I would like to ask about continuous spectrum of unbounded, densely defined closed operator. Let $A\colon X\supset\mathcal{D}_A\to X$, where X is Banach space, $\overline{\mathcal{D}_A}=X$ be a unbounded linear operator. When I read some books I find two a bit different definitions of continuous spectrum:
(a) $\sigma_c(A)=\{\lambda\in\mathbb{C}\, |\,\lambda I-A \textrm{ is injective }, \overline{R(\lambda I -A)}=X,\, R(\lambda I -A)\neq X \}$
(b) $\sigma_c(A)=\{\lambda\in\mathbb{C}\, |\,\lambda I-A \textrm{ is injective }, \overline{R(\lambda I -A)}=X,\, (\lambda I-A)^{-1} \textrm{ is unbounded} \}.$
Could you explain me why that definitions are equivalent?
Let $B$ be a closed operator with domain $\mathcal{D}_B$ which is injective and has dense range.
Suppose $B^{-1}$ is bounded. I claim the range $R(B)$ is closed, so in fact $R(B)=X$. Suppose $y$ is in the closure of $R(B)$, so there exist $y_n \in R(B)$ with $y_n \to y$. Then $x_n := B^{-1} y_n \in \mathcal{D}_B$ converges to some $x$. So we have $x_n \to x$ and $B x_n = y_n \to y$. Since $B$ is closed, this means $x \in \mathcal{D}_B$ and $Bx = y$, so $y \in R(B)$.
Suppose $R(B)$ is closed, so that $R(B) = X$. Then $B^{-1} : X \to X$ is everywhere defined, and is a closed operator since $B$ is. By the closed graph theorem, $B^{-1}$ is bounded.
So $B$ has closed range iff $B^{-1}$ is bounded. Apply this to $B = \lambda I -A$.