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?
 A: 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 closed 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.
