# A question about the multiplication operator

Let $C(\mathbb{R})$ be the set of all continuous functions and $T:C(\mathbb{R})\rightarrow C(\mathbb{R})$ defined by $T(f(x))=xf(x)$. I would like to know if there are subespaces $W_{1}, W_{2}$ of $C(\mathbb{R})$ such that $T|W_{1}$ is nilpotent and $T|W_{2}$ is invertible and $T=T|W_{1}+T|W_{2}$.

-

If $T$ is nilpotent on a subspace $W_1$, then there exists $n\gt 0$ such that for all $f(x)\in W_1$, we have $$T^n(f(x)) = 0;$$ since $T^n(f(x)) = x^nf(x)$, we want function $f(x)\colon\mathbb{R}\to\mathbb{R}$ such that $x^nf(x) = 0$ for some $n\gt 0$ and for all $x$. What can you say about such functions? That will tell you whether $W_1$ can exist, and if so what it should be.
Then whe have $W_{1}=\{0\}$. Am I correct this time? :) –  spohreis Feb 7 '12 at 21:01
Yes, though it takes some argument; you need $f(a)=0$ for all $a\neq 0$; continuity then gives $f(0)=0$, so $f=0$. Thus, $W_1$ will necessarily be the zero subspace, which means that $W_2$ has to be the whole thing. Note that $xf(x)=0$ if and only if $f(x)=0$ (since $f$ must be continuous), so $T$ is already invertible, so setting $W_1=\{\mathbf{0}\}$ and $W_2=C(\mathbb{R})$ will do it (in a trivial way), and it cannot be done in any other way. –  Arturo Magidin Feb 7 '12 at 21:03