TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Closability is equivalent to density of the adjoint domain

Statement

Assume Countable Choice. Let $T$ be a densely defined linear operator on $H$ . Then $T$ is closable if and only if $D (T^{*})$ is dense in $H$ . In that case $\overline{T} = T^{* *}$ and $T^{*} = {\overline{T}}^{*}$ ; moreover if $T \subseteq T^{*}$ , then $T$ is closable.

Facts & Assumptions

[A1]

For densely defined $T$ the adjoint is defined by $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ for $x \in D (T)$ , $y \in D (T^{*})$ , and $Γ (T^{*}) = {(y, w) : ⟨ T x, y ⟩ = ⟨ x, w ⟩ for all x \in D (T)}$ . Putting $W (x, y) : = (- y, x)$ defines a bijective isometry of $H \oplus H$ with $W^{2} = - I$ , and $W$ maps orthocomplements to orthocomplements: $W (M^{⊥}) = (W M)^{⊥}$ (Adjoint of a densely defined operator, Hilbert space, Orthogonality and the orthogonal complement).

[A2]

If $M$ is a linear subspace of the Hilbert space $H \oplus H$ , then $M^{⊥ ⊥} = \overline{M}$ (The double orthogonal complement of a subspace is its closure).

[A3]

For a densely defined $T$ the operator $T^{*}$ is closed. If in addition $D (T^{*})$ is dense, then $T^{* *}$ is defined, is closed, and contains $T$ (The adjoint is well defined, closed, and reverses inclusions, Adjoint of a densely defined operator).

[A4]

$T$ is closable when it has a closed extension; if $T$ is closable then $\overline{T}$ is the least closed extension of $T$ and $T \subseteq \overline{T}$ (Closure of a closable operator, Densely defined, closed and closable operators, and cores).

Proof

technique · direct

Given: A densely defined linear operator $T$ on $H$ .

1.1A1

By [A1] we have $Γ (T^{*}) = W Γ (T)^{⊥}$ : indeed $(y, w) \in Γ (T^{*})$ means $⟨ T x, y ⟩ = ⟨ x, w ⟩$ for all $x \in D (T)$ , which is exactly $⟨ (x, T x), W (y, w) ⟩ = 0$ for all $x \in D (T)$ , that is $W (y, w) ⊥ Γ (T)$ .

2.1step 1.1A1A2

Assume in addition that $D (T^{*})$ is dense, so that $T^{* *}$ is defined. Replacing $T$ by $T^{*}$ in step 1.1 gives $Γ (T^{* *}) = W Γ (T^{*})^{⊥}$ , so by step 1.1 and unitarity of $W$ , with $W (M^{⊥}) = (W M)^{⊥}$ and $W^{2} = - I$ , one has $Γ (T^{* *}) = W (W Γ (T)^{⊥})^{⊥} = W W Γ (T)^{⊥ ⊥} = - \overline{Γ (T)} = \overline{Γ (T)}$ .

2.2A1A2A4step 1.1

Conversely assume $T$ is closable, and let $\overline{T}$ be its closure, a closed densely defined operator with $Γ (T) \subseteq Γ (\overline{T})$ and $\overline{Γ (T)} = Γ (\overline{T})$ by [A4]. Then $D (\overline{T}^{*})$ is dense: if $y ⊥ D (\overline{T}^{*})$ then $(y, 0) ⊥ Γ (\overline{T}^{*})$ , so by step 1.1 applied to $\overline{T}$ and closedness of $\overline{T}$ we get $(y, 0) \in Γ (\overline{T}^{*})^{⊥} = W Γ (\overline{T})^{⊥ ⊥} = W Γ (\overline{T})$ , that is $(y, 0) = W (x, \overline{T} x) = (- \overline{T} x, x)$ for some $x \in D (\overline{T})$ , forcing $x = 0$ and $y = 0$ .

3.1step 2.1A3A4

Under the hypothesis of step 2.1 the set $\overline{Γ (T)}$ is the graph of the operator $T^{* *}$ , which is closed by [A3]; hence $T$ has a closed extension and is closable, and its closure is $T^{* *}$ by minimality in [A4].

3.2A1A2step 1.1step 2.2

With $T$ closable as in step 2.2, apply step 1.1 to $T$ and to $\overline{T}$ : using $\overline{Γ (T)} = Γ (\overline{T})$ and [A2] one gets $Γ (T^{*}) = W Γ (T)^{⊥} = W {\overline{Γ (T)}}^{⊥} = W Γ (\overline{T})^{⊥} = Γ (\overline{T}^{*})$ , so $T^{*} = \overline{T}^{*}$ and $D (T^{*}) = D (\overline{T}^{*})$ is dense by step 2.2.

4.1step 3.1step 3.2∎

The two implications are steps 3.1 (density of $D (T^{*})$ gives closability, with closure $T^{* *}$ ) and 3.2 (closability gives density of $D (T^{*})$ , and $T^{*} = \overline{T}^{*}$ ). If $T \subseteq T^{*}$ , then $T^{*}$ is a closed extension of $T$ by [A3], so $T$ is closable.

Closability is equivalent to density of the adjoint domain

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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