CounterexampleConstruction: AI-adaptedVerification: 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.

Continuous calculus does not contain discontinuous spectral projections

Statement refuted

Assume AC. For the self-adjoint operator $T = M_{t}$ of multiplication by the coordinate on $L^{2} (0, 1)$ , every orthogonal projection commuting with $T$ is of the form $f (T)$ for some continuous $f$ on $[0, 1]$ .

Facts & Assumptions

[A1]

For $T = M_{t}$ on $L^{2} (0, 1)$ one has $σ (T) = [0, 1]$ and $f (T) = M_{f}$ for every continuous $f$ on $[0, 1]$ , where $M_{f}$ is multiplication by $f$ (Functional calculus for a multiplication operator).

[A2]

$L^{2} (0, 1)$ is a Hilbert space of a.e. classes with $⟨ f, g ⟩ = \int_{0}^{1} f \overline{g}$ ; multiplication by a bounded function is a bounded operator on it ( $L^{2}$ with the integral pairing is a Hilbert space, The space $L^{p} (μ)$ as the quotient by null functions, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ).

[A3]

For a measurable set $A$ , pointwise multiplication gives $M_{1_{A}}^{2} = M_{1_{A}}$ and $⟨ M_{1_{A}} u, v ⟩ = ⟨ u, M_{1_{A}} v ⟩$ ; its range is the closed subspace of classes supported in $A$ , so it is the Hilbert orthogonal projection onto that subspace (The Hilbert orthogonal projection onto a closed subspace).

[A4]

AC is the hypothesis of the calculus supplier (The Axiom of Choice).

Counterexample

technique · direct

Given: $H = L^{2} (0, 1)$ , $T = M_{t}$ and the multiplication operator $P : = M_{1_{(0, 1 / 2)}}$ where $1_{(0, 1 / 2)}$ is the indicator of the interval $(0, 1 / 2)$ .

1.1

$P$ is an orthogonal projection commuting with $T$ : pointwise multiplication by an indicator is idempotent and self-adjoint, and multiplication operators commute, so $P T = T P$ .

A2A3

2.1

If $P = f (T)$ for a continuous $f \in C ([0, 1])$ , then $M_{f} = P = M_{1_{(0, 1 / 2)}}$ by the multiplication form of the calculus, so the two bounded functions agree as elements of $L^{2} (0, 1)$ , that is $f = 1_{(0, 1 / 2)}$ almost everywhere.

step 1.1A1

3.1

A continuous function agreeing almost everywhere with an indicator of a proper subinterval is impossible: $f = 1$ a.e. on $(0, 1 / 2)$ and $f = 0$ a.e. on $(1 / 2, 1)$ , so continuity at $1 / 2$ would give $f (1 / 2) = \lim_{t \to 1 / 2^{-}} f (t) = 1$ and simultaneously $f (1 / 2) = \lim_{t \to 1 / 2^{+}} f (t) = 0$ ; the two limits are computed along intervals where the continuous function is a.e. constant, hence constant there.

step 2.1A2algebra

4.1

Hence no continuous $f$ satisfies $P = f (T)$ , so the commuting projection $P$ demonstrates that the continuous calculus does not contain every spectral projection of $T$ .

step 1.1step 3.1A4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

54 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

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3, printed pp.235–245 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem, §4, pp.10–15 (standard reference, not scraped)