Let $A:\mathcal{D}(A)\to\mathcal{H}$ be a densly-defined, self-adjoint and not-necessarily bounded operator on a Hilbert space seperable $\mathcal{H}$, such that $(A+i)^{-1}$ is a compact operator. (The existence of $(A+i)^{-1}$ is clear, since $A$ is self-adjoint and has therefore a totally real spectrum)
I have to show that $\sigma_{\mathrm{ess}}(A)=\emptyset$.
My idea is to use Weyl-characterisation:
A sequence $(u_{n})_{n\in\mathbb{N}}\subset\mathcal{D}(A)$ is called Weyl-sequence for $(A,\lambda)$ where $\lambda\in\sigma(A)$, if $\Vert u_{n}\Vert=1$ and $\Vert (A-\lambda)u_{n}\Vert\to 0$
Then the Weyl-characterisation states:
$\lambda\in\sigma_{\mathrm{ess}}(A)\Leftrightarrow\exists$ a Weyl sequence of $(A,\lambda)$, which is converging weakly to zero.
and
$\lambda\in\sigma_{\mathrm{d}}(A)\Leftrightarrow\forall$ Weyl sequences $(u_{n})_{n\in\mathbb{N}}$ of $(A,\lambda)$, the sequence $(Au_{n})_{n\in\mathbb{N}}$ has a convergent subsequence.
So to proof the claim, I could for example use the 2nd claim: Let $\lambda\in\sigma(A)$. When I am able to show that for every Weyl sequence $(u_{n})_{n\in\mathbb{N}}$ of $(A,\lambda)$ the sequence $(Au_{n})_{n\in\mathbb{N}}$ has a convergent subsequence, then I am done.
Since $(u_{n})_{n\in\mathbb{N}}$ is bounded and $(A+i)^{-1}$ is compact, the sequence $((A+i)^{-1}u_{n})_{n\in\mathbb{N}}$ has a convergent subsequence, but how can I use this to prove the claim?
EDIT: I think I have found the answer by myself: Let $\lambda\in\sigma(A)$ and suppose $\lambda\in\sigma_{\mathrm{ess}}(A)$. Then there exists a Weyl sequence $(u_{n})_{n\in\mathbb{N}}$ of $(A,\lambda)$, which is converging weakly to zero. Since $(A+i)^{-1}$ is compact, it follows that $\Vert (A+i)^{-1}u_{n}\Vert\to 0$. But then
$$1=\Vert u_{n}\Vert=\Vert (A+i)^{-1}(A+i+\lambda-\lambda)u_{n}\Vert\leq \Vert (A+i)^{-1}\Vert\cdot\Vert (A-\lambda)u_{n}\Vert + \vert i+\lambda\vert\cdot\Vert (A+i)^{-1}u_{n}\Vert\to 0$$
which is a contradiction. Is this right?