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Does every self adjoint operator (on a Hilbert space) with essential spectrum={0} is a compact operator ?

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If the operator $T$ is allowed to be unbounded, then this is obviously wrong. For a concrete counterexample take the operator $T$ on $\ell^2$ given by $$ D(T)=\{\xi\in\ell^2\mid \sum_k k^2 \xi_{2k}^2<\infty\},\,T\xi=(0,\xi_2,0,2\xi_4,0,\dots). $$ On the other hand it is true for bounded self-adjoint operators. I use the characterization of the essential spectru as complement of the isolated eigenvalues of finite multiplicity (in $\sigma(T)$).

Since $\sigma_{\mathrm{ess}}(T)=\{0\}$, the set $\sigma(T)\setminus(-1/n,1/n)$ is bounded and has no accumulation points, which means that it is finite. Hence $\sigma(T)\setminus\{0\}$ consists of at most countable eigenvalue (counted with multiplicity), which accumulate only at zero. Thus $T$ is compact.

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