Let $A: D(A) \to \mathscr{H}$ be a densely defined self-adjoint operator on a Hilbert space $\mathscr{H}$ which is bounded from below. For each $n \in \mathbb{N}$, define: $$\mu_{n}(A) := \inf_{\substack{V \subset D(A) \\ \dim V = n}}\max_{\substack{\psi \in V \\ \|\psi\| = 1}}\langle \psi, A\psi\rangle.$$ By the min-max principle, $\mu_{n}(A)$ is an increasing sequence of real numbers $\mu_{1}(A) \le \mu_{2}(A) \le \cdots$ such that $\mu_{n}(A) \to \mu_{\infty}(A)$ as $n \to \infty$, where $\mu_{\infty}(A) := \inf \sigma_{\text{ess}}(A)$, where the latter denotes the essential spectrum of $A$.
I am trying to prove the following result: if $A$ has compact resolvent, that is, if $(A-\lambda)^{-1}$ is compact for every $\lambda$ in the resolvent of $A$, then $\mu_{\infty}(A) = \infty$.
I have some vague thoughts about how to prove it. My idea would be something as follows. If $\mu_{\infty}(A) < \infty$ then maybe I can prove that $\mu_{\infty}(A) \in \sigma_{\text{ess}}(A)$. If this is the case, we could take a Weyl sequence $\|(A-\mu_{\infty}(A))\psi_{n}\| \to 0$ with $\|\psi_{n}\| = 1$ and maybe use this sequence to prove that $(A\pm i)^{-1}$ is not compact.
As you can see, it is just a sketch without something concrete. Any help is welcome.
Theorem. Let $A$ be a self-adjoint operator defined on a complex Hilbert space. If $(\lambda I - A)^{-1}$ is compact for some $\lambda \in \mathbb{C}$, then $\sigma(A)=\sigma_d(A)$.
By the above theorem, we know that $\sigma(A)=\sigma_d(A)$ and so $\sigma_{ess}(A)= \emptyset$. Hence $\mu_\infty(A)=\inf \emptyset =\infty$.