3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $B = \lambda I -A$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.