Suppose $(A\,;\mathcal{D}(A))$ is an unbounded self-adjoint linear operator (obviously, $\mathcal{D}(A)$ must be dense) on a Hilbert space $\mathcal{H}$. Suppose $\mathcal{D}(C)$ is a proper dense subset in $\mathcal{D}(A)$, and consider the restriction $(A_{r}\,;\mathcal{D}(C))$ of $A$ to $\mathcal{D}(C)$.

I would like $(A_{r}\,;\mathcal{D}(C))$ to be closed, and I know that, in order to be so, $(A_{r}\,;\mathcal{D}(C))$ must not be bounded. This is so because a bounded closed linear operator must have a closed domain (see here), and $\mathcal{D}(C)$ is assumed to be a proper dense subspace of $\mathcal{H}$, and thus can not be closed.

Therefore I ask:

Are there general conditions under which $(A_{r}\,;\mathcal{D}(C))$ is closed?


In view of Robert Israel's interesting reply, I would like to expand my question relating it to a specific example. I want to do so because I am new to the theory of linear operators (in the sense that, being a physicist, I had not had the need, unitl now, to pay too much attention to the subleties of domain issues), and hence I am not able to immediately use Robert Israel's answer in a computational approach.

Suppose $\mathcal{H}=\mathcal{L}^{2}(0\,;2\pi)$, $A=-\imath\frac{d}{dx}$ with:

$$ \mathcal{D}(A):=\{\psi\in\mathcal{H}:\psi\in C^{1}(0\,;2\pi)\;\;;\;\;\psi'\in\mathcal{H}\;\;;\;\;\psi(0)=e^{2\pi\imath\gamma}\psi(2\pi)\,\gamma\in[0\,;1)\} $$ and:

$$ \mathcal{D}(C):=\{\psi\in\mathcal{D}(A):\psi(0)=\psi(2\pi)=0\} $$ How can I proceed?

I am sorry if my question is trivial.


1 Answer 1


The graph $G(A)$ of $A$ is a closed linear subspace of $\mathcal H ^2$. Any closed linear subspace of that is the graph of a closed restriction of $A$. So you want to take any closed linear subspace $S$ of $G(A)$ such that $P(S)$ is dense in $\mathcal H$, where $P(u,v) = u$ is the "first coordinate" projection of $\mathcal H^2$ on $\mathcal H$.


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