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

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Closure of a closable operator

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $T$ be a linear operator on $H$ with domain $D (T)$ and graph $Γ (T)$ (Unbounded linear operators: domain, graph and extension). Then the following are equivalent:

$T$ is closable (Densely defined, closed and closable operators, and cores);
whenever $x_{n} \in D (T)$ , $x_{n} \to 0$ and $T x_{n} \to y$ , one has $y = 0$ .

If either condition holds, then the closure of $Γ (T)$ in $H \oplus H$ is the graph of a linear operator $\overline{T}$ , the closure of $T$ ; it is the least closed extension of $T$ , and $T$ is closed if and only if $\overline{T} = T$ . Necessity of the hypothesis is never claimed: both conditions hold automatically for a closed operator.

Facts & Assumptions

[A1]

$Γ (T)$ is a linear subspace of $H \oplus H$ ; $T \subseteq S$ exactly when $Γ (T) \subseteq Γ (S)$ ; $T$ is closed exactly when $Γ (T)$ is closed; and $Γ (T) \cap ({0} \oplus H) = {(0, 0)}$ (Unbounded linear operators: domain, graph and extension).

[A2]

$T$ is closable when it has a closed extension; the closure $\overline{A}$ of a subset $A$ of a metric space is closed, is contained in every closed set containing $A$ , and every point of $\overline{A}$ is the limit of a sequence in $A$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, Densely defined, closed and closable operators, and cores).

[A3]

A sequence in $H \oplus H$ converges exactly when its two coordinate sequences converge in $H$ (Unbounded linear operators: domain, graph and extension).

Proof

technique · direct

Given: A linear operator $T$ on $H$ and the two conditions (1) and (2).

1.1A1A2A3given

If $T$ has a closed extension $S$ , then $Γ (T) \subseteq Γ (S)$ with $Γ (S)$ closed, and for $x_{n} \in D (T)$ with $x_{n} \to 0$ , $T x_{n} \to y$ we get $(x_{n}, T x_{n}) \to (0, y)$ by [A3] with $(x_{n}, T x_{n}) \in Γ (T) \subseteq Γ (S)$ ; closedness of $Γ (S)$ gives $(0, y) \in Γ (S)$ , and by the last clause of [A1] applied to $S$ this forces $y = 0$ . Thus (1) implies (2).

1.2A1A2A3given

Now assume (2), and let $G : = \overline{Γ (T)} \subseteq H \oplus H$ . Then $G$ is closed and, being the closure of the linear subspace $Γ (T)$ , is itself a linear subspace. If $(0, y) \in G$ then by [A2] there are $(x_{n}, T x_{n}) \in Γ (T)$ with $(x_{n}, T x_{n}) \to (0, y)$ , hence $x_{n} \to 0$ and $T x_{n} \to y$ by [A3], so $y = 0$ by (2). Therefore $G \cap ({0} \oplus H) = {(0, 0)}$ .

2.1A1step 1.2

By step 1.2, $G$ determines at most one second coordinate per first coordinate: if $(x, y), (x, y^{'}) \in G$ then $(0, y - y^{'}) = (x, y) - (x, y^{'}) \in G$ since $G$ is a subspace, so $y = y^{'}$ . Hence $D : = {x \in H : there is y with (x, y) \in G}$ is a linear subspace of $H$ , the formula $\overline{T} x : = y$ for $(x, y) \in G$ defines a linear operator $\overline{T} : D \to H$ with $Γ (\overline{T}) = G$ , and $\overline{T}$ is closed with $T \subseteq \overline{T}$ because $Γ (T) \subseteq G$ .

3.1A2step 1.1step 2.1

By step 2.1 the operator $\overline{T}$ is a closed extension of $T$ , so $T$ is closable, and (2) implies (1). With 1.1 this proves the equivalence of (1) and (2), and it shows that whenever either holds the closure $\overline{Γ (T)}$ is the graph of the closed extension $\overline{T}$ .

3.2A1A2step 2.1

If $R$ is any closed extension of $T$ , then $Γ (R)$ is closed and contains $Γ (T)$ , so $G = \overline{Γ (T)} \subseteq Γ (R)$ by [A2], that is, $\overline{T} \subseteq R$ . Thus $\overline{T}$ is the least closed extension of $T$ .

4.1A1step 2.1step 3.2∎

If $T$ is closed then $Γ (T)$ is already closed, so $G = Γ (T)$ and $\overline{T} = T$ ; conversely if $\overline{T} = T$ then $T$ is closed because $\overline{T}$ is closed by step 2.1. Hence $T$ is closed if and only if $\overline{T} = T$ , and the closure of the graph is the graph of the least closed extension.

Closure of a closable operator

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources