The inverse of a compact and self-adjoint operator in Hilbert space is bounded

Question

Let $\;A\;$ be a self-adjoint and compact operator in Hilbert space$\; \mathcal H\;$ and consider $\;μ \neq 0 \;$ a real number. Prove:

1. $\; \exists c \gt 0 \;$ such that $\;\vert \vert (A-μI)x \vert \vert \ge c \vert \vert x \vert \vert\;\;\forall x \in \overline {Ran(A-μI)}\;$
2. $\;Ran(A-μI)\;$ is a closed subset of $\; \mathcal H\;$

My attempt:

2. Let $\;y_n \in Ran(A-μI)\;$ such that $\;y_n \to y\in \mathcal H\;$. There exist $\;x_n \in \mathcal H\;$ such that $\;(A-μI)x_n \to y\;$. $\;x_n\;$ is a Cauchy sequence since $\;\vert \vert x_n- x_m \vert \vert \le \frac{\;\vert \vert (A-μI)(x_n -x_m) \vert \vert }{c}=\;\vert \vert y_n -y_m \vert \vert \to 0\;$ using the first part of the ex. and the fact that $\;y_n\;$ is Cauchy. Then $\;x_n \to x\in \mathcal H\;$ and it follows $\;(A-μI)x_n \to (A-μI)x=y\in Ran(A-μI)\;$

I don't know how to show 1. holds. If I knew $\;μ \notin σ(A)\;$ then I could claim $\;(A-μI)^{-1}\;$ is bounded and the rest comes easily. But I don't have this assumption.

I would appreciate any help. Thanks in advance!!

Answer 1

For (1), assume otherwise. Then we can pick $x_n \in \overline{\operatorname{Ran}(A-\mu)}$ such $\|x_n\| = 1$ for all $n \geq 1$ but $(A - \mu)x_n \to 0$ as $n\to\infty$. Since $A$ is compact, passing to a subsequence we may assume that $Ax_n$ converges as well. Then

$$ x_n = \mu^{-1}(Ax_n - (A - \mu)x_n)$$

also converges. Denoting the limit of $(x_n)$ by $x$, it must satisfy the following three properties:

1. $ x \in \overline{\operatorname{Ran}(A-\mu)}$
2. $\|x\| = 1$,
3. $ x \in \operatorname{Ker}(A-\mu)$, since $(A - \mu)x = 0$.

But since $A - \mu$ is selfadjoint, we must have $\overline{\operatorname{Ran}(A-\mu)} = (\operatorname{Ker}(A-\mu))^{\perp}$ and hence properties 1-3 cannot hold simultaneously, a contradiction!