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

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The adjoint is well defined, closed, and reverses inclusions

Statement

Assume Countable Choice. For a densely defined linear operator $T$ on $H$ the adjoint $T^{*}$ is well defined and linear with linear domain $D (T^{*})$ ; $T^{*}$ is closed; for every $z \in C$ $ran (T - z)^{⊥} = \ker (T^{*} - \overline{z}),$ and if $S \subseteq T$ and both are densely defined, then $T^{*} \subseteq S^{*}$ .

Facts & Assumptions

[A1]

$y \in D (T^{*})$ holds exactly when $x \mapsto ⟨ T x, y ⟩$ is bounded on $D (T)$ , and then $T^{*} y$ is the unique $w$ with $⟨ T x, y ⟩ = ⟨ x, w ⟩$ for all $x \in D (T)$ ; $T^{*}$ is linear and $D (T^{*})$ is a linear subspace (Adjoint of a densely defined operator, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[A3]

$T$ is closed exactly when $Γ (T)$ is closed, and $Γ (T^{*}) = {(y, w) : ⟨ T x, y ⟩ = ⟨ x, w ⟩ for all x \in D (T)}$ (Unbounded linear operators: domain, graph and extension, Adjoint of a densely defined operator).

Proof

technique · direct

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

1.1A1

By [A1] the adjoint is well defined, $D (T^{*})$ is a linear subspace and $T^{*}$ is linear.

1.2A1

Let $y_{n} \in D (T^{*})$ with $y_{n} \to y$ and $T^{*} y_{n} \to w$ . For every $x \in D (T)$ we have $⟨ T x, y ⟩ = \lim_{n} ⟨ T x, y_{n} ⟩ = \lim_{n} ⟨ x, T^{*} y_{n} ⟩ = ⟨ x, w ⟩$ , both limits being scalar limits; hence $x \mapsto ⟨ T x, y ⟩$ is bounded, with $∣ ⟨ T x, y ⟩ ∣ = ∣ ⟨ x, w ⟩ ∣ \leq ∥ x ∥ ∥ w ∥$ . So $y \in D (T^{*})$ and $T^{*} y = w$ by [A1], and therefore $T^{*}$ is closed.

1.3A1

Let $y \in H$ and $z \in C$ . If $y ⊥ ran (T - z)$ , then $⟨ (T - z) x, y ⟩ = 0$ , that is $⟨ T x, y ⟩ = z ⟨ x, y ⟩ = ⟨ x, \overline{z} y ⟩$ , for every $x \in D (T)$ ; this exhibits $x \mapsto ⟨ T x, y ⟩$ as a bounded functional, so $y \in D (T^{*})$ and $T^{*} y = \overline{z} y$ by uniqueness in [A1]. Conversely if $T^{*} y = \overline{z} y$ , then $⟨ (T - z) x, y ⟩ = ⟨ x, \overline{z} y ⟩ - z ⟨ x, y ⟩ = 0$ for every $x \in D (T)$ , so $y ⊥ ran (T - z)$ . Hence $ran (T - z)^{⊥} = \ker (T^{*} - \overline{z})$ .

1.4A3

If $S \subseteq T$ are densely defined, then $⟨ S x, y ⟩ = ⟨ T x, y ⟩ = ⟨ x, w ⟩$ for every $x \in D (S)$ , so every pair $(y, w) \in Γ (T^{*})$ lies in $Γ (S^{*})$ by [A3]; hence $T^{*} \subseteq S^{*}$ .

2.1

Claims collected: well-definedness and linearity from step 1.1, closedness from step 1.2, the kernel-range identity from step 1.3, and the inclusion reversal from step 1.4. ∎

The adjoint is well defined, closed, and reverses inclusions

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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