Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The only thing I know about non-closable operators can be summarised as "they exist, but they're nasty, so let's not talk about them!" This seems to be the case with everyone else I've talked to. I'd appreciate some references or ideas about the following:

What classes of non-closable operators have been studied (if any). For instance have the operators with non-empty, real, positive spectrum been studied? Are there weaker, but similar conditions to closable/symmetric which gaurentee nice properties (e.g. non-empty spectrum, adjoint with non-trivial domain, some form of polar decompostion)?

EDIT: As the comments suggest, the usual definition of spectrum doesn't make sense if the operator's not closable. Is there another way to define it - or is there a similar object which is useful?

share|improve this question
The traditional definition of the spectrum of an operator (set of complex numbers $\lambda$ such that there is no bounded inverse to $\lambda-X$) doesn't make much sense for non-closed operators, as the spectrum is always $\mathbb{C}$ in this case. –  Florian Nov 26 '10 at 13:21
Moreover, a closed operator can have empty spectrum. –  Michael Renardy Nov 26 '10 at 14:18
Yes, you're right, I've edited the question accordingly. Perhaps it would be better phrased in terms of the adjoint (i.e. a closed operator, not necessarily densely defined)? –  Ollie Margetts Nov 26 '10 at 15:56
In liu of the above comments I've done some pretty heavy trawling for non-densely defined operators. There seem to be quite a few papers (e.g. bit.ly/ewaz8Q and bit.ly/gFeNtb) on non-densely defined symmetric operators and generalized resolvents. I'll need to research it further to see if these are helpful –  Ollie Margetts Nov 26 '10 at 17:06
@Ollie Margetts: The adjoint $A^*$ of a non-closable operator $A$, while closed, need not carry very much information about $A$ itself. In the most extreme case, if $A$ has a dense graph (as in my answer here: mathoverflow.net/questions/22189/…), then the domain of $A^*$ is just $\{0\}$. –  Nate Eldredge Nov 26 '10 at 18:59
add comment

2 Answers

Non-closable operators are not necessarily nasty. Consider the usual Hilbert space $L^2([0,1],dx)$ and the dense subspace $\mathcal{D}=\mathcal{C}[0,1]$. Define $T$ on $\mathcal{D}$ by $T(f)=f(0)$, i.e. the constant function on $[0,1]$ with value $f(0)$. This is a densely defined operator, but it is easy to see that its adjoint is not densely defined. It is not hard to show that in the class of densely defined operators, a closed linear extension exists if and only if the adjoint is densely defined.

share|improve this answer
add comment

Here's another simple example of a non-closable operator, from Reed & Simon Vol. I, Section VIII.4, Example 4 (p. 252): Let $D(T)$ consist of those $\psi\in L^2(\mathbf{R})$ such that $\int|f\psi|<\infty$, where $f$ is a fixed function not in $L^2(\mathbf{R})$, and set $T\psi=(f,\psi)\psi_0$ for some fixed $\psi_0 \in L^2(\mathbf{R})$. Then $D(T)$ is dense in $L^2(\mathbf{R})$; however, as they show by a short computation, $D(T^*)$ consists of the vectors orthogonal to $\psi_0$, so it's not dense.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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