0
$\begingroup$

Given an unbounded densely defined operator $D: {\frak dom}(D) \subseteq \mathbb{H} \to \mathbb{H}$, on some Hilbert space $\mathbb{H}$, it graph is the subspace $$ \mathcal{G}(D) := \{(x,D(x)) \text{ such that } x \in {\frak dom}(D)\}. $$ We say that an operator is closed if $\mathcal{G}(D)$ is a closed subspace of $\mathbb{H \oplus H}$.

My question is asking of the definition of closure for $D$. We could say, $D$ is closable if

A) the closure of $\mathcal{G}(D)$ in $\mathbb{H \oplus H}$ is the graph of some operator

OR

B) there exists a closed operator $\widetilde{D}$ such that $\mathcal{G}(D) \subseteq \mathcal{G}(\widetilde{D})$.

Clearly, if $D$ is closable in the sense of A then it is cloasble in the sense of B. Is the opposite inference true? If not what is an instructive example?

$\endgroup$
  • $\begingroup$ I'd start by looking for an example of some $D$ that has several closed extensions $D', D''$ that are incompatible, or that has a sequence $D_1,D_2,\ldots$ of closed operators such that $\mathcal{G}(D_i)\subset \mathcal{G}(D_{i+1})$ but $\cup_i\mathcal{G}(D_i)$ is not the graph of anything. $\endgroup$ – Neal Jun 6 at 18:28
  • $\begingroup$ Your notation of B) is unclear. What does $\widetilde{D}$ mean? what is $D'$? what is the relation between them? $\endgroup$ – uniquesolution Jun 6 at 18:33
  • $\begingroup$ @uniquesolution: sorry that was a typo, it's fixed $\endgroup$ – Max Schattman Jun 6 at 18:36
2
$\begingroup$

Condition $(B)$ is equivalent to condition $(A)$. In this answer I show that $(B) \implies (A)$ since you know how to do the other direction.

The key point is that a subspace $\mathcal{G} \subset H \oplus H$ is the graph of a closed extension of $A$ if and only if $\mathcal{G}(A) \subset \mathcal{G}$ and $\mathcal{G}$ has the single valued property $$(x,y_1),(x,y_2) \in \mathcal{G} \implies y_1 = y_2$$ which is equivalent to $$(0,y) \in \mathcal{G} \implies y = 0$$ since $\mathcal{G}$ is a linear space. In general, $K \subset H \oplus H$ is the graph of an operator if and only if it is a linear subspace with the single valued property.

So we want to check that if $(B)$ holds then $\overline{\mathcal{G}(A)}$ has the single valued property. But this is easy since if $(0,y) \in \overline{\mathcal{G}(A)}$ then $(0,y) \in \mathcal{G}(\tilde{D})$ since $\mathcal{G}(\tilde{D})$ is a closed subset containing $\mathcal{G}(A)$. Since $\mathcal{G}(\tilde{D})$ is the graph of an operator, it has the single valued property and so $y = 0$. Hence $\overline{G(A)}$ has the single valued property and so is the graph of an operator.

$\endgroup$

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.