Invariant subspace problem

Let $H$ be a separable complex Hilbert space and let $T\colon H\to H$ be a bounded linear operator. A closed linear subspace $W\subseteq H$ is invariant under $T$ if $T(W)\subseteq W$, and nontrivial if $W\ne\{0\}$ and $W\ne H$.

Conjecture. If $\dim H\ge2$, then every bounded linear operator $T$ on $H$ has a nontrivial closed invariant subspace.

If $2\le\dim H<\infty$, an eigenvector of $T$ spans a nontrivial invariant subspace, so the content of the problem is the case where $H$ is separable and infinite-dimensional. Nonseparable spaces are excluded because every orbit closure $\overline{\operatorname{span}}\{T^nx:n\ge0\}$ with $x\ne0$ is then a nontrivial invariant subspace. Equivalently, the conjecture asks whether every bounded operator on $\ell^2(\mathbb N)$ has a nonzero vector $x$ that is not cyclic, that is, with

$$\overline{\operatorname{span}}\{x,Tx,T^2x,\dots\}\ne H .$$

The problem is to prove this statement, or to construct a bounded operator on a separable infinite-dimensional Hilbert space with no nontrivial closed invariant subspace. [1, 9, 10]

Background

The problem arose in the 1930s and 1940s from work of von Neumann and others on the structure of operators [9]. In finite dimensions the Jordan form decomposes an operator into invariant pieces. In infinite dimensions there may be no eigenvalues at all, and invariant subspaces are the natural substitute. A positive answer would give every operator a triangular form relative to a chain of invariant subspaces.

The question makes sense on any Banach space. There the answer is negative. Enflo announced a counterexample in the 1970s and published it in 1987 [3]. Read constructed counterexamples independently, including one on $\ell^1$ [4, 5]. The Hilbert space case remains open, as does the case of reflexive Banach spaces. On some Banach spaces every operator has an invariant subspace. Argyros and Haydon built a space on which every operator is a scalar multiple of the identity plus a compact operator, so Lomonosov's theorem applies to all its operators [8].