Definition of the closure of an operator

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I am reading Reed & Simon - Methods of Modern Mathematical Physics book and, in Chapter VII, the notion of an unbounded linear operator is introduced. Let $\mathscr{H}$ be a Hilbert space and $T$ be a densely defined linear operator with domain $D(T) \subset \mathscr{H}$. According to the authors, $T$ is called closed if its graph: $$\Gamma(T) := \{\langle \varphi, T\varphi\rangle: \varphi \in D(T)\}$$ is a closed subset of $\mathscr{H}\times \mathscr{H}$, when the latter is equipped with the product topology.

In addition, if $T_{1}$ and $T$ are operators on $\mathscr{H}$ such that $\Gamma(T) \subset \Gamma(T_{1})$, then $T_{1}$ is said to be an extension of $T$.

So far, so good. Now, consider the following definition.

Definition: An operator $T$ is closable if it has a closed extension. Every closable operator has a smallest closed extension, called its closure, which we denote by $\bar{T}$.

So, my first question is: what is the precise meaning of "smallest"? Does it mean that every other closed extension $T_{1}$ of $T$ is such that $\Gamma(\bar{T})\subset \Gamma(T_{1})$? In other words, every closed extension of $T$ is also a closed extension of $\bar{T}$?

As a second question: how to be sure that a smallest closed extension exists? And is it unique or not necessarily?

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Your intuition on what is meant by the smallest closed extension is correct.

We can identify operators with their graphs which are subsets of $\mathscr{H} \times \mathscr{H}$. As such, the set of such graphs (and hence operators) inherits a partial order from the power set of $\mathscr{H}\times\mathscr{H}$ given by inclusion. This means that a smallest closed extension is a minimal element of the set of closed extensions for this partial order i.e. a set $\Gamma(\bar T)$ which is the graph of a closed extension of $T$ and is such that if $T_1$ is a closed extension of $T$ then $\Gamma(T) \subseteq \Gamma(\bar T) \subseteq \Gamma(T_1)$.

It remains to see that whenever $T$ is closable, a smallest closed extension exists. For this, note that a linear subspace $\Gamma \subseteq \mathscr{H} \times \mathscr{H}$ is the graph of an operator if and only if it has the property that whenever $(0, y) \in \Gamma$, one has that $y = 0$.

In particular, if $\Gamma$ is a linear subspace of $\Gamma(T_1)$ for some linear operator $T_1$ then $\Gamma$ is the graph of a linear operator since it is easy to see that the above property is inherited by subspaces.

As a result, one can check that the smallest closed extension of a closable linear operator $T$ is the operator with graph $\overline{\Gamma(T)}$ which is the graph of an operator since $\overline{\Gamma(T)} \subseteq \Gamma(T_1)$ for any closed extension $T_1$ of $T$.