# 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?

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Are you assuming continuity on the functions? – azarel Nov 2 '13 at 12:42
No, that would imply $f \equiv 0$. I'm only assuming they're bounded. – ScroogeMcDuck Nov 2 '13 at 12:48
What is the space on which $M_f$ acts. Specifically what is the norm you are putting on $g$? – Owen Sizemore Nov 2 '13 at 13:07
@ScroogeMcDuck $f$ can have an uncountable range, make $f(x)=x$ for $x$ in the Cantor set and zero otherwise. – azarel Nov 2 '13 at 13:12
@azarel Ah you're right, I've edited my post. The space of bounded functions on the unit interval, with the supremum norm. – ScroogeMcDuck Nov 2 '13 at 13:26

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.
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The range of $f$ is countable and $f^{-1}(\alpha_n)$ is finite for all nonzero $\alpha_n$ in the range of $f$, therefore the set where $f$ is nonzero, which is $\cup_n f^{-1}(\alpha_n)$, is countable! Have I understood correctly that this is what you're saying? – ScroogeMcDuck Nov 2 '13 at 14:58
That, and the $\alpha_n$ converge to $0$. Another way of formulating it is to say $f^{-1}(\{ z : \lvert z\rvert \geqslant \varepsilon\}$ is finite for every $\varepsilon > 0$. – Daniel Fischer Nov 2 '13 at 15:01
Thanks, that was extremely helpful! The fact that $\alpha_n\rightarrow 0$ is also quite useful for proving compactness, I had already proven it for the countable case, but in a more difficult way. – ScroogeMcDuck Nov 2 '13 at 15:14

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.

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You're right, this was not what I meant though, I formulated my question in the wrong way. I am supposing $f$ has a countable range now. – ScroogeMcDuck Nov 2 '13 at 13:45
Even with the range of $f$ being countable, $M_f$ might still not be compact. Just let the range cluster somewhere other than zero. – Ben W Nov 2 '13 at 14:09