# Definition of a closable operator

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?

• 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. – Neal Jun 6 at 18:28
• Your notation of B) is unclear. What does $\widetilde{D}$ mean? what is $D'$? what is the relation between them? – uniquesolution Jun 6 at 18:33
• @uniquesolution: sorry that was a typo, it's fixed – Max Schattman Jun 6 at 18:36

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.