It is said that a function is linear if if satisfies Cauchy's equation (A):
(A)$$\forall(x,y)\,\in\,\mathbb{R};\quad F(x +y)=F(x)+F(y)$$
And 1 Point Homogeneity $(B)$:
(B)$$ \forall \sigma , x\, \in \mathbb{R};\quad F(\sigma x)=\sigma F(x)$$
Then the function is already linear and continuous. $F(x)=ax\,;\,a=F(1);\, F(x)=F(1)\times\,x$
****Is not $(B)$ alone sufficient already for $F(x)=x$, over $[0,1]$ with $\text{dom(F)}:=[0,1]$, with $F(1)=1$ without $(A)$ or continuity or anything else explicitely mentioned?**
That is, if one If one could magically get to $(B)$, one would not need $(A)$; or the superposition $(C)$:which is the admixture of $(B)$+ $(A)$
$(C)$Superposition principle
$$(C)\forall(x,y) \in \text{dom}(F)=[0,1]; \forall (\sigma_1, \sigma _2)\in \mathbb{R+}\cup {0}, ;\quad F(\sigma_1 x +\sigma_2y)=\sigma_1F(x)+ \sigma_2 F(y)$$
where $ (C)$ is the fusion of the $(B)$ and $(A)$, or convexity and concavity where $\sigma_1$, $\sigma _2$ but these do not need to sum to $1$ so that $F(0)=0$ is fixed, as for either $(A)$ or $(B)$
Under continuity or bounded assumptions Cauchy's equations $(A)$, which already satisfies rational homogeneity $(D)$ and real valued additive rational superposition $(D1)$.
And can extended, given entail $(B)$ under mild almost non-existent regularity requirements apparently; given the non negative domain and range.
$$F:[0,1]\to [0,1]\quad F(1)=1\quad \forall(x,y)\,\in\,\mathbb{R};\quad F(x +y)=F(x)+F(y)$$
to entail $(B)$, homogeneity $\forall \text{real} \sigma $ as well as continuity anmd $F(x)=x$ $(B)$ as well
$(D)$ rational homogeneity
$$(D) \forall (x)\in \text{dom}(F); \forall \sigma\, \in \mathbb{Q}: F(\sigma x)= \sigma F(x)$$
$(D1)$ rational superposition
$$(D1) \forall(x,y) \in \text{dom}(F)=[0,1]; \forall (\sigma_1, \sigma _2)\in \{\mathbb{Q+}\cup {0}\}, ;\quad F(\sigma_1 x +\sigma_2y)=\sigma_1F(x)+ \sigma_2 F(y)$$
Does not $(B)$ entail $(A)$ directly in any case over $[0,1]$ with $F(1)=1$ , One does not need , the superposition principle? $(C)$ above ,or $(B)$ plus $(A)$ to get that $F(x)=x$, $(B)$ alone with $F(1)=1$ with $\text{dom(F)}=[0,1]$ is sufficient?
****Ie, doesn't $(B)$ alone directly entail entail continuity, and the Cauchy's equation as well given $F(1)=1$ $\text{dom}(F)=[0,1]$and thus specify the function as $F(x)=cx$ where $c=F(1)$ directly**? At least under most circumstances and you are careful with the range and the domain. One does not need to specify $(A)$.
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One can resolve all irrational numbers, using the unit event if necessary. It essentially entails Cauchy's equation, in its most general form, and that the function is continuous; if $(B)$if it holds for real $\sigma$ and real $x$,
At least if you fix the domain and range, where $F$ is a function satisfying (1)$, $(2)$, $(3)$
$$(1)\text{dom}(F)=[0,1]$$
$$(2)F(1)=1$$ $$ (3)\forall (\sigma) \in \{ \mathbb{R}^{+} \cup 0 \}; \, \forall (x) \in \text{dom}(f)=[0,1];\, F(\sigma x)=\sigma F(x)$$
$$(1)F(x)=F(\,(x+y)\times\,(\frac{x}{x+y})\,)= (\frac{x}{x+y}) \times F(x+y)$$using real valued homogeneity, these are number in the non-negative reals $$(1a)F(y)=F(\,(x+y)\times (\frac{y}{x+y})\,)= (\frac{y}{x+y}) \times F(x+y) $$ $$ [(1a)\land (1)] \rightarrow (2)$$ usin
$$(2)F(x)+F(y)=[(\frac{y}{x+y}) \times F(x+y)]+[(\frac{x}{x+y}) \times F(x+y)]=(\frac{x+y}{x+y}) \times F(x+y)\quad = 1\,\times F(x+y)=F(x+y)$$ $$(2)\leftrightarrow (3)$$ $$(3)F(x+y)=F(x)+F(y)$$
I presume that it entails Cauchy's equation in general, and can be generalized.
Wont the function= just be $$F(x)=x$$ over $[0,1]$, by definition ? Perhaps I am wrong here?
But I presume that real 1-pt homogeneity $(B)$in and of itself is sufficient,to get $F(x)=x$ without Cauch'ys equation $(A)$ above, being explicitly mentioned or any regularity requirement other then the domain, and $F(1)=1$ and $F$ is a function
Of course cannot usually get to $(B)$ except via Cauchy's equation $(A)$and a mild regularity assumption.
Perhaps one needs $(D1)$ to get to continuity and cauchy's equation as it somewhat specifies a linkage between the irrational $(x,y)$ to other rational or irrational $(x,y)$ via additivity and presumably just is cauchy's equation in an of itself. I presume it an accurate specification of jensen's equation with $F(0)=0$ where $\sigma_1, \sigma _2$ are not restricted to sum to one.
Does $(D)$ only specify additivity for $x,y \in $ irrational which add to an irrational value $x+y\in$irrational, but are rational multiples of each other? So $(D)$ does not specify, real valued additivity where irrational $x,y$ sum to a rational and not, or irrational and are not multiples of each other. Unlike $(A)$, $(C)$, $(B)$ and $(D1)$ despite the fact that some of these only entail rational homogeneity they specify real valued additivity.
Does jensen's equation require any mono-tonicity assumption under the requirements; with $F(0)=0$ or does it grant automatic continuity under non-negative domain and range.
$$F:[0,1]→[0,1]$$. $$F(1)=1, F(0)=0$$ $$(3)∀(x,y)∈dom(F)=[0,1];F[\frac{x+y}{2}]=\frac{F(x)}{2})+\frac{F(y)}{2})$$
is $F(x)=x$ immediate without further assumptions
And are both cauchy's equation and (jensens's eq $(3)$ plus $F(0)=0$) then equivalent to, rational super-position to $D1$, above. Where the $\sigma$ are rational, not need to sum to $1$ due to $F(0)=0$, and
is continuity automatic, without monotonicity or strict monotonicity, assumed, under $(D1)$ with $F(1)=1$ and $F:[0,1]\to[0,1]$
unlike rational homogeneity $(D)$ which would require $$(A)F:[0,1]→[0,1]$$. $$(B)F(1)=1$$
$$(C)F \, \text{where,monotonic increasing or strictly monotone increasing}$$ $$(D) \forall(x) \in \text{dom}(F)=[0,1]; \text{satisfies (D) above}$$
using the fact that $F$ above agrees with $F(x)=x$ over all rationals, a dense set of $[0,1]$ and as $F$ is strictly monotonic increasing, and $F(x)=x$ is continuous, $F$ is continuous and identical to $F(x)=x$ over $(0,1)$,where one then puts in the specified end points as mentioned above.
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