OP-0018DisputedAnalysis

The invariant subspace problem for separable Hilbert spaces

Does every bounded operator on an infinite-dimensional separable Hilbert space have a non-trivial closed invariant subspace?

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Problem statement

Let $H$ be an infinite-dimensional separable complex Hilbert space and $T: H \to H$ a bounded linear operator. Must there exist a closed subspace $M$ with $\{0\} \ne M \ne H$ and $T(M) \subseteq M$ ?

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Background

The Banach-space version is false: Enflo (1987) and Read (1984) constructed operators on certain Banach spaces with no non-trivial invariant subspace. The Hilbert-space case remained open. Per Enflo posted a claimed proof in 2023; as of now the argument has not been accepted as verified by the community, so the entry is recorded as contested rather than closed.

Known partial results

True for compact operators (von Neumann; Aronszajn–Smith), for polynomially compact operators (Bernstein–Robinson), and for normal operators via the spectral theorem.
Lomonosov's theorem gives invariant subspaces for operators commuting with a non-zero compact operator.
The reflexive-Banach-space case and the Hilbert-space case are the remaining headline cases.

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