Ytukyg

When I refer to 'Midpoint Convexity', I mean its form when restricted to the unit interval. I denote this by $(MP)$ .
Where, $text{Dom(F)}=[0,1]$.

$$(MP):forall((x,y)in [0,1]):, Fleft(frac{x+y}{2}right)leq frac{F(x)+F(y)}{2}$$

When I refer to 'Rational Super-Additivity', I mean the equation below, denoted by $(QS)$, which is super-additivity, restricted to rationals $in [0,1]$.
I use $(RQ)$ to denote rational convexity, again restricted to rationals in the domain of the function, $[0,1]$

$$(QS):forall( (x,y,x+y)in ([0,1]cap mathbb{Q})):F(x+y)geq F(x) +F(y)$$

$$text{(RQ)}:,forall ((x,y)in [0,1]):forall(sigma in (mathbb{Q}cap [0,1])): F(sigma x +(1-sigma) y)leq sigma F(x) +(1-sigma)F(y) $$.

As Jensen's midpoint convexity equation, $(MP)$, even if continuity is not assumed, entails: rational convexity $(RQ)$, at least on open intervals.

see Tpt. 7.11 of chapter "Continuous Convex Functions" in http://link.springer.com/chapter/10.1007%2F978-3-7643-8749-5_7

So, my questions are:

1. Is strictly monotonic increasing, mid-point convex $ F$ below, rationally convex, over its entire domain $[0,1]$ (that is over the closed interval, not just $(0,1)$.


2. 
   

   Is $F$, a rationally super-additive Function?
   That is, does $F$ satisfy $(QS)$?.

   
   
   
   
   
   3. Moreover, if $F$ does not satisfy $(QS)$, does it, at least, satisfy it, for rational in the open interval, $(0,1)$. That is, where, $(QS)$, above, is restricted/modified to only hold for rationals in the open interval: $(x,y,(x+y))in (mathbb{Q}cap [0,1])$?
   
   
   
   

   If the answer to any of the above is no, if, $F$ is presumed strictly monotonic increasing, and to also satisfy, $F(0)=0$, $F(1)=1$, does this change this. Or will $F$ still fail to be rationally super-additive?**

   
   

**Note, I know that $F$, if $F$ is presumed to satisfy the additional constraints, above, strictly monotone increasing, and the 2 endpoint fixed points, will satisfy, convexity.
As a result, $F$ will satisfy the above two equations over $[0,1)$, and by consequence, $F$, given these additional constraints, will be rationally convex, in the half open interval, and super additive for reals in $(0,1)$, as, $F(0)=0$.
Ie, $F(0)=0rightarrow F(0)leq 0$, which means that $F$ is star convex.
Let, $y=0$.

$F(sigma x +(1-sigma)0)leq sigma F(x)+ (1-sigma)times F(0)leq sigma F(x)+ (1-sigma)times 0=sigma F(x)$.
Use star convexity twice:
$$(1):text{let},sigma=frac{x+y}{x}$$.
And,
$$(2):text{let},sigma=frac{x+y}{y}$$.

Then,
$$F(x)=F(frac{x}{x+y}times (x+y))leq frac{x}{x+y}times F(x)$$.

And,

$$F(y)=F(frac{y}{x+y}times (x+y))leq frac{y}{x+y}times F(y)$$.

$$F(x)+F(y)leq frac{y}{x+y}times(F(x+y))+frac{x}{x+y}times (F(x+y))=frac{x+y}{x+y}times F(x+y)=F(x+y)$$

That is, $$(RSS): F(x+y)geq F(x)+F(y),text{, where,},, x>0,land,y>0$$.

Ie, as $F$ may not be convex over the entire unit interval (so that one cannot in that case, let $sigma=1$, then in the substitutions above, $(1) :xneq x+y$, and in $(2):yneq x+y$, and thus that, $xneq 0$, and $yneq0$, that given the domain constraints, that, $(x,y)in [0,1]$, that, $x>0$, and $y>0$. (i.e, they are non-negative, but not equal to zero, so they are positive).
It is also satisfied at $0$, because, $$F(0)=F(0+0)=0=0+0=F(0)+F(0)rightarrow F(0+0)=F(0)+F(0)rightarrow F(0+0) geq F(0)+F(0)$$. And given the domain constraints, this is only instance where $x+y=0$.
Does, super-additivity, extend to domain sums of 1, ie, one instance clearly holds, given the $F(0)=0, F(1)=1$:

$$(RSSM):F(1)=F(0+1)=1=1+0=F(1)+F(0)rightarrow F(1+0)geq F(1)+F(0)$$.

I suspect so, given strict monotonic increasing of $F$ and $F(1)$, as follows:

My intuitions are as follows: if it did not hold that $x+y=1rightarrow F(x)+F(y)leq 1$, so that instead $F(x)+F(y)>1$.

Then if, (and this is what it depends on), there is $F(x)+F(y_1)=1$, it would follow from strict monotonic increasing-ness of $F$ that, $y_1<y$, as $$F(x)+F(y_1)<F(x)+F(y)rightarrow,F(y_1)<F(y)rightarrow,y_1<y$$.

This implies, that, (A): $x+y_1<x+y=1rightarrow x+y_1<1$

As, (B): $x,yin text{dom(F)}=[0,1]rightarrow, xgeq0,land,ygeq 0$,
so, that by $(A)text{and}(B)$ give that: $x+y_1in [0,1)$.

As the case, under consideration, however, it must be that, $x+y_1neq 0$, for otherwise, if $x+y_1=0$, that, $x=0, land y_1=0$, which would mean, that as, $F(y)>F(y_1)$, that $F(x)+F(y)>F(x)+F(y_1)=1=F(1)$, and thus, $F(x)+F(y)>1$.
This, not only violate the one instance of super additivity, $(RSSM)$, which has been shown to hold earlier. This is because as $(RSSM)$ requires that $F(x)+F(y)leq 1$, but instead, $F(x)+F(y)>1$. Moreover, $F(x)+F(y)>1$, also contradicts, $F(0)=0wedge F(1)=1$, simpliciter.

This is as, $x=0wedge x+y=1rightarrow y=1$, which means that $F(y)=1$, and thus, that: $F(x)+F(y)=F(0)+F(1)=1$, which contradicts $F(x)+F(y)>0$, above

As a result, a contradiction results. I.e, $0=F(0)=F(x)wedge 1=F(1)=F(y)$, and the one instance of super-additivity (with a sum of 1) above, $(RSSM)$, for that matter.

In any case, as $x+y_1in (0,1)$, this means that the restricted form of super-additivity $(RSS)$, applies. i,e, from the strict monotone increasingness of $F$, and $F(1)=1$, that, $x+y_1<1rightarrow F(x+y_1)<F(1)rightarrow F(x+y_1)<F(1)=1rightarrow F(x+y_1)<1$, which by super-addivity, means that $F(x)+F(y_1)leq F(x+y_1)<1rightarrow F(x)+F(y_1)<1$, which contradicts super-additivity on the open interval $(RSS)$.

So, I believe I have answered the first part, and (I suspect) the last part of own questions.

That is, it is super additive over $[0,1)$, but not necessarily super additive over the entire unit interval, because there are midpoint convex functions defined over closed intervals that are not continuous and thus convex at $1$, i.e the step function:
$$F:[0,1]rightarrow [0,1]$$.
Where,

$$xin (0,1]:2x^2;
$$F(1)=1$$.

Which is discontinuous at $1$, and is not super-additive, at $1$, so $(RSS)$ fails, in precisely the way indicated where neither $x$, nor $y$, are in ${0,1}$. It is however, super additive on $[0,1)$.

(For a counter-example to $(RSS)$, just let, $x=0.7, y=0.3$ wedge x+y=1$.

I.e, $F(0.7)+F(0.3)=0.98+0.42>1=F(1)rightarrow neg (F(0.7)+F(0.3)leq F(0.7+0.3))rightarrow neg (F(x)+F(y)leq F(x+y))rightarrow neg (forall ((x+y)in {1}):F(x)+F(y)leq F(x+y))$.

And as,

$neg(forall((x+y)in {1}):F(x)+F(y)leq F(x+y))rightarrow neg(forall ((x,y,(x+y))in [0,1]):F(x)+F(y)leq F(x+y))$.

which is to say that there may not be $x+y_1in (0,1), $x+y_1<1,text{where} $F(x)+F(y_1)=1$, so as to contradict by strict monotonic increasing of $F$ that $F(x)+F(y)leq F(

See:
mid-point convex but not a.e. equal to a convex function.

Actually I have realized that if $F$ is real valued then its rationally convex, Thus if Are midpoint convex real valued functions rationally or (nearly) rationally super-additive if $F$ is real valued.
$F:[0,1]to [0,1]$ and $F(0)=0$

Such functions should be able to express and for all rational

$$ forall p in { [0,1] cap mathbb{Q} };forall(x) in text{dom}(F):; F(px), leq pF(x)$$

$$F(px) leq p F(x)$$

for all rational $p$ $p$>1 ;

$$F(px) leq p F(x)$$

and by a change of variable.

As such functions satisfy when real valued $$forall (x,y)in text{dom}(F);, forall tin {[0,1] cap mathbb{Q}};F(sigma x +(1-sigma)y ) leq sigma F(x) +(1-sigma)F(y)$$
And so long as $F(0)=0$ and $F$ Strictly monotone increasing and in-jective. Then one can express such claims.

I presume that if the function $F$ not only satisfies $(1)$ but also $(2)$ and $(3)$ and perhaps one of $(4)$ and $(5)$ it will be continuous and convex?

1 $F:[0,1]to[0,1]$, and $F$ is midpoint convex, where $F(0)=0$ and $F$ strictly monotonically increasing,

$$ 2 Ftext{ is strictly quasi concave,strictly quasi convex, quasi concave, strictly quasi concave}$$.

$$3.F(1)=1,quad F(0.5)=0.5$$

$$4.F(2x)=2F(x)$$

$$5.F(1-2x)=1-2F(x)$$

Such midpoint convex functions, will be convex and continuous, or if not, at least will be, if they agree with $F(x)=x$ a continuous function, over a dense set such as the dy-adics rationals, with the same end points. If for example if $F$ also satisfies both $(4)$ and $(5)$ or both of their strengthened variants, $(4.1)$ and $(5.1)$ below.

For example if $F$ also satisfies.
$$(4.1)F(x)= frac{F(2x)}{2} =0.5- frac{F(1-2x)}{2}$$

and

$$(5.1)F(1-x)=1-F(x)=1- frac{2F(x)}{2} = 1-frac{F(2x)}{2}=0.5+ frac{F(1-2x)}{2}$$

Then,
$F$ agrees with $F(x)=x$, a continuous function, over the dyadic rationals from as dense subset of $[0,1]$ and as $F$ is strictly increasing, and agrees with $F(x)=x$ in said sub-set, and $F$ is generally equal to $F(x)=x in (0,1)$ by the following theorem:

and as $F$ is stipulated to also agree on the endpoints $F(1)=1$ and $F(0)=0$ as well, I presume that, $F(x)=x$ and is continuous.

I am not sure if the strict monotonic increasing, , given three fixed points is enough otherwise, unless strict monotonically helps with lebe-sgue measur-ability but I might be wrong?

. On the other hand, I have been recently informed by a scholar, that there are midpoint convex functions that are not super-additive. These are mentioned in <(see the Hille-Phillips monograf, Chapter 7, especially Th. 7.2.5)>.

I have not read into this, but if these functions have the property that $F(0)leq0$, and are defined on domain and on the positive reals$;(which they may not, but I do not know, as I said I only recently found about it) .

Then such mid-point convex functions presumably cannot be convex functions, given convex functions have ( have, or generally have this property, when $F(0)leq0$).

If there exist Midpoint convex functions with $F(0)leq0$, are not super-additive on the positive reals, then presumably such functions cannot be continuous either (I presume). If such mid-p-point convex functions exist, that is. So these, may be a, or a purported/alleged, an example of a non-convex, albeit midpoint convex function. As per the question inquiry.

So I can only suggest that book, unless you have already considered it, and they may not have $F(0)leq0$ in which case, I presume that they would be convex and not an example but I do not know about their other properties, so I am tentative.

I presume that there do not exist continuous, midpoint convex functions of this nature, lest they would be convex unless, they are highly irregular and not strictly, monotonic, which would not make a great deal of sense,in some cases .

I mention this, because unlike Jensen's equation, which in the non-continuous case with $F(0)=0$, is additive (for all reals, or at least rationals I presume), and rationally (or at least dyadic ally rationally homogeneous).

It is not clear that a midpoint convex, strictly monotonic increasing function with $F(0)=0$,

even if $$F:[0,1]to [0,1]$$

$$text{where F is a one -one function, real valued midpoint convex function} $$

$$text{and with}, $F(0)=0,, F(1)=1,, F(0.5)=0.5$$,

are even rationally super-additive, in the not necessarily-continuous case?.

I note that is now wrong. Given that $F(0)=0$ and rational convexity of real valued midpoint convex functions. See below pt. 7.11 in

The midpoint convex function $F$ may at best be only, dyadic ally super-additive, although from what I have read ,and though about since, it presumably can express $F(3x)geq 3F(x)$ and

thus $$F(frac{1}{3} x) leq frac{1}{3}F(x)$$

$$F(frac{2}{3} x) leq frac{2}{3}F(x)$$.

Its homogeneity-inequality properties:

$$F(sigma x) leq sigma F(x)$$;

If at all, may only extend to positive integer multiples such as

$$ sigma;{ sigma in mathbb{N},sigma geq 1 }, text{where} ,x in text{Dom}(F)=[0,1]$$

s.t $$F( sigma x) geq , sigma F(x)$$
$$F(3x) geq 3F(x)$$

Otherwise (if $sigma <1$, it may express, at best $leq$ which are
$F( frac{1}{ sigma} x) leq frac{1}{ sigma}F(x)$, if $F$ can only express dyadic rational inequality fractions such as:

$$F( frac{1}{4}x) leq frac{1}{4}F(x)$$.

If so, then I would have extreme doubts about whether it even express integer multiples of $x$ other then halving and doubling inequalities such as:

$$sigma;{ sigma; sigma>1, ,sigma in mathbb{N} }, text{where},x, in text{dom}(F)=[0,1]; F(sigma x) geq sigma F(x) text{for all positive integers>1, as well}$$

Except, that is, the dyadic type integers integers 2,4,8 etc, for which this clearly holds:

$F(0)=0$, $F(4x) geq 4F(x)$, $F(2x) geq 2F(x)$, $F(0.25x) leq 0.25F(x)$

Which holds as far I can see when $F(0)=0$ if, $F$ is midpoint convex on real interval with $F$ is strictly monotonic increasing.

As convex functions generally have this property, it might be useful to check out this reference, to see what is going on there. These are may be 'allegedly' non- convex (I say this very tentatively, as I have not yet had time to look at the reference).

The scholar did say it was an example of a mid-point convex function that was not super-additive.

If he/she only mentioned instead,that it was a convex function, then I would not have thought much of it; although convex functions are midpoint convex, generally one does not assert the weaker property when it has the stronger. But I might be wrong.

?. Available from: https://www.researchgate.net/post/Are_midpoint_convex_functions_rationally_or_nearly_rationally_super-additive_with_F00_or_under_other_conditions_if_not [accessed Apr 25, 2017].

In any case

I have been informed that there are strictly monotonic increasing, midpoint convex functions, $F$ defined on $F:[0,1]to [0,1]$ which are not rationally super-additive, even with $F(0)=0, F(1)=1$where by rationally super-additive I mean $(QS)$

I note that with symmetry $(S)$, with $F(frac{1}{2})=frac{1}{2}$
$F(0)=0$, $F(1)=1$ ,, and $F$ strict monotonic increasing, where $text{dom(F)}=[0,1]$ can be weakened to midpoint convexity at $1$, and $0$, w

I presume monotonically or strict monotone increasing with $F(0)=0$ and $F(1)=1$entails that the co-domain is $[0,1]$? A

As strictly monotone increasing $F:[0,1]to [0,1]$ is a doubling halving symmetric, function which with $(S)$ $F(0)=0, F(1)=1$.

As these midpoint inequalities due $F(0)=0, F(1)=1,(S)$ collapse into equalities, and thus is a doubling, halving function that is symmetric, and thus $F$ agrees with the identity, $G$,a continuous function, G(x)=x for all dyadic rationals in $in[0,1]$,, over a dense, set of $[0,1]$ and as $F$ strictly monotone increasing, $F=G$ is is the identity function=on $(0,1)$

1. List item

For example , to show that if midpoint convexity entails 'rational convexity',

1.Given $F:[0,1]to[0,1]$ midpoint convexity $(MP)$, $F(1)=1,,F(0)=0,F(0.5)=0.5$.

2. To show that $F(x)=x, forall xin [0.5,1]cap mathbb{Q}$




3. I presume that its generally a consequence of rational convexity
   with $F[0,1]to [0,1],F(1)=1, F(0)=0$ that $forall(x)in {[0,1]cap mathbb{Q}};,F(x) leq x$

That is just by varying the rational $t,(1-t)$, for $x=0,y=1,text{where},F(x=0)=0,, F(y=1)=1$ in the rational convexity equation?

Or at least for all dyadic rationals (which are nonetheless a dense set).

4. Although then one could only use that argument for the second $geq$ which wont give one very much.

Strangely not dyadic multiples of the unit event ,$1$but of $0.5$ such $F(frac{2}{3})geq frac{2}{3}$

5. I note that, the other inequality $forall rational(x)in [1/2,1]F(x) leq x$;

depends really only on rational convexity at $0$.

6. I note that ifs its merely, star convex at $0$ (which is real valued, but only applies to that point) with those three fixed points F(x)=x it appears, for all real $xin [0.5,1]$.




7. Example, $F(0.5)=0.5$,$F(0)=0$ that, $F(0.75)geq 0.75$ as below, and standard midpoint convexity to show that $F(0.75)leq 0.75$ ie $F(0.75)=0.75$ .

$$0.5=F(0.5)=F(frac{1}{3}times 0 + frac{2}{3}times frac{3}{4}) leq frac{1}{3}times F(0)+frac{2}{3}times F(frac{3}{4})=frac{2}{3}times F(frac{3}{4})$$

$$0.5rightarrow frac{2}{3}times F(frac{3}{4})rightarrow frac{3}{2}times 0.5leq F(frac{3}{4})$$
$$rightarrow F(frac{3}{4})geq frac{3}{2} times 0.5= frac{3}{4}$$
$$rightarrow F(frac{3}{4})geq frac{3}{4}$$

• or using midpoint convexity twice without using $F(0.5)=0.5$

using only$F(1)=1 ,F(0)=0$$($x=1,x=0,to get that $$F(0.5)=F(frac{(1+0)}{2})leqfrac{F(1)+F(0)}{2}=frac{1}{2}+frac{0}{2}= 0.5rightarrow F(0.5)leq 0.5$$

$$rightarrow frac{1}{2}times F(frac{1}{2}) leq frac{1}{2} times frac{1}{2}=frac{1}{4}$$

$$rightarrow frac{1}{2} times F(frac{1}{2})leq frac{1}{4}$$

$$rightarrow frac{1}{2} + frac{1}{2} times F(frac{1}{2}) leq frac{1}{2}+ frac{1}{4}=frac{3}{4}$$,
$(MP)$and at$$x=1,x=frac{1}{2}$$

$$ F(frac{3}{4})=F(frac{1+frac{1}{2} }{2}) leq frac{1}{2} times (, F(1)+F(frac{1}{2}),)= frac{1}{2}+ [frac{1}{2} times F(frac{1}{2})]leqfrac{3}{4}rightarrow F(frac{3}{4})leqfrac{3}{4}$$
so $$[F(0.75)leq 0.75 land F(0.75)geq 0.75 ]rightarrow F(0.75)=0.75 $$.

and using $F(1)=1$ and $F(0)=0$ (dyadic convexity)
/ midpoint convexity, twice (without $F(0.5)=0.5$)using $,F(1)=1,,F(0.5)=0.5$ $x=1$,$y=0.5$ @), to get $F(0.75)leq 0.75$

or once, with $F(0.5)=0.5, F(1)=1$
$rightarrow F(frac{3}{4}) leq frac{3}{4}$

Question ## Can use the identity function's limiting behaviour on (0.5,1)to show continuity at $1$.

One already knew that $F$ was continuous in $(0.5,1)$, via monotonic increasing-ness and midpoint convexity (measurablity) and thus, is it continuous in $[0,1)$ due to $F(1)=1,F(0)=0$
but not precise the form of the function in that half interval as one may know know.

1. As $G$ the same end points as $F$, $G(1)=F(1)=1$ and $G$clearly limits toward that value $xto 1, G(x)to 1$,as $G$continuous,


2. As presumably when you take the limit you never actually get to $1$ which is precisely where $F=G$.(well at least on the top half $(0.5,1)$)? Can you use this limit as F's limit? Correct or not.?



3. And other wise can do this if one can show $F=G$ on the entire open interval $(0,1)$ to show continuity at the end points, where both agree $F(1)=G(1)=1$, $G(0)=F(0)=0$ where $G$ continuous and $F$ strictly monotonic increasing and identical to $G$ on $(0,1)$ for example, if $F=G$ all dyadic rational/rationals, a dense set $in[0,1]$, where $F:[0,1]to[0,1]$




4. Can one do this? Or does one have to do it indirectly or does the fact that F is stipulated before hand, to be $F(1)=1$ and $F(0)=0$ give one the desired result that that F=G on the entire interval already given that $F=G$ on $(0,1)cap {1}cap {0}=[0,1]$?

$