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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 $t$.

user109360
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3 Answers3

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$\newcommand{\IR}{\mathbb{R}}$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 $g:I\to I$.

Johannes Hahn
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We cheat a bit, giving a solution that wouldn't work for the space of functions from $X$ to $\Bbb R$:

Of course one possibility is $L(f)=0$ for every $f$. If we assume there exists $f$ with $L(f)\ne0$ then $L(f)=L(1f)=L(1)L(f)$ shows that $L(1)=1$ on the set where $L(f)\ne0$.

If we simplify things by assuming that $L(1)=1$ then there exists $\psi:\Bbb R\to\Bbb R$ with $\phi(f)=f\circ \psi$.

First,

Lemma If $\phi:V\to\Bbb R$ is a homomorphism with $\phi(1)=1$ then there exists $x\in\Bbb R$ such that $\phi(x)=x$ for every $f$.

Proof: Define $i\in V$ by $i(t)=t$. Let $x=\phi(i)$. Then $\phi(i)=i(x)$.

Suppose that $f(x)=0$. Then there exists $g$ with $f=(i-x)g$. So $$\phi(f)=(\phi(i)-x\phi(1))\phi(g)=0\phi(g)=0.$$

Now for a general $f\in V$, let $g=f-f(x)$. Then $g(x)=0$, so the previous paragraph shows that $\phi(g)=0$. So $$0=\phi(f)-f(x)\phi(1)=\phi(f)-f(x).$$QED.

If $L$ is as in the question and in addition $L(1)=1$ then there exists $\psi:\Bbb R\to\Bbb R$ with $L(f)=f\circ \psi$.

Proof: For each $x\in\Bbb R$ define $\phi_x:V\to\Bbb R$ by $$\phi_x(f)=L(f)(x).$$The lemma shows that for every $x$ there exists $y$ such that $$\phi_x(f)=f(y)$$for all $f$. For a given $x$ there is at most one such $y$, since if $y_1\ne y_2$ there exists $f$ with $f(y_1)\ne f(y_2)$. So we can define $\psi:\Bbb R\to\Bbb R$ by $\psi(x)=y$, and we're done.

If you don't want to assume that $L(1)=1$: Since $L(1)=L(1^2)=L(1)^2$ it follows that $L(1)$ takes only the values $0$ and $1$. Let $A$ be the set where $L(1)=1$ and let $B$ be the set where $L(1)=0$. You can modify the above to show there exists $\psi:A\to\Bbb R$ such that $L(f)(x)=f(\psi(x))$ for $x\in A$ and $L(f)(x)=0$ for $x\in B$.

David C. Ullrich
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This paper says that they are all evaluation functions: http://projecteuclid.org/download/pdf_1/euclid.pja/1195524949

(Provided that you consider only continuous functions.)

Elle Najt
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