linearity, continuity, $F(x)=x$, and the Cauchy field automorphism equations and transcendental values

Question

Where $F: [0,1] \to [0,1]$; along with $F(1)=1$. Do the field auto-morphism equations

1. $F(x+y)=F(x)+F(y) \quad\forall x\in\, [0,1]$

2. $F(xy)=F(x)F(y)\quad \forall x\in\, [0,1]$

Uniquely specify that $F(x)=x$, and that $F$ is an involution $F(F(x))=x$

I find that hard to completely believe, do these equations say anything precise about the value $F(x)$ where $x$ is transcendental number,? $\in [0,1]$.

Transcendental values, that is $\in [0,1]$ compatible with the probability calculus Which are presumably, all or, almost all such transcendental values in $[0,1]$.

Which would be entailed by $F(x)=x$) without any further regularity requirements and conversely as well except only monotonicity, (not even strict)mono-tonicity and not even $F(1)=1$.

If, so What about,along with

$$F:[0,1]→[0,1]$$;

$$F(1)=1$$

$$(1) \forall (x,y) \in [0,1];\, F(x+y)=F(x)+F(y)$$

Please see the comment by below.

Where merely (1) holds, without any further continuity, or regularity conditions explicitly added. Also note, that the domain is restricted to $[0,1]$ so when $x+y>1$ but $(x,y)\in dom(F)$ but $x+y \notin dom(F)$. Does this result still under this restriction. Its presumably implicit. The function is not defined for values of x+y>1 (it will nonetheless restrict these function sums, at least if x+y is rational $F(60)+F(70)=1.3$ nonetheless

Note Please see comment below by Mohsen Shahriari

Please see the Azcel (1989) quotation (eq)18 and (2) corollary 9 ;

$$\text{eq}(2)\, \forall(x,y) \in \mathbb{R_{2}^{+}};\,G(x+y)=G(x)+G(y)$$

Where if $G:_\mathbb{R^{+}} \to R$ satisfies $(2)$ and is continuous at a point, or monotonic, or Lebesgue measurable, or bounded from one side on a set of positive measure, then there exists constants, c, such that $$(18): 'G(x)=cx\, \forall x\, \geq 0$$

"in particular if $(2)$ holds with '$g(x) \geq 0$'; then "$(18)$" holds, with $c\geq 0$ (Azcel 1989, page 18)

I presumed that the last sentence was meant to read that given:

'the regularity conditions(in bold) of corollary $(9)$ are satisfied (continuity at a point etc), and $G:_R+ \to \mathbb{R}$ satisfies $(2)$, then if, in addition $g(x) \geq 0$ holds, the general solution is the same, except that $c$ is restricted to be non-negative.

(18)'$$ \forall x \geq 0;\, G(x)=cx ,\quad\text{and} c \geq 0$$

In contrast, to $(19)$ id '$G:_R+\to R$' satisfies '$(2) \, {\&} g(x) \geq 0$' then '$x\geq 0\,;\,G(x)=cx; \,;c \geq 0$' holds full stop.

And that the continuity/regularity conditions are immediately satisfied.

Although, $(2)$ suggests that it needs to hold for all non-negative reals, I am not sure, but I presume that this remain valid if the Function is only defined on a particular non negative interval ie $[0,1]$ and that with relation to that relevant interval $(2)$ holds, ie here $[0,1]$.

If so, that would mean that any function $F:[0,1]→[0,1]$; would already satisfy $F(x)\geq 0$ as its co-domain is $[0,1]$?

Then, as its domain, is a non-negative real interval $[0,1]$ as well,ie $$(A)F:[0,1]→[0,1]$$

Then if $(1)$ and $(A)$ below hold; $(A)$ being Cauchy' equation over the entire domain then $F(x)=x$ immediately, with no further regularity or continuity requirements; not even mono-tonicity having being presumed for the irrational values: $$(1)F(1)=1$$

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

Which are all for all elements of the non-negative real valued domain.

I think I must be missing something here, as this would appear to reduce $$F(x)=ax .;\,a \geq 0$$;

From being the unique continuous solution for a function, $F$ that satisfies Cauchy's equation, over its entire domain, whose domain is $[0,1]$ and co-domain is $[0,1]$, if in addition $F(1)=1$,. to the unique solution full stop.

As continuity is automatic, $F(x)=x$ over $[0,1]$changes to only solution (not merely the only unique solution ) solution>

Thus in the context of an infinite modal probability (canoncal simplex vector space) probability function representation.

That is, infinitely many probability spaces (the entirety of $\triangle^{2}$. The canonical probability 2-simplex)the entire convex hull of its three vertices, as a euclidean triangle $\in \mathcal{R}^{3}$ with vertices #(0,0,1), (1,0,0), (0,1,0)# etc) .

The set of all and only three outcome probability triples.

all ,and only all triples $\in [0,1]^3$, whose elements are $3$ non negative reals $\in [0,1]$ which sum to 1,$ \sum_{t=1}^{t=3}=1$ .

$\langle x_{v_{i}},y_{v_{i}},z_{v_{i}},\emptyset_{v_{i}}=0, \Omega_{v_{i}}=1 \rangle_{v_{i}};\quad \forall (v_{i} \in \mathcal([0,1]^{3}\cap \triangle^{2}):$.

$\forall(v_{i}\in [0,1]^3): x_{v_{i}}+ y_{v_{i}}+z_{v_{i}}=1$. $\forall(v_{i}\in [0,1]^3): x_{v_{i}},y_{v_{i}},z_{v_{i}}\in [0,1]$.

which gives $\triangle^{2}$.

and the function $F$ being such that: $\forall(v_{i}\in triangle^{2}):F(x_{v_{i}})+F(y_{v_{i}})+F(z_{v_{i}})=1$.

$\forall(v_{i}\in triangle^{2}): F(x_{v_{i}}),F(y_{z_{i}}),F(z_{v_{i}})\in [0,1]$.

$$\forall(v_{i} \in \triangle^{2}):\text {in the entire 2-probability canonical simplex} :F(\Omega_{v_{i}})=1 \land F(\emptyset_{v_{i}})=0$$.

$$\forall (v_{i}\in \triangle^{2}):\, \forall(e^{j}_{i} \in v_{i}): 0<e^{j}_{i}<1 \iff 0<F(e^{j}_{i})<1 $$.

$$ \land \text{ on vertices,e.g}:.$$

$v_{b1}=\langle0,0,1\,\rangle_{v_{b1}}=$. $\langle\,e^{1}_{v_{b1}}=0,e^{2}_{v_{b1}}=0,e^{3}_{v_{b1}}=1\,\rangle_{v_{b1}}$.

$v_{b3}=\langle0,0,1\,\rangle_{v_{b3}}=$. $\langle\,e^{1}_{v_{b3}}=0,e^{2}_{v_{b3}}=0,e^{3}_{v_{b3}}= 1\,\rangle_{v_{b3}}$. $v_{b2}=\langle0,1,0\,\rangle_{v_{b2}}=$. $\langle\,e^{1}_{v_{b2}}=0,e^{2}_{v_{b2}}=1,e^{3}_{v_{b2}}=0\,\rangle_{v_{b2}}$ :

$F(e^{2}_{v_{b2}}=1)=F(e^{1}_{v_{b3}}=1)=F(e^{3}_{v_{b1}})=1)=1$ .

$\land F(0)=0\text{for the other 6dom value=0 basis elements/events}$ .

whose individual events, are globally ordered, vector independent(everywhere). mo-dally (between vector), locally (within a vector), vector regardless of whether $x_{v_{j}} , x_{v_{k}} ,\, y_{v_{i}},\,z_{v_{m}}$ are elements of the same vector or not.

And event independent,regardless of whether the events are both are not both $x$'s etc $( x,y,z)$ connected.

And a probability function $F$ whose domain is the probability simplex.

Or numerical elements /coordinates,in the vectors in the simplex) and is strictly totally ordered numerically by the equivalence classes and domain values of each event in each vector and between each vector, probability vectors $v_{i}$. where the events are denoted by their point wise numerical domain value in the simplex .

The function only has sum to unity within a vector (vector dependent_) for all vectors) but the rank is context free, (basis independent) which will lead to :.

$\forall(v_{i},v_{j})\in\,\triangle^{2}:\forall(e^{t}_{j})\in \,v_{i}\,,t=[1..5]:\,\forall(e^{t}_{j})\in\,v_{j}\,,t\in [1..5]:$.

$$ \text{all probability vectors in simplex}.$$

$$\text{each element, in each vector denoted/defined by their probability/numerical coordinate value}.$$
$$e^{t_{[1..5]}}_{j}<e^{t_{[1..5]}}_{i}\,\iff\,F(e^{t_{[1..5]}}_{j}) < F(e^{t_{[1..5]}}_{i}).$$ $$e^{t_{[1..5]}}_{j} > e^{t_{[1..5]}}_{i}\,\iff\,F(e^{t_{[1..5]}}_{j}) < F(e^{t_{[1..5]}}_{i}).$$ $$e^{t_{[1..5]}}_{j}=e^{t_{[1..5]}}_{i}\,\iff F(e^{t_{[1..5]}}_{j}) = F(e^{t_{[1..5]}}_{i}).$$

$$\text{ie}$$. $x_{v_{i}}=e^{1}_{i},x_{v_{j}}=e^{1}_{j}$ $y_{v_{j}}=e^{2}_{i}, y_{v_{i}}=e^{2}_{j}\,,z_{v_{i}}=e^{3}_{i}$ $\,z_{v_{j}=e^{3}_{j}},\emptyset_{v_{i}}=e^{4}_{i}=0\,,$ $\emptyset_{v_{j}}=e^{4}_{j}=0$, $\Omega_{v_{i}}=e^{5}_{i}=1, \Omega_{v_{i}}=e^{5}_{j}=1).$

$$\text{ elements, atomic events + unit, and empty set)}.$$

$$\text{st. function is globally ordered by these values,regardless of whether those number/event, are members of the same vector} i=j\or \text{or not},\, i\neqj \text{or the same type of event super-scripts equal or not}$$.

Where'= '(means elements with the same numerical values s must have the same function value .

The elements can be distinct, but their domain value \in [0,1] is the same real number (these may be be a different event, and may be on a different vector but have the same simplex domain probability value and thus function range probability value).

$\if F(x)<F(y) $ are globally, locally and modall-y ranked point-wise, by the values in the canonical probability simplex, somewhat like QM ,event and vector independent.

but which are locally finite $dim\geq3 $ where $\text{dim=the same finite number of atomic events in each vector}.

Where the natural unit and bottom element on each space and on the vertices is:

$$\forall(v):F(\Omega_v)=1 \,\land F(\emptyset_{v})=0$$.

$$ \text{on vertices eg} \,, \langle1,0,0\rangle\,,F(1)=1,\land\, F(0)=0$$ .

And the domain and range of $F$ is $[0,1]$ a derivation of Cauchy's equation would grant linearity. Due to rank equalites disjoint nomramlization of both entities and global rank equalites vector independtnnoramlization of $F$ for disjoint events on the same vectors becomes bvector independent normalization between events $(1)$above becomes global and (to any arbitary three events, same vector or not)) $$(1)F(1-x-y)+F(x)+F(y)=1$$ $$F^{-1}(1-x-y)+F^{-1}(x)+F^{-1}(y)=1$$ $$x+y=z+m\iff F(x)+F(y)=F(z)+F(m)$$ $x+y>z+m\iff F(x)+F(y)>F(z)+F(m)$$ $$x+y1\iff F(x)+F(y)+F(z)>1$$

$$x+y = \frac{2}{3} \iff F(x)+F(y)=\frac{2}{3}$$ $$F(\frac{1}{3})=\frac{1}{3}$$ $$F(\frac{1}{2})<\frac{13}{24}$$ $${2}{3}>x>\frac{1}{3}\iff \frac{2}{3}<F(x)<\frac{1}{3}$$ $$\frac{1}{3}>x\geq\frac{1}{4} \iff frac{1}{3}>F(x)\geq\frac{11}{48}$$ \frac{2}{3}\geq x \iff frac{1}{3}\leq F(x)<\frac{1}{6}$$ etc

rational homogeneity and Cauchy's equation (deriving , and continuity (complete linearity) in one move, without further ado?

See Chapter 2 section 1 corollary 9,eq (18 on page(18) in

Aczel, Janos, Dhombres, J, Functional equations in several variables with applications to mathematics, information theory and to the natural and social sciences, Encyclopedia of Mathematics and its Applications 31. Cambridge: Cambridge University Press (ISBN 978-0-521-06389-0). xiii, 462 p. (2008). ZBL1139.39043.

Answer 1

In fact the first equation is enough. First use induction and prove that $ F \left( \frac 1 n \right) = \frac { F ( 1 ) } n $ and then note that since the value of $ F $ is always nonnegative, hence by the first equation, it's increasing which proves what was desired.

Answer 2

Is this the case for a Jensen's function where $(1)$ is replaced with Jensen's equation and $F(0)=0$ For example

$$(A)F:[0,1]→[0,1]$$ $$(1) F(0)=0,\, F(1)=1$$ $$(2)\forall (x,y)\, \in [0,1]; F(\frac{x}{2}+\frac{y}{2})=\frac{F(x)}{2}+\frac{F(y)}{2}$$

If 'Jensen's equation when combined with $F(0)=0$ where $0$ is an element of the domain and co-domain', over $F:[0,1]\to[0,1]$, is literally identical to 'Cauchy's equation' before continuity is applied, then one would presume that continuity and $F(x)=x$ would be likewise automatic, as above without any further requirement.Is it?

By this I mean not just indirectly (or for all extents and purposes) but that continuity should be automatic, $F(x)=x$ automatic, without so much as mentioning monoton-icity?. Due to the non-negative domain and range, and $F(1)=1$ bounding the function.

I believe that monotonic increasing-ness, however, is required to get to continuity for Jensen's equation in the form expressed above. Is that correct?

• whilst both functional equations, (Cauchy and Jensen's equation) are expressed in their non-continuous form. (that is not, via '1-point homo-geineity', or the 'linearity superposition principle', or 'concavity plus convexity for Jensen's equation' without the restriction that $\sum \lambda=1$ due to $F(0)=0$ or the restriction that $\lambda \in \mathbb{Q}$

Whilst 'Jensens equation with $F(0)=0$', and 'Cauchy's equation' both can only, and only express 'rational homogeneity'.

Cauchy's equation applies additivity to all reals or at least to an appropriate subset of reals, in the domain of interest, nonetheless.

so is Cauchy's equation, even in the restricted form above, restricted as( $x+y>1; x+y \notin text{dom}(F)=[0,1]$ and Cauchy' equation cannot be directly applied?. Nonetheless is it stronger than Jensen's function described above.

Where remember that with $F(0)=0$; $0\in text{dom}(F)$, in some sense or not? (before continuity, being applied)

Unless that was the error, that the result above does not hold for cauchy as stated, because it only applies when one can extend the function; its unclear as I stated it, whether Cauch'ys equation applies to $x+y>1 $ (it says for all elements of the domain $(x,y)$ with no mention of$ x+y$, one one hand, and restricted the domain to $[0,1]$ f

its nonetheless real valued additive (for all, $(x,y)$ in the domain) whether $x$, $y$ are irrationals, that sum to an irrational, or a rational, or whether or not they are rational multiples of their sum or each other. $$\forall \lambda \in \mathbb{Q^{+}\cup 0}, \forall(x)\in dom(F)=[0,1]:F(\lambda x)=\lambda F(x)$$ so long that $\lambda \times x\in [0,1]$

Is Jensen's equation "real valued additive" in the same way? That is, before continuity is applied, and if not, what this explain the difference.

that Cauchy' equation bounds the irrationals values, in a stronger sense, to between the closest two rational function values, to abuse terminology. That is Cauchy equation above' essentially' implies strict monotonic increasing-ness even for irrational values?

That is, in the above context, which is over above non-negativity, as it ensure that their values will be approximately correct already.? perhaps, due to the interconnections with other rational values, and irrational values 'vis a vis' additivity, and rational homogeneity over even irrational domain values?

$$x>y \leftrightarrow F(x)>F(y), (x,y)\in \mathbb{IR} \land (x,y)\text{ are NOT rational multiples of each other}$$

.

Which is generally implied by Cauchy's equation.

that is via real valued additivity, which may connect the function values of irrational elements of the domain of the function indirectly to each other and some the rational values which are clearly strictly monotonic increasing?.

$F(x)=x$ clearly for all rational values in both cases I think before continuity is applied

(which apparently it already is, above for Cauchy's equation).

That is when Jensen's equation applies to all elements of a real valued domain, then if $F(0)=0$ does Cauchy's equation apply as to real reals in the domain, so long that $x+y \in dom(F)$.?

Apparently Jensen's equation only implies restricted rational convexity and concavity.

$$(A)\text{(restricted convexity and concavity, jensens generalized, without F(0)=0)}:\forall t \,\in\, \{\mathbb{Q} \cap [0,1]\}\,;\,\forall (x,y)\in \text{dom}(F)\,;\, F(tx +(1-t)y)=tF(x) + (1-t)F(y)$$

$$(B)\text{(convexity and concavity, Jensen's generalized, with F(0)=0)}:\forall (t_1,t_2)\,\in\, \{\mathbb{Q} \cap [0,1]\}\,;\,\forall (x,y)\in \text{dom}(F)\,;\, F(t_1x +t_2y)= t_1F(x) + t_2F(y)$$

where $t_1 +t_2$ need sum equal $1$ in $(B)$; that is,the requirement that $\sum_i t_i=1$ in $(A)$ is removed in $(B)$ given $F(0)=0$.

Are these generalizations of Jensen's equation correct?

That is before continuity is applied, but when $F$ is real valued or applies whose domain and co-domain, are real valued intervals such $[0,1]$

When it holds over the entire domain or only over the real line or open intervals.

But does Jensen's equation really imply real valued Cauchy add-itivity, in the non-continuous case, as does Cauchy's equation?

I know that midpoint convexity may have issues with rational convexity as opposed to dyadic rational convexity on closed intervals, which is somewhat related to the issue of continuity of the endpoint.

presume that jen-sen equation would not have this issue because its an equality equation.

And I presume that the 'convexity and concavity ' combined equation can made less restrictive when $F(0)=0$ . That is so that $F(x+y)=F(x)+F(y)$ the rational superposition principle and $F(0)=0$ can be recovered from it, via lifting the restricting that $\sum \lambda =1$

midpoint convexity- implies $$\forall t\,\in\, \{\mathbb{Q} \cap [0,1]\}\,;\,\forall (x,y)\in \text{dom}(F)\,;\, F(\sigma x +(1-\sigma)y) \leq \sigma F(x) + (1-\sigma)F(y)$$.

Ff $F$ is real valued on an open interval

Does it imply this, without continuity on a closed interval. Or only

$$\forall t\,\in\, \{\ \mathbb{D_{Q}} \cap [0,1]\}\,;\,\forall (x,y)\in \text{dom}(F)\,;\, F( \sigma x +(1- \sigma) y) \leq [\sigma F(x) + (1- \sigma)F(y)]$$.

where \mathbb{D_{Q}} stands for the dyadic rationals; with $F(0)=0$ and $F(1)=1$, and strictly monotonic, ?

It can express convexity for $$ \sigma \in\mathbb{Q_D} \in [0,1],\ \sum\sigma=1$$

$$ F( \frac{a}{2^{n}} x) \leq \frac{a}{2^{n}}F(x)$$ with $F(0)=0$where $n \in \{\mathbb{N}\cup 0\};\forall(a) \in \mathbb{N};1 \leq a \leq 2^{n}$

$$F( \frac{a}{2^{n}}x) \leq \frac{a}{2^{n}}F(x)$$ with $F(0)=0$

$$F( \frac{2^{n}}{a}x) \geq \frac{2^{n}}{a}F(x)$$ with $F(0)=0$ where $a>1$\in where $n \in \{\mathbb{N} \cup 0\};\forall(a);1 \leq a \leq 2^{n}$

or dyadic rational super-additivity (otherwise sub-linear functions would be linear full stop) with $F(0)=0$ and maybe not even that that.

Some of this, has to do with its mid-point convexity being an inequality-case which makes things difficult.One often uses $F(2x)=2F(x)$ on midpoint convexity to get sub-additivity AND $F(2x)=2F(x)$ on midpoint concavity to get superaddivity (despite convexity being most often associated with super-additivity).

Is the same nonetheless the case with Jensen's equation? That the equations $(A)$ , $(B)$ above only hold for dyadic $t, t_1,t_2$ over closed intervals?

I presume not, as the equality just as it makes it make it easier to get to derived real valued additivity (even in the non continuous case).

It makes it much easier to progress from mid-2pt-to dyadic and all all the way rational homogeoneeity/rational convexity concavity?.

I know that one can trivially 'derive Cauchy's equation' from jensens equation with $F(0)=0$, buts it unclear to me that this is derivation in full generality or just a pairwise result

Answer 3

I note that its unclear, in the formulation of my question above, whether Cauchy's equation applies when $x+y>1$?

$$(1)\forall ((x,y)\in \text{Dom(F)=[0,1]};x+y>1): F(x+y)=x+y$$.

I presume not, as above I said that;

$$\forall ( (x,y)\in \text{Dom(F)=[0,1]}): F(x+y)=F(x)+F(y)$$.

Whilst, there is no explicit mention of a restriction to only those domain elements,$\,(x,y);\,\,(x,y)\in \, \text{Dom(F)}=[0,1]$, whose sum: $x+y;\,\, x+y \in \, \text{Dom(F)}=[0,1]$ in the question above. Clearly, Cauchy' equation, as I used it, is restricted to domain elements: $(x,y)\in [0,1]=\text{Dom(F)}$, whose sum: $x+y;\,\, x+y\in [0,1]$ s.t $x+y\,\leq1$ ?

I presume that:$(A):\, \forall (x \in [0,1]):\,F(x)=x$, still holds, regardless of this restriction? Is that correct?

This being, due to non-negativity of $F:\, F(x)\geq 0\,,\,\text{CoDom(F)=Dom(F)}=[0,1]$,which gives, from Cauchy' equation above, that $F$ is monotonic increasing over $[0,1]$.

And, $F(1)=1$,which gives:

$$\forall(x\in [0,1]\cap \mathcal Q): F(x)=x$$.

And these collectively give $(A)$?

So I presume that restricted Cauchy function $F$ still preserves rational sums $x+y>1$ in the sense of: $(1)$, $(2.1$), or $(2.2)$ below?

$$(1):\forall ((x,y)\in \{\text{Dom(F)=[0,1]}\cap \mathcal Q\};x+y>1): F(x+y)=x+y$$.

$$\forall ((x,y,z)\in \text{Dom(F)=[0,1]};\forall (x_1, y_1,z_1)\in \text{Dom(F)=[0,1]};\,x+y+z>1):$$ $$(2.1): \, x+y+z=x_1+y_1+z_1 \iff F(x)+F(y)+F(z) =F(x_1)+F(y_1)+F(z_1)$$.

$$(2.2):\, F(x+y+z)=F(x_1+y_1+z_1) \iff F(x)+F(y)+F(z) =F(x_1)+F(y_1)+F(z_1)$$.

Is that correct?