Linear operators on the functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that distribute over multiplication

Let $V$ denote the vector space of all functions $f:\mathbb{R}\rightarrow\mathbb{R}$. What are the linear operators $L:V\rightarrow V$ such that $L[fg]=L[f]L[g]$ for all $f,g\in V$?

I made a bit of progress by considering the functions $$\chi_t(x) = \begin{cases} 1 & x=t \\ 0 & x\neq t \end{cases}.$$ For fixed $x\in\mathbb{R}$, the value of $L[\chi_t](x)$ is either $0$ or $1$ for each $t\in\mathbb{R}$. If there exists some $t$ such that $L[\chi_t](x)=1$, then $L[f](x)=f(t)$. I was unable to do the case in which $L[\chi_t](x)=0$ for all $t$.

Solution 1:

$\newcommand{\IR}{\mathbb{R}}$You are asking about endomorphisms of the algebra $A:=\IR^I := \prod_{i\in I} \IR$. All of them are of the form $x \mapsto (f_i(x))$ where $f_i$ are arbitrary homorphisms $A\to\IR$.

In particular we have to find certain maximal ideals of $A$. Now luckily there is a nice(ish) classification of all ideals of products of fields: They are all of the form $V_\phi := \{ x \mid \{i | x_i=0\} \in\phi\}$ where $\phi$ is a filter on $I$. The maximal ideals correspond to ultrafilters. (This is a nice exercise)

There are two cases of ultrafilters: Principal ultrafilters, i.e. $\phi=\phi_x=\{P\subseteq I \mid x\in P\}$ for some fixed $x\in I$, and non-principal ultrafilters.

The principal ones are easy: They are exactly the kernels of the projection maps. Also there are no non-identity ringhomomorphisms $\IR\to\IR$. Therefore any $f: A\to\IR$ with kernel $\phi_x$ is equal to the $x$-th projection map.

The non-principal ultrafilters are more tricky. The quotients $A/V_\phi$ are called ultraproducts of $\IR$. These are ordered fields as well, but interestingly they are not isomorphic to $\IR$, they don't even embed into $\IR$, because they have infinitesimal and infinite elements. Therefore they cannot occur as kernels of our maps $f_i: A\to\IR$.

Summary: All algebra homomorphisms $A\to A$ are of the form $(x_i)\mapsto (x_{g(i)})$ for some map $g:I\to I$.