Is there an operator have for the essential spectrum the unit disk?

Question

Is there a bounded linear operator $T \in l^2(\mathbb{N})$ have for the essential spectrum the unit disk of $\mathbb{C}$; i.e, such that $\sigma_{e}(T)=\textbf{D}(0, 1)$; where $\textbf{D}(0, 1):=\{\lambda \in \mathbb{C} : |\lambda|\leq 1\}.$

Answer 1

Given any compact $K\subset \mathbb C$, there exists $T\in B(\ell^2(\mathbb N))$ with $\sigma_e(T)=K$.

Let $\{P_n\}$ be a pairwise orthogonal sequence of infinite projections that add to $I$ (construction at the end). Take $\{q_n\}\subset K$ to be a countable dense set. Now define $T$ such that $$ TP_n=q_nP_N. $$ That is, $T$ is $q_nI$ on each subspace $P_n\ell^2(\mathbb N)$. First, note that $\sigma(T)=K$. Indeed, each $q_n$ is an eigenvalue. If $r\in\mathbb C\setminus K$, let $d=\operatorname{dist}(r,K)$. Define $S$ to be given by $$ SP_n=\frac1{q_n-r}\,P_n. $$ Then $S$ is bounded (because $|q_n-r|\geq d$ for all $n$) and $S(T-r I)=I$; this shows that $\sigma(T)\subset\overline{\{q_n\}}=K$. As the multiplicity of each $q_n$ is infinite, $\{q_n\}\subset \sigma_e(T)$. As the essential spectrum is closed and contained in $\sigma(T)$, we get that $\sigma_e(T)=K$.

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Construction of the projections. Let $\{e_k\}\subset\ell^2(\mathbb N)$ be an orthonormal basis. Let $\{G_n\}\subset \mathbb N$ be a partition of $\mathbb N$ into infinite sets. Now let $P_n$ be the projection onto the span of $\{e_k:\ k\in G_n\}$.

Answer 2

Consider $T$ on $\ell^2(\mathbb N \times \mathbb N)$, with $T(a)_{m,n} = a_{m,n+1}$. Then for $|\lambda| < 1$, elements of the form $(b_m \lambda ^n)_{m,n}$ are in the kernel of $T - \lambda I$. So the kernel has infinite dimension.

Also $T$ on $L^2(D(0,1))$ (which is isomorphic to $\ell^2(\mathbb N)$) given by $Tf(z) = z f(z)$. Then the cokernel of $T - \lambda I$ is infinite dimensional for all $\lambda \in D(0,1)$.