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I am considering an arbitrary measure space $(X, \mathcal{A}, \mu)$ and the Hilbert space $H = L^2(\mu)$. The space $B = L^{\infty}(\mu)$ of essentially bounded functions can then act on $H$ as bounded linear operator by multiplication, i.e. if $g \in B$, then $M_g(f) = gf \in H$ for all $f \in H$.

I've managed to prove the following if $\mu$ is either $\sigma$-finite or locally finite.

For any $g \in B$, the spectrum of $M_g$ is equal to the essential range of $g$ and the point spectrum of $M_g$ is equal to $\{ \lambda \in \mathbb{C} \mid \exists A \in \mathcal{A} \text{ with } \mu(A)>0: f(A) = \lambda\}$.

I also know this doesn't hold for general measure spaces, because there is the following counter example: If we have $X = \{0,1\}$, $\mu(0) = 1$, $\mu(1) = \infty$, $g(0) = 0$ and $g(1) = 1$. Then $M_g=0$ but the essential range and point spectrum of $g$ is $\{0,1\}$.

My questions are the following then:

  • Is there a more general class of measures for which this would hold? I know locally finite doesn't imply $\sigma$-finite and vice versa, so preferably it would hold them both.
  • If such a class doesn't exist, are there maybe counterexamples less trivial than the one I mentioned?
  • Are there any non-trivial counterexamples where a part of the statement still holds, meaning the essential range isn't the spectrum but the point spectrum is still the same set described in the statement? Or maybe the other way around.
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  • $\begingroup$ Do you have any non-trivial measure that is neither $\sigma-$finite nor locally finite. $\endgroup$
    – C. Ding
    Commented Jun 14, 2017 at 1:08
  • $\begingroup$ @C.Ding I don't really know much measure theory, so I don't know much about such special examples. Most measure spaces I know are both. But I guess the counting measure on $\mathbb{R}$ would be neither $\sigma-$finite nor locally finite. $\endgroup$
    – Demophilus
    Commented Jun 14, 2017 at 1:29
  • $\begingroup$ Your conclusion holds for any counting measure space. $\endgroup$
    – C. Ding
    Commented Jun 14, 2017 at 2:47
  • $\begingroup$ The course of proof of your conclusion shows that $\sigma(M_g)\subset \operatorname{ess-ran}(g)$ for any measure space, but for the converse we need $\sigma-$ finite. If there exists some point $x\in X$ with infinite measure, then $\sigma(M_g)\neq\operatorname{ess-ran}(g)$ (see the answer). $\endgroup$
    – C. Ding
    Commented Jun 14, 2017 at 2:57

1 Answer 1

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Let $(X', \mathscr M', \mu')$ be an arbitary measure space and $x_\infty\notin X'$. Set $X=X'\cup \{x_\infty\}, \mathscr M=\mathscr M'\cup \{A\cup \{x_\infty\}|A\in \mathscr M'\}$, and $\mu(A)=\mu'(A)$ if $A\in\mathscr M'$ and $ \mu(A)=\infty$ if $x_\infty\in A$.

Let $g(x)=1\forall x\in X'$ and $g(x_\infty)=0$. Then $0\in\operatorname{ess-ran}(g)$ but $M_g $ is invertible.

$\sigma_p(M_g)=\{\lambda\in \mathbb{C}|\exists A\in X~ \text{with} ~ \mu(A)>0:f(A)=\lambda\} $ for an arbitary measure space.

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  • $\begingroup$ Ah yes of course, so in order for the statement to hold at every point there should be some measurable set containing that point with finite measure. And of course locally finite and $\sigma$-finite are measures that have this feature. And your example shows that if there is a point at which the measure is always infinite, the statement can never hold. Thank you, surprised I couldn't come up with that myself. $\endgroup$
    – Demophilus
    Commented Jun 14, 2017 at 8:54

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