What is the difference between linear and affine function

A linear function fixes the origin, whereas an affine function need not do so. An affine function is the composition of a linear function with a translation, so while the linear part fixes the origin, the translation can map it somewhere else.

Linear functions between vector spaces preserve the vector space structure (so in particular they must fix the origin). While affine functions don't preserve the origin, they do preserve some of the other geometry of the space, such as the collection of straight lines.

If you choose bases for vector spaces $V$ and $W$ of dimensions $m$ and $n$ respectively, and consider functions $f\colon V\to W$, then $f$ is linear if $f(v)=Av$ for some $n\times m$ matrix $A$ and $f$ is affine if $f(v)=Av+b$ for some matrix $A$ and vector $b$, where coordinate representations are used with respect to the bases chosen.

An affine function is the composition of a linear function followed by a translation. $ax$ is linear ; $(x+b)\circ(ax)$ is affine. see Modern basic Pure mathematics : C.Sidney

An affine function between vector spaces is linear if and only if it fixes the origin.

In the simplest case of scalar functions in one variable, linear functions are of the form $f(x)=ax$ and affine are $f(x)=ax +b$, where $a$ and $b$ are arbitrary constants.

More generally, linear functions from $\mathbb{R}^n$ to $\mathbb{R}^m$ are $f(v)=Av$, and affine functions are $f(v)=Av +b$, where $A$ is arbitrary $m\times n$ matrix and $b$ arbitrary $m$-vector. Further, $\mathbb{R}$ can be replaced by any field.

More abstractly, a function is linear if and only if it preserves the linear (aka vector space) structure, and is affine if and only if it preserves the affine structure. A vector space structure consists of the operations of vector addition and multiplication by scalar, which are preserved by linear functions:

$$f(v_1+v_2) = f(v_1) + f(v_2), \quad f(kv) = k f(v).$$

An affine structure on a set $S$ consists of a transitive free action $(v,s)\to v + s$ by a vector space $V$ (called the associated vector space). Informally, $v+s\in S$ is the translation of the point $s\in S$ by the vector $v\in V$, and for any pair of points $s_1, s_2\in S$, there is an unique translation vector $v\in V$ with $v + s_1=s_2$, also written as $v = \overrightarrow{s_1 s_2}$. Then a function between two affine spaces $S$ and $T$ (i.e. sets with affine structures) is affine if and only if it preserves the structure, that is

$$f(v + s) = h(v) + f(s)$$

where $h$ is a linear function between the corresponding associated vector spaces.

$(1)$Linear continuous Functional equations of the Form: $$F(\alpha x +\delta y)= \alpha F(x) + \delta F(y)$$;

or rather 1- point homogeneity

$(1a)$ (often with Cauchy's equation as well), as in the above post. However, sometimes Cauchy's equation is not needed in addition to $(1a)$ if one can get to it $(1a)$ (for all reals) directly, which is not generally the case.

$(2)$Affine of the form;

Continuous Function Form: $$\forall (x,y)\in \mathbb{R}F((1-t) x + t y)= tF(y) +(1-t)F(y); t\in [0,1]$$ ie:'concave and convex' .

*Or is it $\forall (x,y)\in $ $\text{dom(F)}$ or $\forall (x,y)\in$[0,1]$?

For(2) Function Form: $$F(x)=Ax+B$$

Where $A$ is an arbitrary constant

(1a)$$\forall x\in \mathbb{R},\forall \delta \in \mathbb{R}:F(\delta x)=\delta F(x)$$ (1/1a) Linear Function: $$F(x)=Ax$$(in this case $A=F(1)$)

One always (or nearly always) needs to derive Cauchy's equation beforehand, to derive $(1a)$ for all rational numbers. And then a weak regularity requirement, generally, given Cauchy's equation, to extend homogeneity to all reals ie to get to equation $(1a)$)

Sometimes on can extend it to the algebraic irrationals given the field auto-morph-ism equations, although they 'apparently' already specify the function and grant continuity I have some issues with that (but that another story; vis a vis- the trans-transcendental numbers).

In both cases, I may have accidentally may be restricted domain to $[0,1]$ but are in their continuous forms, so that in some sense they are now function rather than functional equations . Or rather in a functional form from which the function should be directly derivable; as opposed to 'the general (often continuous) solution to Cauchy equation or Jensen's equation etc).

Although that may not be quite correct.

I think that sometimes convex functions (presumably not convex and concave ones but I might be wrong) can have trouble with continuity at the end points . However, I presume they are ruled as degenerate or not possible in the current system, ie not Lebesgue measurable.

And often that is when one is speaking of midpoint convexity or Jensen's equation rather than their,already, or (possibly) allegedly already, continuous versions $(1)$ and $(2)$.

I am just being tentative about $(2)$ here. I am not disputing anything, I just want to be careful.

Generally unless it is not restricted to an interval (or real line version).

interval (there is presumably a real line generalization). Notice that delta is restricted here, one cannot solve for the origin directly.

I presume that $(2)$ is just the real line version $(2)$ of convexity and concavity perhaps. once the origin has been fixed.

In (2) One cannot solve for $F(0)=0$,or directly solve for $F(1)=a$; for $F(x)=ax$, but only that $F(0)=b$ for $F(x)=ax+b$.

Although I think that both $(1)$ and $(2)$ are the restricted versions to the unit range, but with $F(0)$ being directly incorporated into $(1)$.

As one can set $a=1$ and $b=1$,$x=0$, $y=0$

To get $[F(0)=2F(0)]\,\Rightarrow\,F(0)=0$ in $(1)$

But one cannot do so in $(2)$.

As there is only one free parameter, so that one always gets $F(0)=F(0)$

I suppose one gets that (I think/perhaps)

(2a)$$F(tA)= t(F(A)-F(0)] +F(0)$$

Where $$F(A)$$ is the F(max element of domain; when $A$ positive or increasing) or $$F(A)=F(1)-F(0)$$ here.

I am not sure if $(2)$, in this form, $F$ is defined only the domain $[0,1]$ .

However, its unclear whether $F(x)=ax+b$ falls out/(is derivable) of $(2)$ As directly as $F(x)=Ax$ (is derivable)/falls out of $(1)$.

I am probably using the wrong version confined to that domain $[0,1]$ domain).