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Let $\mathcal{H}$ be a complex, separable, infinite-dimensional Hilbert space. Given an operator $T\in\mathcal{B(H)}$ and $\pi:\mathcal{B(H)}\rightarrow\mathcal{B(H)}/\mathcal{K(H)}$ the canonical quotient map, we define the essential spectrum of $T$ by $\sigma_e(T):=\sigma(\pi(T))$. It is easy to see that $\sigma_e(T)\subseteq\sigma(T)$.

I know that if $T$ is a normal operator, then $\sigma_e(T)$ contains exactly the elements of $\sigma(T)$ that are not isolated eigenvalues of finite multiplicity. In contrast, a lot can change when $T$ is not normal. The forward unilateral shift $S:\ell^2(\mathbb{N})\rightarrow\ell^2(\mathbb{N})$ has spectrum $\sigma(S)=\overline{\mathbb{D}}$ (the closed unit disk) but essential spectrum $\sigma_e(S)=\mathbb{T}$ (the unit circle).

What I would like to know specifically is which elements in $\sigma(T)$ may disappear in $\sigma_e(T)$ when $T$ is a non-normal element in $\mathcal{B(H)}$. Does the essential spectrum always contain the boundary of the spectrum? What happens when $\sigma(T)$ has empty interior or is totally disconnected? If $\sigma(T)=\overline{\mathbb{D}}$ can multiple holes appear in $\sigma_e(T)$? What can be said about the spectral radius of $\pi(T)$ in terms of that of $T$?

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If $T$ is compact, then $\sigma( \pi(T))=\{0\}$. Hence if $T$ is not quasi-nilpotent, then $\sigma( \pi(T))$ is a proper subset of $ \sigma(T)$

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  • $\begingroup$ This is sort of a trivial setting, but I guess it does shine some light on a couple of the questions raised. In particular, it seems that there is no relation between the spectral radius of $T$ and the spectral radius of $\pi(T)$ (beyond the obvious $r(T)\geq r(\pi(T)$), and there's nothing that can be said when $\sigma(T)$ is totally disconnected. I'd still like to know what happens when $\sigma(T)=\mathbb{T}$ or $\sigma(T)=\overline{\mathbb{D}}$. $\endgroup$ – Zack Cramer Nov 30 '16 at 16:57
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A result due to Putnam and Schechter states the following: if $\mathcal{H}$ is a Hilbert space and $T\in\mathcal{B(H)}$, then for $\lambda\in\partial(\sigma(T))$ we have that $\lambda$ is isolated in $\sigma(T)$ or $\lambda\in\sigma_e(T)$. As a consequence we have that $\sigma(T)=\sigma_e(T)\cup\Omega$ where $\Omega$ is a (possibly empty) set consisting of some bounded components of $\sigma(T)$ and sequences of isolated points in $\sigma(T)$ converging to elements in $\sigma_e(T)$.

This gives us a better understanding of some of the situations above. In particular, if $\sigma(T)=\mathbb{T}$ (the unit circle) then no point of $\partial(\sigma(T))$ is isolated and hence $\sigma_e(T)=\mathbb{T}$ as well. As for an operator whose spectrum is $\overline{\mathbb{D}}$, it must be the case that the essential spectrum contains $\mathbb{T}$ but it could remove multiple holes in the disk or the whole interior.

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