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?

  • $\begingroup$ Are you sure they didn't assume $A$ is a closed? $\endgroup$ – DisintegratingByParts Jun 18 '17 at 14:11
  • $\begingroup$ You are right - A is closed. $\endgroup$ – akap Jun 21 '17 at 16:24

Let $B$ be a closed operator with domain $\mathcal{D}_B$ which is injective and has dense range.

  1. 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 $y \in R(B)$.

  2. 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 $B = \lambda I -A$.

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  • $\begingroup$ Let me ask only one question: in the second part of the proof, when we want to use closed graph theorem we need to know, that $\mathcal{D}_B$ is a Banach space. Is it a consequence of the assumption $\overline {\mathcal{D}_B}=X$? $\endgroup$ – akap Jun 22 '17 at 13:49
  • $\begingroup$ @akap: Sorry, I didn't write that very well. Indeed, we should think of $B^{-1}$ as an operator from $X$ to $X$ (whose range happens to equal $\mathcal{D}_B$). In particular, I am asserting that the graph of $B^{-1}$ is closed in $X \times X$, not merely in $X \times \mathcal{D}_B$. $\endgroup$ – Nate Eldredge Jun 22 '17 at 14:42
  • $\begingroup$ ok, now it is clear. Last but not least question: In the first part part of the proof we know, that $x_n:=B^{-1}y_n$ converges to some $x$, because $x_n$ is a Cauchy seqence, so that exist such $N\in\mathbb{N}$ that for $n,m\geq N$ we have $\parallel x_n-x_m \parallel=\parallel B^{-1}y_n-B^{-1}y_m\parallel=\parallel B^{-1}(y_n-y_m)\parallel\leq\parallel B^{-1}\parallel \parallel (y_n-y_m)\parallel\to 0$, because $\{y_n\}$ is convergent and $B^{-1}$ bounded. $\endgroup$ – akap Jun 23 '17 at 5:53
  • $\begingroup$ @akap: Correct. At its root, it's the general metric space fact that $B^{-1}$, being bounded and hence uniformly continuous on $R(B)$, has a unique continuous extension to the closure $\overline{R(B)}$. (This of course uses the fact that the codomain $X$ of $B^{-1}$ is complete.) $\endgroup$ – Nate Eldredge Jun 23 '17 at 17:48

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