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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

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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.

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    @Acton: The spectrum of$a$compact self-adjoint operator typically consists of a countable sequence of eigenvalues converging to zero, or else a finite set of eigenvalues, with all nonzero eigenvalues having finite multiplicity. The spectrum of the "multiplication by $g$" operator equals the essential range of $g$, and $\lambda$ is an eigenvalue of $g$ iff $g = \lambda$ on a set of positive measure. So the spectrum of the operator $A$ is the interval $[a,b]$, an uncountable set, and it has no eigenvalues.2018-05-30