Strictly Monotonic, Continuous Sub-linear functions with F(1)=1; Any real difference between these and linear functions?

Is there a great deal of difference between sub (or super-linear functions) Functions F, and linear functions, ie, when a sub-linear function $F$, is also strictly monotonic increasing and continuous , where $F$, such that $F(1)=1$, and $F:[0,1] \to [0,1]$?

Won't $F$ be both super-additive and sub-additive? That sub-linear $F$, are both sub-additive and mid-point convex, with $F(0)=0$, along with homogeneity properties such as $F(nx)=nF(x)$ where $n$ is generally a positive integer.

So when such a function is continuous, presumably it be convex and super-additive (and sub-additive), over $[0,1]$ given convexity and $F(0)=0$ ? And with strict monotonic increasing and $F(1)=1$, presumably $F(x)=x$?

See the following: for more on these functions.

Bingham, N.H.; Ostaszewski, (1)A.J., [Automatic continuity: subadditivity, convexity, uniformity] (http://dx.doi.org/10.1007/s00010-009-2982-x), Aequationes Math. 78, No. 3, 257-270 (2009). ZBL1209.26009.

For example, where $F$ is a Strictly monotonic increasing, continuous, sub-linear function , with, $F(1)=1$ from the unit interval to the unit interval.

$$(1.1)F:[0,1] \to [0,1]$$, .

$$(1.2)F(1)=1, F(0)=0 \quad \text{ where F(0)=0, as result of (3) below}$$

$F$ is Sub-additive:

$$(2). \forall (x,y) \in [0,1]:F(x+y)\, \leq\, F(x)+F(y)\,$$

$F$ is positive homogeneous with regard to integers

$$(3)\forall x \in [0,1];\, \forall (n) \in \mathbb{N}\,;n \geq 0;\, F(nx)=nF(x)$$

$(4),(5)$, $F$ is Continuous $(4)$and midpoint-Convex$(5)$

$$(5)F(\frac{x}{2}+\frac{y}{2}) \leq \frac{F(x)}{2} + \frac{F(y)}{2}$$

All or most sub linear $F$ functions are midpoint convex, as a result of positive homogeneity for $n=2$ and and sub-additivity. As $F$ here is presumed continuous $F$ will be convex, and sub-additive, (and super-additive over the positive reals)?

See Category-measure duality: convexity, mid-point convexity and Berz sub-linearity>

What is the nature of such a function,$F$?.But it looks like its more linear then one may think over the unit interval, its just linear?

It would appear to be weaker then a real valued, Jensen function, with $F(0)=0$, $F(1)=1$ and $F$ strictly monotonic increasing from the unit interval to the unit interval,at least in the non continuous case?

Especially if eq $(3)$ only holds for $n=2,F(2x)=2F(x)$?

I do not think the continuous case will be any different as will have more then three fixed points, and be strictly monotonic, increasing, sub-additive, doubling, and continuous,convex and super-additive.

In both cases the function is strictly monotonic increasing $F:[0,1] \to [0,1]$, and $F(0)=0$ and $F(1)$, and in some sense both super-additive and sub-additive** A convex function, real valued with $F(0)=0$ is generally super-additive over the positive reals.

However, the continuous results do not appear greatly different; particular if strict monotone continuity holds and $F(nx)=nF(x)$ for all positive integers.

And thus effectively all rationals by substitution, if one can express a 3 times, and 2 times, one express halves and third, and thus two thirds.

Generally if $F$ is strict monotonic increasing function entails quasi-convexity and strict quasi convexity which also entail, convexity given midpoint convexity with milder constraints, as well, which is weaker than continuity.

Even if $(3)$ only holds for $F(2x)=2F(x)$, such models will be sub-additive, iff midpoint convex, and conversely (midpoint convex, if sub-additive) .

Thus such functions, if continuous, will be convex, and thus will presumably be both, sub-additive and super-additive over the positive reals, as $F(0)=0$ given

$F(2x)=2F(x)$, and$F$ i s convex, https://en.wikipedia.org/wiki/Convex_function.
Given $F(1)=1$, and strict monotonic increasing; and $F(2x)=2F(x)$ There could not much of a difference.

Midpoint convexity which generally follows from F(2x)=2F(x), sub-additivity) then F is generally super-additive on the positive reals as wells, I presume the same or a similar result would follow if (instead of sub-additivity $F(2x)=2F(x)$ and midpoint convexity were posited instead (of sub-additivity and $F(2x)=2F(x)$

The issue being that once can express this claim for all integers one can do it for a great deal if not all rationals (and sometimes homogeneity is taken to include for all positive real valued sigma; $F( \sigma x)=\sigma F(x)$.

Just given the unit event $F(1)=1$, continuity and strict monotonic increasing, a weak sub-linear model, that is sub-additive and satisfies only, $F(2x)=2F(x)$, and $F(0)=0$ will be $F(x)=x$?.

Maybe only with $F$ monotonic, except for the irrationals values, it will be strictly monotonic for all extents and purposes.

Se eLapidot, Eitan, On generalized mid-point convexity, Rocky Mt. J. Math. 11, 571-575 (1981). ZBL0503.26007.

Bingham[pdf, ps, other] " Cauchy's functional equation and extensions: Goldie's equation and inequality, the Gołąb-Schinzel equation and Beurling's equation "

and: Xiv:1405.3948 >[pdf, ps, other]

$(2)$ "Bunce, L.J.; Wright, J.D.Maitland, Non linearity and approximate additivity, Expo. Math. 12, No.4, 363-370 (1994). ZBL0844.46030. by,

$(3)$N. H. Bingham, A. J. Ostaszewski: Companion paper to: Cauchy's functional equation and extensions: Goldie's equation and inequality, the Goldie-Schinzel equation and Beurling's equation."

$(4)$. see> [pdf, ps, other]

$(5)$N. H. Bingham, A.J. Ostaszewski, "Cauchy's functional equation and extensions: Goldie's equation and inequality, the Gołąb-Schinzel equation and Beurling's equation:Companion paper to: Additivity, subadditivity and linearity: automatic continuity and quantifier weakening"/

• integers, dyadic rationals or all rationals). – William Balthes Apr 23 '17 at 19:48
• And I can only suppose that such models differ, only because its more difficult to derive continuity, or are considering different domains from [0,1] to [0,1]; or are not strictly monotonic or in-jective. Is this only a conditional cauchy equation which holds for arbitrary pairs of values. – William Balthes Apr 23 '17 at 19:49
• Not generalizing to factors x,y,z,which either do not sum to the unit event,F(1), and/onlyapplies to pairs (and/or sets of four, eight) x,y,z etc in the domain) and thus, not generalizable to F(x+y+z)=F(x)+F(y)+F(z) unless such factors are divisible by 2 ie only in virtue of such events being expressed as having a common factor (or divisible by 2). – William Balthes Apr 23 '17 at 19:49