It was also unclear to why anything more then $f(λx)=λf(x)$; is not cauchy's equation just a means to that end. cant one derive cauchy's equation; that that the function is of the form $F(x)=ax$, and $F$ is continuous from , $f(λx)=λf(x)$ alone it holds:
Unless a restricted domain is required, I know that needs cauchy's equation to get to $f(λx)=λf(x)$ $\forall$ real$\lambda$ but I am more having a dispute that Cauchy's equation is just a means to that end.
That is Cauchy equation, is a mean to get to real valued homogeneity for rational numbers,(which does not imply cauchy's equation, ie real valued additivity) but then cauchy equation provides enough structure with regard to the non-rational numbers (which is not implicit in rational homogeneity, but implicit in cauchy' equations) , as in monotonicity, which often automatically entails that $F(x)=ax$, and automatic continuity when when the domain and range are specified to be non negative and , and $F(1)=a$ is specified.
On other hand real valued homo-geniety implies real valued additivity cauchy equation generally, and automatically specifies the function when it holds for all reals. One needs to compare apples with apples and homogeneity is generally stronger when defined over the domain (for all reals) then add-itivity.
One can also see that in the sense that real valued sub-additivity, can hardly, by itself can anthying more then integer-inequality homogeneity (and not even all rationals, or dyadic even dyadic rational-inequality homogeneity)
$\forall x\in \text{dom}(F), \forall \lambda \in \mathbb{R}; F(λx)=\lambda F(x)$
If $F$ is a function, defined on a real interval, then can't $\lambda$ be equated with $x$ in the follow sense? That is, any element of the domain , only has one function value, so when $x=\lambda$, then, $(x, \lambda)$ are one and the same element of the domain of $F(x)=F( \lambda)$, so have the same function value?
So long that for every domain value, $\exists \sigma =x $
For example;
$$\forall(x\in \text{dom}(F); [x=\lambda\,]\rightarrow [F(x)=F( \lambda)]\,\rightarrow[ F(x)=F(\lambda )=F(\lambda \times (1)) \,=\, \lambda \times F(1)=x\,\times F(1)]\rightarrow [F(x)=x \times F(1)]$$
I hardly , real valued homogeneity to be something that needs to be added with additivity, to get the super-position, it presumably entails both of them, in and of itself.
So I and hardly see real valued 1-pt homogeneity, in most contexts as a functional equation,if it holds for all reals. I rather see it as the function itself, just $F(x)=ax$ stated in another way. It literally just is $F(x)=ax$,for the most part.Stated in another way. Unless I am mistaken.
So long as one specifies that $F(1)=1$, and the domain of the function, then $F(x)=x$ is automatic (its continuous for definition, as it literally just states what the function value is, is for every real number, in the domain).