CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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A symmetric operator need not have a self-adjoint extension

Statement refuted

Let $H$ be a complex Hilbert space and let $T : D (T) \to H$ be a densely defined symmetric linear operator. The false assertion is that there is a self-adjoint operator $A : D (A) \to H$ on the same Hilbert space with $D (T) \subseteq D (A)$ and $A f = T f$ for every $f \in D (T)$ . By Stone's theorem, such an $A$ would generate a strongly continuous one-parameter unitary group $U (t) = e^{i t A}$ on $H$ .

Statement references

Symmetric, self-adjoint and essentially self-adjoint operators

Stone's theorem: unitary groups and self-adjoint generators

Facts & Assumptions

Given: The half-line operator and Hilbert space of A symmetric half-line momentum with no self-adjoint extension.

[A1]

The Axiom of Choice, Countable Choice, and Dependent Choice are the hypotheses of the half-line and integration-by-parts suppliers (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[F1]

The closure of $- i D$ initially on $C_{c}^{\infty} (0, \infty)$ has deficiency indices $(1, 0)$ and no self-adjoint extension on $L^{2} (0, \infty; C)$ (A symmetric half-line momentum with no self-adjoint extension).

[F2]

For compactly supported smooth $f, g$ on $(0, \infty)$ , clause (d) of Integration by parts and the endpoint-vanishing space on intervals (or integration by parts on a bounded interval containing their supports) gives $⟨ - i D f, g ⟩ = ⟨ f, - i D g ⟩$ .

[F3]

Stone's theorem says a strongly continuous one-parameter unitary group has a self-adjoint generator, and every self-adjoint operator generates such a group (Strongly continuous one-parameter unitary group, Stone's theorem: unitary groups and self-adjoint generators).

[F4]

A self-adjoint operator is closed, and every closed extension of a closable operator contains its least closed extension (Symmetric, self-adjoint and essentially self-adjoint operators).

Counterexample

Given: Let $S_{c} = - i D$ on $C_{c}^{\infty} (0, \infty)$ , with values in $L^{2} (0, \infty; C)$ .

1.1A1F2

By [F2], for all test functions $f, g$ , $⟨ S_{c} f, g ⟩ = ⟨ f, S_{c} g ⟩$ ; thus the test-domain operator is symmetric.

1.2A1F1

The same operator's closure has unequal deficiency indices $(1, 0)$ , so by [F1] it has no self-adjoint extension on the same Hilbert space. Thus symmetry on the core does not guarantee a self-adjoint realization.

2.1A1F1F3F4step 1.2∎

If a strongly continuous unitary one-parameter group had a self-adjoint generator $T$ extending $S_{c}$ , Stone's theorem [F3] supplies that self-adjointness. By [F4], $T$ is closed and contains the least closed extension ${\overline{S}}_{c}$ : explicitly, every point of the graph of ${\overline{S}}_{c}$ is a limit of graph points of $S_{c}$ , and the closed graph of $T$ contains each such limit. This contradicts [F1], which says ${\overline{S}}_{c}$ has no self-adjoint extension. Thus the proposed general claim fails for this operator while its integration-by-parts symmetry hypothesis holds.

A symmetric operator need not have a self-adjoint extension

Statement refuted

Statement references

Facts & Assumptions

Counterexample

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