Multiplication operator with a function non-vanishing on the cantor set Let $M_f$ be the multiplication operator, which acts on bounded functions $g$ on the unit interval as $g\mapsto fg$, with $f:[0,1]\rightarrow \mathbb{C}$ such that $f$ is nonzero only on the Cantor set in $[0,1]$, but with countable range. I am trying to find out whether $M_f$ is compact (using the supremum norm).
I've already proved $M_f$ would be compact if the set of $\{x \in [0,1]| f(x) \neq 0\}$ were countable, but my argument (involving sequences) cannot be generalized to the uncountable case. Also, if $f$ had an interval on which it is nonzero, I can prove $M_f$ would not be compact (using unit ball arguments). 
However, on the Cantor set (or the irrationals or any other uncountable set not containing an interval), I have no idea where to start... Any ideas?
 A: Without any regularity assumptions on the functions, the talk about the Cantor set is a red herring. Leaving aside trivial identifications, you have a projection operator $\pi \colon \ell^\infty(S) \to \ell^\infty(T)$, where $T \subset S$ are sets of cardinality $2^{\aleph_0}$, followed a multiplication operator $\mu_f \colon \ell^\infty(T)\to\ell^\infty(T)$, and the question is when $\mu_f$ is compact.
The spectral theorem for compact operators yields the necessary conditions


*

*the range of $f$ must be countable, and has no other accumulation point than $0$,

*for each $\alpha \neq 0$, the set $f^{-1}(\alpha)$ must be finite,


so we have that either


*

*$f(t)\neq 0$ for only finitely many $t$; then the range of $\mu_f$ is finite-dimensional and $\mu_f$ is compact, or

*there is a sequence $0 \neq \alpha_n \to 0$, and there are finite sets $T_n \subset T$ with $T_n = f^{-1}(\alpha_n)$, and $f(t) = 0$ for $t\notin \bigcup\limits_{n\in\mathbb{N}} T_n$. Then $\mu_f$ is the norm-limit of a sequence of operators with finite-dimensional range, hence compact.

A: Azarel has already answered.  $M_f$ need not be compact if its range is uncountable, which is easily constructed by taking $f(x)=x$ if $x\in\mathcal{C}$ and $f(x)=0$ otherwise.  Observe that this leads to uncountably many eigenvalues.
