TITLE: Do "exotic vectorfields" exist?
QUESTION [11 upvotes]: By an exotic vectorfield on $\mathbb{R}^n$, I mean a non-zero derivation on the algebra $C^0(\mathbb{R}^n)$.
Do such things exist?

REPLY [15 votes]: No, they don't.  Suppose $d:C^0(\mathbb{R}^n)\to C^0(\mathbb{R}^n)$ is a derivation.  First, suppose $f\in C^0(\mathbb{R}^n)$ is nonnegative everywhere.  Then the nonnegative square root $\sqrt{f}$ is continuous, and $$d(f)=d(\sqrt{f}^2)=2\sqrt{f}d(\sqrt{f}).$$  It follows that if $f(p)=0$, then $d(f)(p)=0$ (since $\sqrt{f}(p)=0$).
Now let $f$ be any continuous function and suppose $f(p)=0$.  Then $d(f)=-d(|f|-f)+d(|f|)$.  Since $|f|-f$ and $|f|$ are both nonnegative and vanish at $p$, it follows that $d(f)(p)=0$.
But now for any $f$ and any $p$, we can consider the function $g=f-f(p)$.  Then $g$ vanishes at $p$, so $d(g)$ vanishes at $p$.  But $d(g)=d(f)$ since they differ by a constant.  Thus $d(f)$ vanishes at $p$.  Since $f$ and $p$ were arbitrary, this means $d=0$.