Compactness of  Multiplication Operator on $L^2$ Suppose we have an bounded linear operator A that operates from $L^2([a,b]) \mapsto L^2([a,b])$.  Now suppose that $A(f)(t) = tf(t)$.  
Is A compact?
Edit: I know $A = A^*$ but I'm not really sure how to start on this.  It's not homework, just summer fun :D
 A: Do you know what the spectrum of a compact self-adjoint operator looks like?
Do you know what the spectrum of a multiplication operator looks like?
Edit: Here's a second approach, that doesn't rely on the self-adjointness.
If $0 \notin [a,b]$, then show that $A$ has a bounded left inverse (i.e. a bounded operator $B$ such that $BA = I$).  Show that a compact operator on an infinite-dimensional space cannot have this property.
If $0 \in [a,b]$, show there is an infinite-dimensional closed (added on edit) subspace $K \subset L^2([a,b])$ such that the restriction of $A$ to $K$ has a bounded left inverse.  Show that a compact operator cannot have this property, either.
A third approach would be to explicitly find an $L^2$-bounded sequence $\{f_n\}$ such that $\{A f_n\}$ has no $L^2$ convergent subsequence.  Actually, the "second approach" above would help with this.
Edit 2: Using the second approach above, one could prove the following generalization:

Proposition.  Let $(X, \mu)$ be a measure space without atoms, and let $h : X \to \mathbb{C}$ be a bounded measurable function.  Let $Af = hf$ be the corresponding multiplication operator on $L^2(X, \mu)$.  Then, except in the trivial case that $h = 0$ $\mu$-a.e., $A$ is not compact.

