CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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An everywhere-defined closed operator on a Banach space is bounded

Statement refuted

Assume Dependent Choice. The inference that a closed linear operator defined on all of a Banach space can nevertheless be unbounded is false. Indeed, if $X, Y$ are Banach spaces and $T : X \to Y$ is linear, defined on all of $X$ , and has closed graph in $X \oplus Y$ , then $T$ is bounded. Consequently an unbounded self-adjoint operator on a Hilbert space $H$ cannot have domain $H$ : its domain is a proper dense subspace.

Facts & Assumptions

[A1]

Under Dependent Choice, an everywhere defined linear map between Banach spaces is bounded if and only if its graph is closed (Closed graph theorem, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[A2]

A self-adjoint operator is densely defined and closed, being equal to the adjoint of a densely defined operator (Symmetric, self-adjoint and essentially self-adjoint operators, The adjoint is well defined, closed, and reverses inclusions, Unbounded linear operators: domain, graph and extension).

[A3]

A Hilbert space is a Banach space, so the closed graph theorem applies to everywhere defined operators on it (Hilbert space, Closed graph theorem).

Counterexample

technique · direct

Given: Banach spaces $X, Y$ and an everywhere defined linear $T$ with closed graph.

1.1

The closed graph theorem gives that $T$ is bounded: an everywhere defined linear map between Banach spaces with closed graph is bounded.

A1given

2.1

Let $H$ be a Hilbert space and let $S$ be a self-adjoint operator on $H$ . A self-adjoint operator is closed, being equal to the adjoint of a densely defined operator; if in addition $D (S) = H$ , then step 1.1 applied to $S : H \to H$ shows that $S$ is bounded.

A2A3step 1.1

3.1

Therefore a self-adjoint operator that is unbounded must have $D (S) \neq H$ , and its domain is dense by the definition of self-adjointness; the claimed impossibility of an unbounded everywhere-defined self-adjoint operator follows.

A2step 2.1

4.1

Both conclusions are steps 1.1 and 2.1, and the hypothesis used is exactly Dependent Choice, through the closed graph theorem. ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 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)