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Let $a\in \mathcal{L}(L^2([0, 1], \mathbb{R}))$ be a multiplication operator. I wonder whether there is any work on calculating its essential spectrum. Is there any way to explicitly compute its essential growth bound and elements of its discrete spectrum?

What about the $n$-dimensional case, i.e., $a\in \mathcal{L}(L^2([0, 1], \mathbb{R}^{n}))$? (Note that the $n$-dim multiplication operator may not be self-adjoint.)

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  • $\begingroup$ @WillieWong Thanks for the comments. I know of this note. But I'm asking to compute the 'essential spectrum' of a multiplication operator, not the spectrum decomposition of `point, continuous and residual spectra'. $\endgroup$ – potionowner Jan 22 at 20:10
  • $\begingroup$ Ah, sorry, I didn't read carefully. $\endgroup$ – Willie Wong Jan 22 at 20:14
  • $\begingroup$ The notes that you linked for the definition of the essential spectrum only refer to self-adjoint case. So are you mainly interested in the case where $a$ is self-adjoint (i.e., where the symbol of $a$ is real)? $\endgroup$ – Jochen Glueck Jan 22 at 20:29
  • $\begingroup$ @JochenGlueck Thanks for pointing this out! I'm interested in the case where $a$ is real. But this does not necessarily mean that $a$ is self-adjoint when $a$ is not $1$-dimensional. $\endgroup$ – potionowner Jan 22 at 20:37
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    $\begingroup$ @potionowner: Thanks for your reply! In the self-adjoint case the essential spectrum of $a$ coincides with the spectrum. This follows from a spectral projection argument and from the fact that $[0,1]$ does not contain any atoms. I'd guess that it is true for the non-self-adjoint case, as well, but I'm not quite sure right now. $\endgroup$ – Jochen Glueck Jan 22 at 20:41
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The spectrum of $(Af)(x)=a(x)f(x)$ is the essential range of $a(x)$. As usual in this context, essential here means basically (resisting a silly pun) "ignore what happens on null sets." More precisely, $y\in\mathbb R$ is in the essential range if $\{x: |a(x)-y|<\epsilon\}$ has positive (Lebesgue) measure for all $\epsilon>0$.

Next, $y$ is an eigenvalue if and only if $M=\{ x: a(x)=y \}$ has positive measure. Since $\chi_N$, $N\subseteq M$, is an eigenfunction, all eigenvalues have infinite multiplicity. In particular, they are in $\sigma_{ess}$. To find all of $\sigma_{ess}$, add the accumulation points of $\sigma$ to this set.

Another useful fact to keep in mind here is that the spectral projections $E(M)$ of $A$ are multiplication by $\chi_{a^{-1}(M)}$.

(All these facts have straightforward proofs once you have them.)

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  • $\begingroup$ I'm not sure I understand the sentence "Too find all of $\sigma_{ess}$, add the accumulation points of $\sigma$ to this set." As pointed out in a comment above, the essential spectrum is simply the spectrum (and hence equal to the essential range of $a$). $\endgroup$ – Jochen Glueck Jan 22 at 21:21
  • $\begingroup$ @JochenGlueck: That is a good point. Still, my statement is correct (though unnecessary here); I was simply following the identity (definition?) $\sigma_{ess} = P\cap A$, with $P =$ eigenvalues of infinite multiplicity, $A=$ accumulation points of $\sigma$. $\endgroup$ – Christian Remling Jan 22 at 22:27
  • $\begingroup$ Thanks for your reply! Yes, for sure your statement is correct, just a bit more complicated than necessary. By the way, the set $P \cup A$ (which you probably mean, rather than $P \cap A$), is equal to the essential spectrum in the self-adjoint case, but it can be larger than the essential spectrum in the non-self adjoint case. (If we use that definition that $\lambda$ is in the essential spectrum of $T$ iff $\lambda - T$ is not Fredholm). $\endgroup$ – Jochen Glueck Jan 22 at 22:35
  • $\begingroup$ @JochenGlueck: Yes, $\cup$, thanks. And I indeed had in mind the self-adjoint case exclusively. $\endgroup$ – Christian Remling Jan 22 at 22:43
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In the following answer I'll focus on the case for general $n$.

Let $m: [0,1] \to \mathbb{C}^{n \times n}$ be measurable and bounded. Let $a \in \mathcal{L}(L^2([0,1]; \mathbb{C}^n))$ be the multiplication with $m$.

Proposition 1. Each value in $\partial \sigma(a)$ is an essential spectral value of $a$.

Proof. Let $\lambda \in \partial \sigma(a)$ and assume that $\lambda$ is not an essential spectral value. Since $\lambda-a$ can be approximated by invertible operators, it follows from analytic Fredholm theory that $\lambda$ is an isolated spectral value of $a$ and a pole of the resolvent with finite-dimensional spectral space.

Let $p$ denote the corresponding spectral projection; it has finite rank. Since $p$ can be written as a contour integral of the resolvent of $a$, it follows that $p$ is a multiplication operator, too; let $q: [0,1] \to \mathbb{C}^{n \times n}$ denote its symbol. After changing $q$ on a set of measure $0$ if necessary, we may assume that the matrix $q(x)$ is a projection for each $x \in X$.

The set of $x$ for which $q(x) \not= 0$ has non-zero measure (since $p \not= 0$), and thus it follows (similarly as in Christian Remling's answer) that the range of $p$ is infinite-dimensional - a contradiction. $\square$

Corollary 2. If $m(x)$ is self adjoint for (almost) every $x \in [0,1]$, then the essential spectrum of $a$ coincides with the spectrum.

Proof. Under the assumption of the corollary, the operator $a$ is self-adjoint, so every spectral value of $a$ is in the boundary of the spectrum. $\square$

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In Biberdorf, E. A. and Väth, M., On the spectrum of orthomorphisms and Barbashin operators, Z. Anal. Anwendungen 18, 1999(4), 12-31 it is shown that even in the more general case of an orthomorphism, the essential spectrum (for various definitions of “essential spectrum”) is the same as the spectrum and coincides with the “essential range” of the operator (for an appropriate definition of ”essential range”). Unsurprisingly, in case of a multiplication operator, the essential range of the operator is the essential range of the mulitplication function.

Note: To avoid a misunderstanding: This is a different sort of generalization than the original question in case $n=1$; the multiplication operator is not an orthomorphism in case $n>1$ (at least not with any order that I am aware of), except if it is diagonal a.e. (with the natural order) or, more genreal, diagonalizable a.e. (with respect to a measurable basis transformation). The latter is the case, in particular, if the multiplication operator is normal a.e.

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  • $\begingroup$ The link to the article seems to be broken. (+1 anyway, interesting article!) $\endgroup$ – Jochen Glueck Jan 25 at 10:08
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    $\begingroup$ Thanks. Should be fixed now. $\endgroup$ – Martin Väth Jan 25 at 18:12

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