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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Range criterion for self-adjointness

Statement

Assume Countable Choice. Let $T$ be a densely defined symmetric operator on $H$ . Then the following are equivalent:

$T$ is self-adjoint;
$T$ is closed and $\ker (T^{*} - z) = {0}$ for every $z \in C ∖ R$ ;
$T$ is closed and $\ker (T^{*} - i) = \ker (T^{*} + i) = {0}$ ;
$ran (T - z) = H$ for every $z \in C ∖ R$ ;
$ran (T - i) = ran (T + i) = H$ ;
$σ (T) \subseteq R$ .

In particular a closed symmetric operator is self-adjoint if and only if $ran (T \pm i) = H$ , and the same criterion holds with $i$ replaced by $i λ$ for any real $λ > 0$ .

Facts & Assumptions

[A1]

$T \subseteq T^{*}$ , $D (T)$ is dense, and $T$ is self-adjoint exactly when $T = T^{*}$ ; a self-adjoint operator is closed, and for $x$ in the domain of a symmetric operator $S$ the number $⟨ S x, x ⟩$ is real (Symmetric, self-adjoint and essentially self-adjoint operators, The adjoint is well defined, closed, and reverses inclusions, Real and complex inner-product spaces and their induced length).

[A2]

$ran (T - z)^{⊥} = \ker (T^{*} - \overline{z})$ for every $z \in C$ (The adjoint is well defined, closed, and reverses inclusions).

[A3]

For a linear subspace $M$ of a Hilbert space, $M^{⊥ ⊥} = \overline{M}$ ; in particular if $M$ is closed and $M^{⊥} = {0}$ then $M = H$ (Orthogonality and the orthogonal complement, The double orthogonal complement of a subspace is its closure).

[A4]

For self-adjoint $T$ one has $C ∖ R \subseteq ρ (T)$ , hence $σ (T) \subseteq R$ ; conversely every $z \in ρ (T)$ gives that $z - T$ is a bijection $D (T) \to H$ (Resolvent of a self-adjoint operator: nonreal resolvents and the estimate, Resolvent and spectrum of an unbounded operator).

[A5]

If $ρ (T) \neq \emptyset$ then $T$ is closed (Resolvent and spectrum of an unbounded operator).

Proof

technique · direct

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

1.1A1

Preparatory identity. Let $S$ be densely defined and symmetric, let $z = a + i b$ with $b \neq 0$ and let $x \in D (S)$ . Expanding $∥ (S - z) x ∥^{2}$ and using that $⟨ S x, x ⟩$ is real by [A1] gives, as in the proof of Resolvent of a self-adjoint operator: nonreal resolvents and the estimate, $∥ (S - z) x ∥^{2} = ∥ (S - a) x ∥^{2} + b^{2} ∥ x ∥^{2} \geq b^{2} ∥ x ∥^{2} .$

1.2A1

(1) implies (2): if $T = T^{*}$ then $T$ is closed by [A1]. If $z$ is nonreal and $v \in \ker (T^{*} - z)$ , then $v \in D (T^{*}) = D (T)$ and $z ∥ v ∥^{2} = ⟨ T v, v ⟩ = \overline{z} ∥ v ∥^{2}$ by [A1], so $v = 0$ .

1.3given

(4) implies (5) is immediate, since $\pm i$ are nonreal.

1.4A1A2

(5) implies (1): let $v \in D (T^{*})$ . Since $ran (T - i) = H$ , choose $w \in D (T)$ with $(T - i) w = (T^{*} - i) v$ . Then $(T^{*} - i) (v - w) = 0$ , because $T^{*} w = T w$ for $w \in D (T)$ ; and by [A2] with $z = - i$ the kernel of $T^{*} - i$ is $ran (T + i)^{⊥} = {0}$ . Hence $v = w \in D (T)$ . So $D (T^{*}) \subseteq D (T)$ , and with $T \subseteq T^{*}$ this gives $T = T^{*}$ .

1.5A4

(1) implies (6) by [A4]. Conversely (6) implies (4): if $σ (T) \subseteq R$ then every nonreal $z$ lies in $ρ (T)$ , so $z - T$ is surjective and $ran (T - z) = H$ .

2.1A1step 1.1

Closed range for closed symmetric $S$ . If in addition $S$ is closed, then $ran (S - z)$ is closed for nonreal $z$ : given $(S - z) x_{n} \to y$ , step 1.1 makes $(x_{n})$ Cauchy with limit $x$ , so $S x_{n} \to z x + y$ , and closedness of $S$ gives $x \in D (S)$ with $S x = z x + y$ , that is $y = (S - z) x$ .

2.2A2A5step 1.1

(5) implies (3): by 1.1 with $S = T$ and $z = - i$ the map $T + i$ satisfies $∥ (T + i) x ∥ \geq ∥ x ∥$ , so its inverse on its range is bounded by $1$ ; since the range is $H$ by (5), $- i \in ρ (T)$ because $- i - T = - (T + i)$ , and $T$ is closed by [A5]. The two kernels vanish by [A2] and (5).

3.1A2A3step 2.1

(2) implies (4): for nonreal $z$ , [A2] and (2) give $ran (T - z)^{⊥} = \ker (T^{*} - \overline{z}) = {0}$ , while (2) and step 2.1 show $ran (T - z)$ is closed; hence the range is all of $H$ by [A3].

3.2A2A3step 2.1

(3) implies (5): by [A2], $ran (T \mp i)^{⊥} = \ker (T^{*} \pm i) = {0}$ , and by step 2.1 applied to the closed $T$ with $z = \mp i$ the two ranges are closed; hence they equal $H$ by [A3]. So (3) and (5) are equivalent.

4.1step 1.2step 3.1step 1.3step 1.4step 2.2step 3.2step 1.5∎

Collecting: $(1) \Rightarrow (2) \Rightarrow (4) \Rightarrow (5) \Rightarrow (1)$ by steps 1.2, 3.1, 1.3 and 1.4; $(5) \Rightarrow (3)$ and $(3) \Rightarrow (5)$ by steps 2.2 and 3.2; and $(1) \Rightarrow (6) \Rightarrow (4)$ by step 1.5. Thus all six statements are equivalent. The final clause follows because only nonreality of the parameters was used, so $i λ$ with $λ > 0$ may replace $i$ .

Depends on

Used by

Dependency tree · two levels

33 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)
Gerald Teschl, Mathematical Methods in Quantum Mechanics, second edition (standard reference, not scraped)
Theo Buehler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)