The essential spectrum of a bounded linear operator $A$ on a separable Hilbert space $\mathcal{H}$ is defined as $$ \sigma_{\mathrm{ess}}(A) \equiv\left\{z\in\mathbb{C}\left.\right|A-z\mathbb{1}\text{ is not a Fredholm operator}\right\} \tag{1}\,. $$

This way to define the essential spectrum immediately implies that it is stable under compact perturbations, using Atkinson's characterization of Fredholm operators and the fact compact operators form an ideal.

Let now $A$ be a normal operator, so we may apply the spectral theorem on it to obtain a spectral measure. We know that the absolutely-continuous spectrum of $A$ has a characterization in terms of the Lebesgue's decomposition (it is the support of the absolutely-continuous (w.r.t. the Lebesgue measure on $\mathbb{R}$) part of the spectral measure. Furthermore we also know that the ac-spectrum is stable under trace-class perturbations. Furthermore trace-class operators also form an ideal.

So this leads one to attempt to define operators of class $\mathcal{T}$ as those which are invertible up to trace-class. One should note that there is a paper by Murphy characterizing Fredholms as invertible up to finite rank, and trace-class operators are of course limits of finite rank in trace class norm, so this already seems suspect.

Anyway, then one would attempt to try to prove $$ \sigma_{\mathrm{ac}}(A) \stackrel{?}{=}\left\{z\in\mathbb{C}\left.\right|A-z\mathbb{1}\text{ is not }\mathcal{T}\right\} \,.$$

But this cannot possibly be true, because I don't believe it would be possible to bypass the Lebesgue decomposition theorem to define the ac-spectrum.

Anyway, does anything vaguely in this direction exist? Defining invertibles up to other ideals of operators instead of compact? What kind of spectrum do such definitions induce?


1 Answer 1


You wrote "normal" earlier, but I assume (from your use and description of $A_{ac}$) that you want to consider self-adjoint $A$. Then your set (denote it by $T$) is simply $T=\sigma_{ess}(A)$ again.

If $A-z$ isn't invertible modulo compacts, then it won't be invertible modulo trace class, so $\sigma_{ess}(A)\subseteq T$. Of course, it's also clear that $T\subseteq\sigma(A)$. If $z\in\sigma\setminus\sigma_{ess}$, then $z$ is an isolated eigenvalue of finite multiplicity, and we can invert the part of $A-z$ in $N(A-z)^{\perp}$. In other words, we can invert $A-z$ modulo finite rank, and thus $z\notin T$.

  • $\begingroup$ Thanks, of course! In other words, as far as the spectrum is concerned, the whole Schatten ideals collapse into the compacts... $\endgroup$
    – PPR
    Feb 5, 2021 at 16:34

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