Spectrum of a closed operator Could someone please explain this fact: if $A$ is a closed operator and $A^{-1}$ is a compact operator, then spectrum of $A$ consist only of eigenvalues?
I forgot to mention that operator $A^{-1}$ is bounded, but $A$ isn't continuous.
 A: I am assuming you mean that $A$ is a continuous linear bijection on a Banach space .Let $s$ belong to the spectrum of $A$.Let $(v_n)_n$ be a sequence   with  $||v_n||=1$ for each $n$ and with $(||A v_n-s v_n||)_n$ converging to $0$.Now $A^{-1}$ is compact, and $\{A v_n-s v_n\}_n$ is a bounded set ( because the sequence converges to $0$ ) so there exists a subsequence of $(v_n)_n$,which we will call $(w_n)_n$, for which $(A^{-1}(A w_n-s w_n))_n$ $=(w_n-A^{-1} s w_n)_n$ is  convergent  to some $p$ . The continuity of $A$ implies that $(A(w_n-A^{-1} s w_n))_n =$ $(A w_n-s w_n)_n$  converges to $A p$.But $(A w_n-s a_n)_n$ converges to $0$ so $A p=0 , $  and $A^{-1}$ exists, so $A p=0\implies p=0$.Hence $(w_n-A^{-1} s w_n)_n$ converges  to $0$.We use the compactness of $A^{-1}$ again: The bounded sequence $( w_n)_n$ has a subsequence ,which we will call $(x_n)_n$, for which $(A^{-1} s x_n)_n$ converges to some $q$, and since $(x_n-A^{-1} s x_n)_n$ converges to $0$, the sequence $(x_n)_n$ must also converge to $q$.Note that $q\neq 0$ because $||x_n||=1$ for each $n$.Finally we have $A q-s q=$ $\lim_{n\to \infty} A(x_n-A^{-1} s x_n)=0$. So $s$ is an eigenvalue of $A$. Footnote: We note that $q\neq 0$ in case $s=0$, else the sentence $A q-s q=0$  may be equivalent to $A0=0$.
