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).

1.1

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).

A1A2A3given

1.2

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 $G \cap ({0} \oplus H) = {(0, 0)}$ .

A1A2A3given

2.1

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 $Γ (T) \subseteq G$ .

A1step 1.2

3.1

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 $\overline{T}$ .

A2step 1.1step 2.1

3.2

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 $T$ .

A1A2step 2.1

4.1

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.

A1step 2.1step 3.2∎

Depends on

Used by

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Dana P. Williams, Lecture Notes on the Spectral Theorem (standard reference, not scraped)
Theo Buehler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)