# Is this true? A strict monotonic function F satisfying Jensen equation with F(0)=0, only expresses dyadic rationals multiples?

. Is is really true that a function satisfying jensen equation F(x/2+y/2)=F(x)/2+F(y)/2, F(0)=0, and F strictly monotonic, where F is a function from F:[0,1] to [0,1] can only express dyadic fraction of integer multiples of x?.

Perhaps what is meant, is that it cannot directly express these facts without additional constraints F(0)=0=min, in-jective F(1)=max value F, non negative;F:[0,1] to [0,1] etc

1. F is a function from F:[0,1] to [0,1] tion 2.F strictly monotonic increasing, and injective. 3.F(0)=0 and F(1) =1
1. F(x/2+y/2)=F(x)/2+F(y)/2 for all x in [0,1] domain.

Can it express F(2/3x)=2/3F(x) and that also via its form of functional additivity (F(sigma x + sigma y)= sigma F(x) +sigma F(y) where sigma is rational but not a dyadic rational number or integer multiple theirof.

That is Jensen's equation (before continuity is applied) can only express that F(p/q x)=p/qF(x); when p/q is in {a/2^n:a\inZ,a\in N U {0}} but not F(2/3x)=2/3F(x) for example. Which is unlike cauchy's equation. For example It cannot express that an element of the domain that is is two thirds of the way between two other elements of the domain, is two thirds of the between said function values.

Even when F(0)=0 and F(1); or that F(2/3x)=2/3F(x).For example, F(3/2^(0)x)=F(3x)=3F(x) F(3/2^5x)=3/2^5F(x); F(2x)=2F(x) but not F(1/3x)=1/3F(x) or F(2/7x)=2/7F(x) .

Or that F cannot express (F(2/3) is two thirds of the way between F(0)=0 and F(1)=max (=1 in this case) . It may be indirect, but it appears that it can in a finitary manner, without limiting/continuity arguments) express thatdirectly without.

Presumably it can express F(3x)=3F(x) (using midpoint on F(4x)=4F(x) and F(2x)=2F(x) which follow from Jensens equation with f(0)=0, then it can express F(1/3x)=1/3F(x)? for example just given F(0)=0, and if injective and increasing (strict if necessary) .

See this post, which alludes to this. It is important.

I know that its not precisely the same as Cauchy equation for all x in [0,1]. The route is more 'round about' with jensens equation but it appears that can be easily done. But I maybe wrong.

Is it the case tat F cannot directly express rational homogeneity but only dyadic rational homogeneity. That is before the continuity or some other regularity is applied (to ensure its linearity). This being in contrast to cauchy's equation, F(x+y)=F(x)+F(y) for all x in [0,1] which can express F(sigma x)=sigma F(x) for all rational sigma.

ie is it the case that a function satisfying Jensens equation (before continuity is applied), with F(0)=0, and F injective and strictly monotonic) can only express that F(p/q x)=p/qF(x); when p/q is in {a/2^n:a\inZ,a\in N U {0}}; a dyadic rational fraction multiple of an integer a.

ie the only sigma=p/q such that F(sigma x)=sigma F(x) explicitly, are those sigma \in{a/2^n:a\inZ,a\in N U {0}) ; ie 3/2, 1/2. 3/4.1/4 etc.That is, only those x, which are integer multiples of (a), and dyadic fractions of x; a/2^(n)x;1/2^(n)x,1x,2x,3x, 3/*(2^4)x where n is a non-negative integer.

It may not be able to directly express other fractions.

But that seems peculiar, because that when F(0)=0, it can express F(2x)=2F(x) and 1/2F(x)=F(1/2x), and given the argument below, (A), using midpoints, it can express F(3x)=3F(x), using F(2x)=2F(x) and its consequences F(4x)=4F(x), It can express, then ia jensens equation, that F(3x)=3F(x) and likewise, as an indirect substitution that F(1/3x)=1/3F(x) .

And then as a result F(2/3x)=2/3F(x) even though its indirect. It seem odd, that an (injective) Jensen function on [0,1] to [0,1], with F(0)=0 can odd multiples, integers such as 3 or 7 but not 1/3 or as a result 2/3

. It appears peculiar that it express F(3/2x)=3/2F(x) but not F(2/3x)=2/3F(x).

. For example can presumably express F(2x)=2F(x) and F(3x)=3F(x) ie (for all integers); and that 3F(0.33)=1; and that F(3x)=3F(x) and express these as thirds F(1/3x)=1/3F(x)and halfs.

, So that F(2/3x)=F(2\times1/3x) 2F(1/3x)=2/3F(x) as F(3x)=3F(x)

To get (1) F(3x)=3F(x),

just consider that and F(2x)=2x, F(4x)=4F(x), and that the midpoint of 6x; F(6x/2)=F(3x)=F(4x/2 +2x/2)=F(4x)/2+F(2x)/2= 4F(x)/2+2F(x)/2= 3F(x); is it invalid then by by careful substitution that given F(3x)=3F(x); to let x=1/3y F(y)=3F(1/3y);F(1/3y)=1/3y, and then use the doubling induced by f(0)=0 to entail F(2/3.x)=2/3F(x)

For example, that F(3x)=3F(x); when F(0)=0; and F, is strictly monotonic increasing cant one do this:

And that (1)F(2/3\times1)=2/3F(1)=2/3 when F(0)=0, via

F(x)=3F(1/3x); and thus F(2/3x)=2/3F(x) as it can express F(3x)=3F(x); 3/2^0 (dyadic in that sense), and F(1/2x)=1/2F(x) which follows from F(2x)=2F(x) from jensen equation with F(0)=0,

And thus in this indirect manner, at least express all rational multiples of x; ie rationallly homegeneous, as is cauchy's equation. Perhaps not as easily but if F(p/qx)=p/qF(x). For instance F(3.333x or 10/3x etc)= F(1/3 10x)=1/3F(10x)=10/3F(x); but if itexpress F(10x)=10F(x);F(3.333x)= F(1/3 10x)=1/3F(10x)=10/3F(x).

((A)maybe injectivity or (A).1 strict monotone increasing is required, or (B) F(1)=1.

but I don't think the latter, (B) F(1) is necessary, so long that F:[0,1] to [0,1]) or perhaps some of the former (A).1 is unnecessary, if f(1) is just designated as the max value, F(0)=0 and is the min and F injective). THis not not appear to be the case, at least so long as f:[0,1] to [0,1] and F is strictly monotonic

• what do you mean by " can only express" and "can express" ? – Mirko Apr 23 '17 at 16:45
• Can a Function (with the above constraints)and which satisfys Jensens equation, with F(0)=0; injective and F(1)=1; directly prove that F(sigmax)=sigmaF(x), for all real x in [0,1] and all rationals sigma in that F(p/qx+l/k)y=p/qF(x)+L/qF(y); or so much as F(2/3x)=2/3F(x), directly without having to use some kind of continuity requirement, or approximation technique; in the sense that a function satisfying cauchy's equation can do so directly for all rationals – William Balthes Apr 23 '17 at 16:51
• By express; I mean can said the functional equations (in full generality) or rather can the same of some (arbitrary) singular non-dyadic rationals F(p)=p, where p is a non-dyadic irrational,be read off/inferred the structure, direcltly one at time, if you as it were , ie ; precisely (without some kind of approximation of squeezing technique;). One might need that to get them all collectively – William Balthes Apr 23 '17 at 16:58