*strong text****Is this symmetric, doubling,**strictly monotone increasing function $F:[0,1]\to[0,1]$ below, continuous and $F(x)=x$ over the unit interval?

Let $F: [0,1]\to [0,1],F(1)=1,F(\frac{1}{2})=\frac{1}{2}\, F(0)=0\, \,:\text{be a strictly increasing Function}$ that in addition satisfies,

$(2)$halving/doubling and $(3)$ symmetry, inverse symmetry; oddness/left relfection right symmetry around $\frac{1}{2};F(1-x)+F(x)=1$


$$(1)F:[0,1]\to [0,1]\,,,F(1)=1,F(\frac{1}{2})=\frac{1}{2}, F(0)=0$$

$$(2).(x)∈[0,1];∀(n)∈Z;[F(\frac{1}{2^{n}}\times x)=\frac{1}{2^{n}}\times F(x)]$$

$$(3)\text{ (symmetric,and in-jective/bi-conditional symmetric)}\,\forall (x,y)\in\text{dom(F)=[0,1]}\,[x+y=1] \leftrightarrow F(x)+F(y)=1$$

$$(4)\text{F is a Strictly monotone Increasing function over it domain=[0,1]}$$ $$\forall(x,y)\in\text{dom(F)}=[0,1]$$:

  1. $$(a)x>y \leftrightarrow F(x)>F(y)$$
  2. $$(b)x=y \leftrightarrow F(x)=F(y)$$
  3. $$(c)x<y \leftrightarrow F(x)<F(y)$$

## Heading ##1. is :$F(x)=x$? over the unit interval $[0,1]$?


I originally had this which can only apply when $2x\in [0,1];x\leq 0.5$ so I changed the conditional $\forall(x)\in \text{dom(F)=[0,1]}:\,F(2x)=2F(x)$

Thanks to one of the answers for point thing out.

  1. Symmetry $(3)$ is to be read as if and only if, any $2$ elements of the domain sum to $1$, the corresponding $2$ images, function values sum to $1$, and if any $2$ elements of the domain do not sum to $1$ , then their respective $2$ function values do not sum to $1$,ie ie $F(1-x)+F(x)=1$, and $F^{-1}(x)+F^{-1}(1-x)=1$

    The second consequence is a result of $F$ being injective and the inequalities version due to strict monotony which also gives this.

2.$(2)$ Where Doubling is to be read as$$F(\frac{1}{2} \times x)=\frac{1}{2}\times F(x)$$ and generalizes to $4$,$8$ and$\frac{1}{2^{n}}$ etc. dyadic powers although not necessarily fractions,(it gives those values, but that is numerical) and the values will respect these appropriate mutliplicative relations, insofar as its solvable . I presume that this a little weaker

Just as rational homogeneity $F:[0,1]\to[0,1]$ with $F(1)=1$, is stronger then $F(x)=x,\forall x\in[0,1]\cap \mathbb{Q}$ which does not say anything but the rational relations between irrational values when they apply unlike the first unless it does, given symmetry (it gives those results numerically due to the range cod-domain and unit). one also hs

  1. $$\forall(x)\in[0,1];\,F(1-x)=\frac{1}{2}+\frac{F(1-2x)}{2}$$ $$F(x)=\frac{1}{2} - \frac{F(1-2x)}{2}$$ 2.$F(1-x)=1-\frac{(F(2x)}{2},F(1-2x)=1-F(2x)=1-2F(x)$

    Moreover $(4)$strict increasing-ness also entails the symmetry conditions and inequality variants as bi-conditionals and likewise for the inverse, function,essentially one and the same statement when expressed in this form as far as I can see.

That is, for any two elements of the image of $(F)$, which sum to $1$, their corresponding domains elements must to $1$ etc), as a result of the bi-conditional of LHS direction for = in $(4)$, $<, \land\, >$ in the inequalities above (the other conditional is presumably irrelevant once has specified that if the domains elements $=1$, $<1$, or$ >1$,given that any two domain elements will always satisfy onone and only one of these three. inequalites

$ F(x)+F(1-x)=1$and $F^{-1}(x)+F-1(1-x)=1$$

1. $F(x)=x$ on all of $[0,1]$ Or only on $(0,1)$ (not sure what that would mean here)

*strong textI presume that this function $F$ will be $F(x)=x$ on $(0,1)$ and continuous as it agrees with $G(x)=x$ ,a continuous (the identity) function over a dense set of $[0,1]$,and as $F$ is strictly monotone increasing,on $[0,1]$ which apparently entails that $F(x)=x$ is continuous and $F=G$ is the identity on $(0,1)$

2. is $F$ symmetric doubling mean 'group-oid' function?

3. Is,there,a,way,of,imbedding the equations$(2),(3),(1)$,into,one equation?

In,particular,doubling $(2)$,and,symmetry $(3)$ given the domain and co-domain the $,F(\frac{1}{2})=\frac{1}{2},F(0)=0\,F(1)=1$

  1. As an example $(2),(3)$ entails by doubling $(2)F(2x)=2F(x)$, and symmetry $(3)F(1-x)+F(x)=1$that $F(x)=x$ for all dyadic rationals,$\in[0,1]$ and a great deal of non-dyadic rationals?

Is this correct.?

Note that $F(0)=0$,$F(\frac{1}{2})=\frac{1}{2}\, F(1)=1$ as consequence. where $F[0,1]\to [0,1]$

(at least up to about $$< 17$$. I am not sure can it capture all, rationals, before monotony is applied?

I not sure if it can go beyond this for prime numbers (before one applies strict monotone increasing-ness) and dyadic multiples of these, as far as I can, as far as I can see, before (strict) monotonicity $(4)$ is applied?.

.5 I presume it will be continuous at $1$ and $0$?

$$4.a.\forall(x,y) \in\text{dom(F)=[0,1]};\,x<y \leftrightarrow F(x)<F(y)$$ $$4.b..\forall(x,y)\in\text{dom(F)=[0,1]};x=y \leftrightarrow F(x)=F(y)$$ $$4.C.\forall (x,y)\in \text{dom(F)=[0,1]};\,x<y \leftrightarrow F(x)<F(y)$$ $$(4,)(3),(a)\forall(x,y) \text{dom(F)=[0,1]}\,\,x+y>1 \leftrightarrow F(x)+F(y)>1$$, $$(4),(3)(b) \forall(x,y)\text{dom(F)=[0,1]};\,x+y<1 \leftrightarrow F(x)+F(y)<1$$ $$((4),(1)(3))b \forall(x\in (0,\frac{1}{2}) \leftrightarrow 0<F(x)<\frac{1}{2}$$

$$(4,(1),(3))a \forall(x\in (\frac{1}{2},1) \leftrightarrow \frac{1}{2}<F(x)<1$$ (where $(2)$ can be read as this)

  1. Example

ie for example (3) from symmetry $$F(\frac{1}{3})+F(\frac{2}{3})=1$$, which,due,to,doubling$(2)$,gives,

$$F(\frac{2}{3})=2 \times F( \frac{1}{3})$$, which by symmetry $(3)$ $$(3)F(\frac{1}{3})+F(\frac{2}{3})=1 \rightarrow 2 \times F(\frac{1}{3})+F(\frac{1}{3})=1 \rightarrow 3 \times F(\frac{1}{3})=1$$ $$\rightarrow F(\frac{1}{3})= \frac{1}{3}$$ $$F(\frac{2}{3})=2 \times F(\frac{1}{3})=2 \times\frac {1}{3}=\frac{2}{3}$$


  1. Is there a way to embed the and doubling and symmetry equation into one constraint without losing any content

strong textIs there a way to embed the symmetry constraints, and doubling/halving constraints, and strict monotonically increasing $(1)$ $(3)$ and $(2)$ into one equation? That is into one functional equation, given this range and co-domain and the three (initial) fixed point that they entail $F(1)=1, $F(0.5)=0.5, $F(0)=0$ .

I presume it does not entail $F(nx)=F(x)+F(n-1)x$?

  1. I presume that strict monotony is required rather than monotony

strong text? Otherwise it would appears that Jensen's equation/Cauchy equation is over-determined on the unit interval, as one only appears to require Jensen' equation at $(1)$ $0$,(doubling)and $F(1)=1$ $F(0)=0$ given symmetry , and given symmetry one can get to $(1)$ from doubling (I only have these as both e-qualities fall out midpoint convexity at $1$ and $0$, I do not think anything weaker would do it.

  1. Only actually require midpoint convexity at $1$ and $0$ for symmetry to entail doubling as far I can see I have shown that. I presume that is not an error (the derivation below is not the whole thing of it).


I presume these are the minimal conditions for the identity functions (except perhaps monotony instead of strict monotony)

This is due to the fact that are im-bedded in the symmetry and doubling/halving constraints $(2)$ and symmetry $(3)$ and the range and co-domain constraints given the symmetry equations.

(although one cannot express it as dyadic-ally homogeneous\forall (x)$[0,1]$ $$F(k\times x)=k\times F(x)$$, $\forall$ dyadic rationals $(k) \in [0,1]$ F(, but only as a function of the $1$, unlike, doubling, which holds every where,Which gives the desired values over the domain and range of interest.

I presume that the former stronger constraint as an equality with symmetry may roughly equivalent to rational homogeneity by a change of variable if one can expresses multiple homgoneity such as $\frac{3}{4}$ is this not one also saying by change of variable $\frac{4}{3}$ and then one use doubling or to get to $\frac{1}{3}$ etc, unless that is not allowed?



  • 1
    $\begingroup$ The second condition doesn't make sense at all, because if $x\in [0,1]$, with $x>0.5$, then $2x \notin [0,1]$ $\endgroup$ – Mercy King Jun 22 '17 at 15:04
  • $\begingroup$ Correct. I should be a bit more careful; obviously is should state that $2x $must be in the domain $[0,1]$ , thanks for this.Otherwise, i should write it as : $ \forall(x)\in[0,1];\,\forall(n) \in \mathbb{Z} ;[F( \frac{1}{2^{n}} x)=\frac{1}{2^{n}} F(x)]$ $\endgroup$ – William Balthes Jun 22 '17 at 15:29

Note that one is also has this can be derived from doubling or symmetry or in any case is a result of the mid point inequalities convexity at $1$ and $0$ collapsing into equalities. however, once has $F(x)=F(2x)$ (as a result of $F$ midpoint convexity at $0$ collapsing into Jensen at $0$) with $F(0)=0$, and $F(1)=1$,

then the fact that one uses midpoint convexity at $1$ and that is also collapses into an inequality becomes redundant, as the given the symmetry one can effectively get that result from $F(1)=1$, $F(2x)=2F(x)$ $F:[0,1]\to[0,1]$ and symmetry $(3)$to get that result becomes redundant fr

$$F(1-x)+F(x)=1, and $F^{-1}(x)+F-1(1-x)=1$$

One can get to Jensen equality with midpoint convexity at everywhere given symmetry $F:[0,1]\to[0,1]$, and presuming as $F(0)=0$ $F(1)=1$ and non negative one can get to Cauchy equation over the unit triangle I presume then that non-negativity or is only required or positivity (rather then monotony or strict monotone increasing).

I am trying to avoid the fact that with the third fixed point the function will , be presumed convex and continuous and monotonic increasing/strict monotone with the three fixed points the third being $F(0.5)=0.5$ that symmetry will would have been redundant so i presumed only mid-star convexty at $0$ and $1$ so as it not rendering symmetry redundant in some sense, and that it appears that given strict monotony, midpoint convexity at points is more than what is required. I do not think any thing less then (midpoint convexity at top and bottom would work) I am not sure. one really needs to only get doubling although even that may be sufficient given monotonotony (as opposed to strict) and does not just get all of the dyadic rationals (i do not think it will get all rationals directly) ie, for non-dyadics prime rations, just those <17,I think. I think if prime, and otherwise, it will get the remainder, (or most of them) I think (perhaps not the multiples of some of these such 1/34 but i do not know or otherwise non prime ie midpoint convexity where the first term is x=0 and the midpoint convexity where the first term is 1, where $F(0)=0$ and thus by symmetry $F(1)=1$

$(1)$ midstar/convex,at $0$,with $F(0)=0$ $(2)$ is mid-convex at $1$ with $F(1)=1$

$$(1)F(\frac{1}{2} x) \leq \frac{F(x)}{2}$$ $$(2)F(\frac{(1+x)}{2})\leq \frac{1}{2}+\frac{F(x)}{2}$$

where I express $(2)$ as $(2a)$ $$F(1-x)=F(\frac{2}{2} \times (1-x))=F( \frac{2-2x}{2})= =F(\frac{1+(1-2x)}{2})\leq \frac{1}{2}+\frac{F(1-2x)}{2}$$

$$(2a)F(1-x)\leq \frac{1}{2}+\frac{F(1-2x)}{2}$$ and $(1)$ as $(1a)$ $$(1a)F(x)\leq \frac{F(2x)}{2}$$

symmetry$$ 1-2x+2x=1 \leftrightarrow F(1-2x)+F(2x)=1\rightarrow \frac{F(1-2x)}{2}+\frac{F(2x)}{2}=\frac{1}{2}$$ symmetry $$1-x+x=1 \rightarrow F(1-x)+F(x)=1$$

These collapsed into eq-ualities given symmetry $(3)$ as you will notice that the sums of the RHS +RHSof $(1a)$ and RHS$(2a)$=LHS$(1a)$+and LHS$(2a)$ are both 1 and thus equal so its not possible for one terms to be strictly smaller then the other, if the other is forced to less than or equal. else the sums would be different. There is a formal way of deriving it of course, its not very difficult but one can easily show that that the other inequalites $\geq $fall out giving (1a) and (2a) as equalities. Likewise for $(1)$ and $(2)$ I think. In any case on has

$F(2x)=2F(x)$,$F(x)=\frac{1}{2}-\frac{F(1-2x)}{2}$ $F(1-x)=\frac{1}{2}+ \frac{F(1-2x)}{2}=1-\frac{F(2x)}{2}$ and

$F(1-x)+F(x)=1$ $F^{-1}(1-x)+F^{-1}(x)=1$

and their halving four times, 8 times varianes etc from here its quite easy to show that F(x)=x for all dyadic rationals.and rationls which all factors of five 3,7 roughtly to about 17 unlesss non prime or some kind of multiple of it

with $F:[0,1]\to[0,1]$ $F(0)=0$, $F(\frac{1}{2})=\frac{1}{2}$ $F(1)=1$

where F is strictly monotone increasing


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