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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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up vote 9 down vote accepted

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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Thanks; I see what you're saying. Marking your answer as the answer since it was key to my realizing what was going on. – Aspar T. Ame Jul 16 '11 at 14:40
Careful: On $L^2([a,b])$, multiplication by a function $h$ with a finite range is not a finite-rank operator. (Indeed, think about the constant function $h=1$.) – Nate Eldredge Jul 16 '11 at 22:07
@paul: I've added a generalization of this statement to my answer. – Nate Eldredge Jul 16 '11 at 22:16
Oop! A gross mis-statement (my earlier incorrect comment suggesting that finitely-many-valued functions give finite-rank multiplication operators). Thx for correcting this! But/and identification of the finite-rank operators (if any) is desirable if one is hoping to identify compact operators. If there aren't any (as here), it is suggestive. ... after mac's following comment, I added a further (clarifying?) comment below it. – paul garrett Jul 17 '11 at 0:02
@paul: I don't know what you mean. There are certainly finite rank operators on $L^2[a,b]$; for example, the map sending $f$ to the constant function $\int f\,d\mu$ has rank one, and is of the form $\langle\cdot,1\rangle 1$; more generally, if $g,h\in L^2[a,b]$ then $\langle\cdot,g\rangle h$ has rank one. – mac Jul 17 '11 at 1:24

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