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


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?

  • $\begingroup$ 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

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.


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