Compactness of Multiplication Operator on $L^2$

Solution 1:

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.