Determine when operator is compact Let $B$ be the Banach space of bounded complex functions on $[0,1]$ with sup-norm. For $q \in B$, define the (multiplication) operator $T_q : B\rightarrow B$ by $(M_q f)(t) = q(t)f(t)$. Which $q$ makes $M_q$ compact? 
The first part was to show that $\sigma(T_q) = \overline{\{q(t) : t \in [0,1]\}}$ for each $q \in B$, which I have done. In view of this, I thought that the spectral theorem for compact operators by Riesz was the way to go. That is, if $T_q$ is compact then $\sigma(T_q)$ is compact, countable and has $0$ as the only limit point. Although it is easy to show that $\sigma(T_q)$ is compact for any $q$, I can't seem to find a way to characterize $q$ satisfying the other conditions. 
By the spectrum it seems natural that $T_q$ only is compact for $q \equiv 0$, but I am struggling to show this. I would appreciate any suggestions on how to proceed. I am not even sure if this is the right approach. The main problem for me is really how to know what theory to use and how to apply it in a concrete example such as this one. Since the book is quite lacking in examples, I could also do with any general tricks to determine the compactness of given operators.
 A: The answer is that $T_q$ is compact iff $q$ satisfies (i) $q([0,1]) \cap B(0,\epsilon)^c$ is finite for all $\epsilon>0$ and (ii) $q^{-1} \{ y \}$ is finite for all $y \neq 0$.
Since $|T_q f (t) | = |q(t)f(t)| \le \|q\| \|f\| $, we have $\|T_q\| \le \|q\|$. You have shown that $\sigma(T_q) = \overline{q([0,1])}$.
($\Rightarrow$): Condition (i) follows from the fact that $\sigma(T_q) = \overline{q([0,1])}$ and the Riesz spectral theorem for compact operators.
To see (ii), let $y\neq0$ and suppose $X=q^{-1} \{ y \}$ is not finite. Then the subspace $S = \{ f \in B \| f(t) = 0 \, \forall t \notin X \}$ is infinite dimensional, and $T_q$ restricted to $S$ is just multiplication by the constant $y$, and hence $T_q (B(0,1) \cap S)$ is not relatively compact. Hence $q^{-1} \{ y \}$ is finite for all $y \neq 0$.
($\Leftarrow$): If $q=0$ then it is clear that $T_q$ is compact, so we may assume that $q \neq 0$.
Let $q_k$ be an enumeration of the non-zero elements of $q([0,1])$, and let $A_k = q^{-1} \{ q_k \}$ for all $k$. By assumption, $A_k$ is finite for all $k$. Also, the $A_k$ are disjoint and if $\{q_k\}$ is not finite, we have $q_k \to 0$.
We can write $(T_q f)(t) =\sum_k q_k 1_{A_k}(t) f(t)$. Let $T_q^N$ be defined by $(T_q^N f)(t) =\sum_{k=1}^N q_k 1_{A_k}(t) f(t)$. We see that the range of $T_q^N$ is finite dimensional, and so it is a compact operator.
Suppose we choose $N$ large enough so that $|q_k| < \epsilon $ for all $k > N$, then
$|(T_q f)(t) - (T_q^N f)(t)| = |\sum_{k>N} q_k 1_{A_k}(t) f(t)| \le \epsilon \|f\| \sum_{k>N} 1_{A_k}(t) \le \epsilon \|f\|$. Hence $T_q^N \to T_q$, and hence $T_q$ is the limit of a sequence of compact operators. It follows that $T_q$ is compact.
