I presume that given the average ratio or rate of change is 1 between 0, and 0.5 ie F(0.5)-F(0)=0, that if the function is going to attain F(0.5)=0.5 and its derivatives continuous, whilst being convex at the same time, that its unless its first derivative just is one, it will never attain F(0.5)=0.5. Its average rate of change is going to be F(0)=0 so its of the increasing which are generally  equal to the identity function; for similar reasongs

I presume that F is merely montone increasing (even if not presumed to be strictly montone increasing that F(x)=x.You can my post in maths stack exchange originally i had that F was twice differentiable strictly monotone increasing with a strictly positive first derivative. I have been since informed that those added conditions are not required. I presume that F has to be at least monone increasing if is is not twice differentiable

You can direcly read off the value values if x>0.5 In fact, one can even if F is star convex. Though I presume that a star convex function at ie star convex at 0 F(tx)

Alternative proof by contradiction:

Let F be continuous on [0,1] with F(0)=0, F(1)=1 and F(m)=m for an m∈(0,1).

If F is convex on [0,1] and F(x)-x is not 0 on [0,1], then there exists b∈(0,m) or b∈(m,1) such that F(b)

Inside the interval between the outermost two points F must be linear, outside that interval all bets are off, even if f is strictly monotone. Like for instance F(x)=x for -inf

Thanks for these responses . Is monotonic increasing even need just the three fixed point and the domain and co-domain being defined F:[0,1]\to[0,1], with F(0)=0 F(m) =m\in (0,1) and F(1)=1 with F continuous convex (but not necessarily differentiable). What if F is midpoint convex, would monotone or strict monotonicity generate this result and thus continuous ; i note that if midpoint convex functions are rationally convex on closed intervals you can essentially see that F(x)=x for all rational x >m,  and this is precisely where such functions have difficulty with continuity convergence towards 1. but then it would agree with the identity function on a dense set of [0.5, 1] arguably and if F is strictly monotone that generally entails that as G is continuous that F=G on all of (0,1), midpoint convex strictly monotone F, on a closed interval is already continuous everywhere except at 1. I wonder if the fact that now knows if functional form and the fact that both F and G have same rhs end point 1, would allows you get that F is continuous at 1, and thus convex, but as a result of the proceeding F(x)=x on [0,1]

@ vierra I presume that monotonic increasing need not be specified, as F(0)=0implies it, F(0)

(0)=0  is non negative which shoud imly imply monotone increasingness, in the presence of super-additive, i am not sure if injectivity is implied, but perhaps its not needed, one could suggest positive )which i presume would imply strict monotony. But however, it works, if monotone increasing does the job,

I could have just said dom F(=[0,1] and codom (F)=[0,1] where F is convex continuous F(1)=1 F(m)=m and F(0)=0 and so F(0)=0

implies F(0)\leq 0 implies (1-t)F(0)=0 and y>=0 implies  x+y>=0 and  1/x+y>0, as1>0

as either x+y=0 ,which implies as x>=0,y>=0 x+y=0 implies x+y=0, 0=0  implies y>0 implies x+y>0

as x=0implies x>=0

one of x,y\neq 0 and as x,y\geq 0 implies one of x,y>0 an>0 and and thus x+y>0

,t=[y+x]/x\in  F(x)=F([x/[y+x]*[ x+y]) \implies by convexity F(x)=F([x/)(x+y)\times [x+y]) z,x>=0, so if z>=x, then z=y or z>y, if z=y,nothing to show as F a function implies  F(z)=F(y) implies F(z)>=F(y),

otherwise z>y,

et z=z-y+y , F(z)=F(z-y +y)>= F(z-y)+F(y)

z>y, then z-y >y-y=0 implies z-y>0,

and as z-y=z-y+0

0=z;

@William, sorry, I do not read your long questions-answers-texts. Here follows my reply to your:

@ vierra I presume that monotonic increasing need not be specified, as F(0)=0implies it, F(0)

Dear Peter thanks for this,I presumably a proof by contradiction it would have to imply that F(m)>m  . Although what you proved was that F(x)=x for F concave (not that it makes any difference) the result would be the same.. once just change the inequalities. 

I wonder what happens on occasion that F is only star convex strictly monotonic increasing and continuous or monotonic and convex.

Note below that if one can get to continuity at 1 then midpoint convex monotonic F will be convex and continuous and hence linear.

hink you proved it for concavity but the details would be the same.  you can direct solve for all reals that F(x)=x  for x>m even if F is only star convex, F(x)=1, x

@Peter, in your first line

"F(y) = F((1-y/x)0+(y/x)x))>=(1-y/x)F(0)+(y/x)F(x)>(y/x)x=y"

is a mistake, the inequality F((1-y/x)0+(y/x)x))>=(1-y/x)F(0)+(y/x)F(x) is true if F is concave.

I do not continue reading.

-------------------------------------

Continuity of F is important.

@William, this excellent thread might be useful: https://www.researchgate.net/post/Is_the_function_satisfying_this_condition_convex

@ vierra and peter, one can actually directly solve for the values for all x>m for midpoint convex m which is precisely where it has difficulty toward the rhs end point 1 with continuity even if strictly monotonic increasing , and measurable as its on a closed domain it will be continuous and convex everywhere except at 1.

howeber,, if Midpoint convex finctions are rationally convex, then they will be rationally linear for all x>m Fx)=x for all x in dom (F), =[0,1]such x>m and x rational, 

before its proven continuous, with those three fixed points given monotonicity midpoint convex m is measurable and continuous and convex everywhere except at 1. However, give midpoint convex F is monotonic is agrees with G(x)=x a continuous function over a dense set,in [0.5,1], for example with m=0.5, then F(xy monotonicity is not only continuous in (0.5,1) but is the identity function. ie F(x)=xall real x in the open interval in (0.5,1) for midpoint convex and monotonic F,

 i

the rational >0.5.x>=0.5

https://en.wikipedia.org/wiki/Convex_function

Not at all @Peter, I am directing nobody. It is not my style.

Since William continued with midpoint convexity etc, I enclosed the link to a recent thread, where similar questions were discussed and solved.

@ peter that is correct as i mentioned if you are wrong in the definition of convexity, then you were consistently so. Just by swapping the inequalities on both you get the proof for for convex function. I think that offically, the definition does not have the inequality other way. But in any case. It does not matter. I guiess you have shown proven if for f concave and convex.  I am wondering whether a midpoint convex strictly monotonic function will be continuous at 1 with those three fixed pts.

The reason why I ask this is I noticed, or had an intuition, that my model 2-outcome (as I call it ) that my symmetry conditions

F(1-x)+F(x)=1, F-1(1-p)+F(p)=1, F(0)=0 and F strictlly monotone increasing F;[0,1]\to [0,1]. Originally on a post where Peter mentioned something about 2 power boundaries. However, it would not have fone aware unless restricted midpoint convexity is emptloty

 so when I added +jmidpoint  entail jensen equality, I derived noticed that one can derived jensen equality, , Although i have had independent confirmation,.

What bothered me, however, was that given the three fixed pts inherent in those equations F(0)=0, F(1)=1 F(1/2)=1/2

was that had I gotten rid of symmetry and used just its fixed pts F(0)=0, F(1)=1 F(1/2)=1/2 i I presumably would have gotten the same result given midpoint convexitty and strict monotonicity .Hence this question.

Rendering my symmetry constriants redundant

and (2) making midpoint convexity too strong

Thus I weakened the mid point convexity constraints to midpoint convexity at 1 and midpoint convexity at 0. Given F(0)=0  so that I can keep and Symmetry, inverse symmetry and strict monotonicity increasing.

F(1/2x)

These still collapse into equalities given symmetry F(1-x)+F(x)=1 and F(0)=0 over F:[0,1] to [0,1] ie jensen equality at 1 and 0

where F is strictly monotone increasing. I hope they would not given continuity and midpoint convexity unless F is symmetric. I am not sure what a three fixed pt continuous jensen convex function only at 1 and 0 that is strictly monotonic increasing looks like.

F(1/2x)=1/2F(x)

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

Where F(0)=0, F(1)=1, x+y=1 iff F(x)+F(y)=1,  and F strictly increasing F:[0,1] to [0,1]

ie F(1-x)+F(x)=1 and F-1(p)+f-1(p)=1

I am not sure what the fused form of this set of three functional equations are. I presume that its stronger then a jense function equality function at 1, and 0 as these are not necessarilt symmetric

F(1/2+x/2)=1/2+F(x)/2=1/2+F(x/2);

they can say this but may be not this as I can

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

F(1/2^nx)=1/2F(x)=1-F(1-1/2x)=1/2-F(1-x)/2

]F(3/4-x)+F(3/4+x)=1.5. F(7/8-x)+F(7/8+x)=1.75, F(1/8-x)+F(1/8+x)=1/4, F(3/8+x)+F(3/8-x)=3/8........................................

and the inequalites 1 etc.

It also oddsness at all dyadic rational pts and not just their values

It also entails that F(x)= x for all dyadics and most other rationals all primes up to 15,

SO i presume that becaues it agrees on a dense set that its the identity given monotonicity

[

In my case is a jensen function at 1 and 0 , though, and is symmetric at alll  dyadics pits.

One still has F(1)=1 F(1/2)=1/2 as result of symmetry i call it mid-star convexity at 0. and mid star convexity at 1, However, those equations. midpoint convex at 1, and o still collapse into jensen equality at 1 and 0. and generate more equations given symmetry F(x)=x equal for all dyadics and most rationals so presumably its continuous given strict monotonicity due it agrees with the identity over a dense set. and once use symmetry to show continuity at one, due tot he concavity ,that convexity at o gives, (and thus continuity at 0) unless it already entailed

I dont k

symmetric function that is strictly monotone satisfy jensens equality at 1 and 0, with

it gives halving dounbling, qujartering

F(2x)=F(2x) F(4x_)=4F(x),F(1/2x)=1/2F(x) F odd on all dyadics points in [0,1]F(3/4-x)+F(3/4+x)=1.5.

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

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

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

F(x/2)

@peter and vierra my main question then is

1If a function agrees F with a G(x)=x , the identity, over a dense set, in [0,1]all dyadics and as well almost all rationals then as F is considered strictly monotonic increasing F:[0,1] to [0,1]  then it agrees  with F is continuous and identical to G on (0,1), now given F(1)=1=G(0)=G(0)=0 does it automatically agree with G everywhere  or does continuity at 1 and zero have to be shown separetely, If so I given the 0 equation it will be continuous at 0, and given symmetry and thus concavity towards

2. what is the general and strictly increasing continuous form of this kind of jensen equality function at  1 and 0 F;[0,1] to [[0,1] F(0)=0, F(1)=1,  is some kind of doubling  function with some extra structure

and 3. What about a midpoint convex function at 1 and 0 that uniformly continous and strictly increasing and F(0)=0, F(1)=1, F(1/2)=1/2

I

note that given star convexity alone my function with symmetry is the identity with symmetry F(1-x)+F(x)=1 and F(0)=0 I dont thinkt hat monotonicity or any else is required

ie F(0)=0, gives F(1)=1 from symmetry F:[0,1]tp [0,1]

so star convexity at o with F(0)=0 gives F(tx)=1-x

but of course star convexity applies to 1-x as well

for t=1-t\in [0,1] , star convexity F(1-tx)=x, means F(x)=1-F(1-x)>=x, soF(x)>=x and we already have F(x)

In my cases its a symmetric doubling function that is  odd at all dyadics plus that extra structure jensen equalioty at 1,

zero function equality function at 1, and zero, without symmetry ,and of a jensen convex function at 1 and 0 that

as one use mid star convexity to show that F(1/2^n)=1-1/2^n

and due to monotonicity

1>=F(1-1/2^n)>=1-1/2^n  and moreover these limits exist, and appears to agree.

so as n goes to infinity  x goes to 1 F(1) limits to one. technically its is an equality, and on the rhs end points due to monotonicityu F(1)

@William, thank you for insisting on details :)

I take back the continuity: my claim that "it is important to suppose continuity on [0,1]" is wrong. 

A third fixed point m∈(0,1) together with convexity of F on [0,1] leads to F(x)=x on [0,1].

Sincerely, Viera (unique r in my name)

That is correct,if you don't have F(0)=0 and F(1)=1 given in advance , and F strictly monotonic where dom [0,1] and co-domain

As F's is co-domain is [0,1] and F(1)=1 and F(0)=0 have beenassigned their values in advance, ,thgough, already I am not trying to infer the function value  end points, ie F(1)=1 or F(0)=0,  from continuity but that its continuous at the end points that it limits to those values.

However, the main point to show that F(x)=x is over [0,1] not so much continuity which is a necessary condition, If F(x)=x, over (0,1) and the function has already been specified in advance of this, (for whatever reasoning to satisfy) F(1)=1 F(0)=0 domain =[0,1] and F strictly monotone

hen maybe it could fail to be continuous at zero or one, but its already specified that F(1)=1 amd F(0)=0. Unless you are talking about continuity, apparently its some relatively trivial theorem. However, what I am asking is generally if its strictly monotone, and F(0)=0 and F(1) you only have to show that limit from below at one at 1 takes on the function value at 1, and the limits from above at zero. Where I presume that monotonicity at the boundaries F(0)=0, and F(1)=1 being specified in advances, ensures that thjat as for all x in the domain [0,1],0F(0)=0, and F(1)=

I

presume one can define function values through monotone limits in the interior(whose limits cannot agree in the interior, on points in the domain unless that is the function value), as its monotone those limits are defined (cant have essential discontinuities), and if they so happens that these  agree  with each other(so it is not a jump continuity), it cant be  removable discontinuity on the interior for a monotone function, and if its F is defined there it cant be a removable singularity either, so I presume that it is the function value at that point modulo something or other about issues with infinity.

presume that this would not be an issue if its limits agree, we were talking about then the interior and the function is defined at those points. I was informed be that when dealing when the boundary one just looks at the relevant one sided limits, there is a one relevant limit in each case, so I presume you have to have show that it limits to the already defined function value.

presume that is what you mean by restriction ie montononcity or strict monotonic increasing  with dom=[0,1] with F(0)=0 and F(1)=1 being pre-specified in advance and essentially specify the co-domain, to be all those points  as x1=0

Dear Peter,

I agree, I presume you mean by a restriction you one deliberatel restricts the codomain of F to be [0,1] F(1)

@ peter just some short points I presume that you provided works regardless of whether F is strictly montonic increasing or merely monotone increasing

Hello peter, were you talking about inferring that F(1)=1, F(0)=0 and that F is continuous at those points in addition, as a consquence that its the identity, that the identity is continuous as with the non dyadic reals, in the earlier posts, or just continuity at 1 , at 0 given F(1)=1 and F(0)=0 which is done after those values have established in most cases. because I can see that restricting the domain may give you both, I presume though that if you already have F(1)=1 and F(0)=0 and that the function is identical F(x) on (0,1) one can either use those pts to help show continuity at them

or (2) its irrelevant s the function would be identity as 1 cup 0 cup (1,0)= [0,1], and as a consequence its continuous as an after thought (i presume though one has to be careful in general and cannot presume this, even if F=G on the open interval and F(1)=G(1)=G(0)=F(0)=0. I am being pedant, BUt i presume that there might be some technicalities with presuming that is, due almost undeterctible inifnitesimal events  unless in real analysis when consider real interval rather than sets on can just say that [0,1] is (0,1)cup {1},cup{0}.

indeed, its interesting. Ironically, i wish it were not true!

But wishing it were not so,  does not change the facts.

I am not sure what caused me to have the intuition, with  the three fixed pts and convexity alone. Originally i thought that it held only under monotoni-city., or differentiablity,

. I am trying to ensure that this  result only obtains as a result having' symmetry: F(1-x)+F(x)=1 , F(0)=0,

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

F monotone/injective and monotone increasing /(or strictly monotone increasing

So i have to weaken midpoint convexity  to midpoint convexity at 1, and 0 ,where because,  of the function has three fixed pts' due to F(0)=0,  and symmetry F(1-x)+F(x)=F(1)=1 and F(1/2)=1/2, the result obtains, F(x)=x, but not merely because I stumbled upon the aforementioned unpleasant fact/intuition,  that given F(0)=0, F(1)=1, F(1/2)=1/2 and convexity said function would bo the identity,  on [0,1]  symmetric or not,

making symmetry redundant or nearly so monotonic midpoint convexity , ogiven convexity or monotonic midpoint convexity F:[0,1] to F[0,1]

with  those points, if it already entails continuity at 1, or F(x)=x without symmetric,. OR if strictly monotonic midpoint convex F does,  I note that symmetry gives the result with star convexity, but I also note that star convex functions are almost linear as well due to the above; for all real x>0.5,in [0,1] or wherever that fixed points is F(x)=x

if F(1)=1 F(0.5)=0.5 and F(0)=0, and F star convex at o F(tx)