# A self-adjoint operator with essential spectrum={0} is compact

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

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.