9
$\begingroup$

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

1 Answers 1

11

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.

  • 0
    Thanks; I see what you're saying. Marking your answer as the answer since it was key to my realizing what was going on.2011-07-16
  • 1
    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 $h=1$.)2011-07-16
  • 0
    @paul: I've added a generalization of this statement to my answer.2011-07-16
  • 0
    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.2011-07-17
  • 0
    @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.2011-07-17
  • 0
    Sorry for opaque comments: ... Ok, one point is that it is bad to mistake the multiplying function for a _kernel_ for an operator. True, $Tf(x)=\int_a^b K(x,y)f(y)dy$ can easily have finite rank. But/and the kernel for multiplication-by-$h$ has kernel $K(x,y)=h(x)\delta(x-y)$, which is fine, but quite far from giving a Hilbert-Schmidt or other compact operator. It is true that finitely-many-valued $L^2$-functions $K(x,y)$ give finite-rank operators, but $K(x,y)=h(x)\delta(x-y)$ isn't in $L^2$, even if we argue that it is finitely-many-valued.2011-07-17
  • 0
    As Nate has shown, it is interesting when we consider $[a,b]=[0,1]$, then, the spectrum of $T$ is equal to the continuous spectrum of $T$, which is $[0,1]$. While Fredholm's alternative2012-12-17
  • 0
    theorem says that a compact operator can only have point spectrum $a$, when $a\not =0.$2012-12-17
  • 0
    @NateEldredge Sorry for digging up this question from 7 years ago but could you please elaborate on your first approach, which you've currently given in the form of two hints - I.e when you asked, do you know the spectra of multiplication operators and compact self-adjoint operators?2018-05-29
  • 0
    @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