will always be in [0,1] if x,y are in [0,1], so its designed in some sense for intervals and not the real line, So I presume that such a function will be F(x)=x on [0,1] as well

I presume that F is rationally homogeneous in the sense that the above function is such that over [0,1] F(x)=x for all rationals. That F(x)=x for all rationals in the equation above;

But generally that requires a monotonicity argument to extend it to saying that F(x)=x for all x in (0,1). Why is not the case with cauchy equation? (in this restricted form).

Is a cauchy equation considered conditional if it does not all apply to all reals (in some sense its about as unrestricted as one can cauch'y equation, wit tat restricted domain and range, function on [0,1],

So by an additive extension I presume you mean a function cauchy's function applies to all reals whetherx,y, or x+y are in [0,1] or not?

I

its just arguably not rationally homongeous; only in the sense x in F(\sigma x)=\sigma F(x) might be irrational, or because

dom(F)=[0,1];F(x+y)=F(x)+F(y)

Dear William,

I am not sure, whether I understood correctly the problem. But let's do a very easy trick: consider interval [0, 1/2] instead [0.1]. Then if x, y are from the smaller interval, then x+y is in [0,1] and apply the original equation. Therefore, we have unconditional additivity on [0,1/2].

There is some problem with Jensen, especially when 0 is outside the domain. But I feel there are more competent people to answer this. I will try to find some articles on the topic.

Best regards, Wlodek

Dear Professor Fechner;

That is correct insofar as we consider pairs of point,

. Insofar as the function,was stated, it would appear that Cauchy's equation's only applies to pairs of points in the functions domain, although of course, it generalizes to  than 2 pts.

Nonetheless, what confuses me, 'is the automatic continuity'; one generally still has to presume some mild,  'yet not trivially  mild'

regularity assumption' (by which I mean,  over and above  'non-negative domain and co-domain, F(1)=1. To get F(x)=F(x) even with an 'unconditional cauchy equation', so I presume so much the more is required, for this weak restriction (which is 'as un-restricted cauchy equation' as it could possibly make sense, given the domain and range). Generally one requires monotonic increasing-ness (if not strictly increasing),  even if F(x)=x for all rationals as above. But that was not specified. I cant see how the above entails this for irrational x,y in [0,1]

For example In the above equation, F(x)= x for all rationals over [0,1]. So that one generally appeals to the fact, that F agrees with G(x)=x for rationals in [0,1] , where G is the identity function  where G(x)=x over all reals in [0,1] from which it follows, that, given' G' is continuous by definition, that:

on  the assumption that F is monotone increasing,  (or strictly increasing), I am not sure which one it has to be), F=G, reals over, (0,1) . But F was not specified to be monotone or strictly monotone increasing.

F could fail those constraints for some of the irrational values (as far as I can see), For example, F(x_1), x_1 \in (0,1) and x_1 is irrational,  is only bounded to be some number, in (0,1) at best.

It appears possible prima facie, that   x_1>y, where ,x_1,y, in dom(F)=(0,1), x_1 is irrational or and y either rational or irrational,  that F(x_1)

F could fail those constraints for some of the irrational values (as far as I can see), For example, F(x_1), x_1 \in (0,1) and x_1 is irrational,  is only bounded to be some number, in (0,1) at best.(non -negative) - unless its positivity that is required for x in (0,1) not just non-negativity, but even still

. I cant see how strict mono-tonicity much less strict monotonicity' is entailed for the irrational function values' of irrational x in domain of (F) the equation as specified above, in the post.

As the irrational values are only bounded (at best to bigger than zero and smaller then onet) or other at worst (greater then or equal to zero, and smaller then or greater than 1 )unless somehow injectivity and monotonicity are entailed by the above. unless the additivite structure connects with with rational values given F(1)=1 and the boundary constraints.

For example, It appears possible prima facie, that   x_1>y, where ,x_1,y, in dom(F)=(0,1), x_1 is irrational or and y either rational or irrational,  that F(x_1)

I presume that Cauchy's equation must indirectly specify that larger irrational values. x1>x, which are not rational multiples of each other, must have larger function values, indirectly, through

(A) both rational homogeineity  of form (2) F(\delta x)=\delta F(x), for irrational,x and irrational,y,  and rational delta, as well as form (1) F(x)=x for rational x in [0,1]

And otherwise indirectly through real valued additivity (even if restricted ) when these irrational x,y, where

(A) x+y is rational (A)  but x,y are not rational multiples of each other

or maybe partially  (B) where x+y is irrational,(x,y) are both irrational, and x,y are not rational multiples of each other ,but x+y in [0,1], through some indirect connection via additive to rational sums, or other irrational values through rational homogeniety.

If not, I presume then that strict mono-tonicity or monotonic increasing ness would have to added, to ensure their values come out partially right. Over and above that 0F(x), for x rational (when they are connected by rational multiples)

would not be rational multiple of x_1 in x=1 xx=1-sqrt(2)/2 +x_1=sqrt(2)/2=1, ths.

Generally one needs to demonstrate cauchy's equation and continuity,  to get to F(x)=x, .

As these functions are often used in the context of 'probabilistic like representations where the non-negative domain and range F(1)=1,is often taken as given'. In these case,  then it appears, that the requirement to show continuity, is is not necessary, if cauchy's equation has been shown. And what is said above is correct?.

And that, therefore, that in these cases, it is un-necessary to show that ,the field auto-morphism equations,  hold. That is once, F(1)=1 and this 'cauch'y's equation' has been demonstrated to hold with F:[0,1] to [0,1].

That is, If the only purpose of the field auto-morphism equations:

The auto-morphism equation, F(xy)=F(x)F(y), it appears, cannot say anything about transcendental x in dom (F); ie that F(1/pi)=1/pi,, is to demonstrate non-negativity of domain and co-domain, then they appear have little use in the probabilistic case (as generally this is taken as a given in a probabilistic perspective).

And in any case, I cant see how they entail continuity, even under the presumption of monotonic increasing-ness in addition); that a different question though.The auto-morphism equation, F(xy)=F(x)F(y), it appears, cannot say anything about transcendental x in dom (F); ie that F(1/pi)=1/pi,

The auto-morphism equation, F(xy)=F(x)F(y), it appears, cannot say anything about transcendental x in dom (F); ie that F(1/pi)=1/pi,

even when combined with F strict monotone increasing, and all of the aboven Cauchy's equation as above,

As it might be hard to show continuity at 1, from midpoint concavity, monotonic increasing, F with and F(0)=0 and F(1)=1

By a Jensen function I mean (A)

 (A)

if (1) F:[0,1] to [0,1]

(2) F(0)=0 and F(1)=1

(3) F is monotonic increasing (or strictly monotonic increasing). Is this necessary,

 (4)\for all (x,y)\in dom(F)=[0,1];F(x/2+y/2)=F(x)/2+F(y)/2

I presume at least monotone (3)increasing is required here. x/2+y/2 is will always be in [0,1].x+y may not be.

I presume that if 0 \in dom(F) as above,  and F(0)=0, then this Jensen function should be roughly equivalent to the Cauchy function above

But if  Jensen F requires monotonic increasing,(3) to get to F(x)=x over [0,1] ,then there is a difference between this Jensen function, and the Cauchy function in the post, though the Jensen function has 0\in the domain, F(0)=0 . Whilst else is equal excep,t that monotonic increasing (3) was not specified in the Cauchy function,  above, and it apparently does not need to specificie?).

 Generally one can show continuity at 0, for monotonic Jensen functions  via’ midpoint convexity, with F(0)=0, F(1)=1, and mono-tonicity increasing’ where F:[0,1] to F{0,1].

I presume the same is case, for midpoint concave functions for continuity @1 and that they only have an issue with continuity at 0).

Whilst midpoint convex functions have an issue with continuity at 1? Or is this not correct. If not, is there no issue when it’s a jensen’s equality function.

 I presume one can show that F(1-1/2^n)=1-1/(2)^N for Jensen’s equation even if cannot do this for strictly monotonic increasing midpoint concave F:[0,1] to [0,1] with F(0)=0 and F(1)=1,

with which it entails and I presume continuity at 1, but I presume that only works given monotonicity

.

There is a distinction between the two equations, Cauchy’s equation F:[0,1] to [0,1], and the ‘F(0)=0, F:[0,1] to [0,1] form of  Jensen's equation, where 0 is in the domain’, w

Both have the same domain and co-domain , and F(1)=1. In both cases.

Usually it is said that one derive (A) Cauchy's equation from (B)‘Jensen equation with 0 in domain and F(0)=0’,  And are passed off as equivalent, in this context, but is that completely accurate in the strongest sense of the word given the above extra requirement of monotonicity ?

So  are the two functions below (A- monotonicity- the jensen function above, without monotone increasing) and (B) equivalent -in the strongest sense of the word.

When in neither case, x+y >1 is in the domain of F. Nor is 'monotone increasing mentioned'.  And so F cannot say anything directly about additivity If (B) leads to F(x)=x, over [0,1]and the jensen (A-monotonicity) does not, as monotonicity is required. Then there is a difference? What is the difference? It is often said that the (A) Jensen equation, with F(0)=0 and 0 in the domain and (B) cauchy''s equation over the same domain and range,  are essentially equivalent?

jensen (A-monotonicity)

(1)if F:[0,1] to [0,1]; F(0)=0 ; F(1)=1

(2)for all (x,y)\in dom(F)=[0,1];F(x/2+y/2)=F(x)/2+F(y)/2

And

(cauchy (B)

(1)F:[0,1] to [0,1], F(1)=1

\(2)forall (x,y) \in dom (F)=[0,1] F(x+y)=F(x)+F(y)

 s.

 Generally one can show continuity at 0 via’ midpoint convexity, F(0)=0, F(1)=1, and mono-tonicity increasing’ where F:[0,1] to F{0,1]

 I presume the same is case, for midpoint concave functions for continuity @1 (and that they only have an issue with continuity at 0). Whilst midpoint convex functions have an issue with continuity at 1? Or is this not correct.

with which it entails and I presume continuity at 1, but I presume that only works given monotone increasing F.

But things might be different with the equality case (Jensen equation).

. x/2+y/2 is will always be in [0,1].

I presume that if 0 \in dom(F) as below and F(0)=0, then this Jensen function should be roughly equivalent to the Cauchy function above

. But if it jensen F requires monotonic increasing to get to F(x)=x, over [0,1] then perhaps not ?