Invariant subspace problem

A rule that sends a pair $(a,b)$ to $(2a,3b)$ keeps every pair $(a,0)$ on the horizontal axis. That axis is a smaller space preserved by the rule. Must a preserved smaller space always exist in the infinite-dimensional setting of this problem?

The vectors here have complex coordinates and a complete notion of length, with a countable set of approximations dense in the space. This is a separable complex Hilbert space. The rule respects addition and scaling of vectors, and has a fixed bound on how much it can amplify length. The sought subspace must be nonzero, smaller than the whole space, and closed: limits of its convergent vectors stay in it. The two-coordinate example illustrates preservation; it does not settle this infinite-dimensional question.[1] [3]

A rule that transforms vectors may keep a smaller collection of vectors stable under repeated application. Must every bounded linear rule on a separable, infinite-dimensional complex Hilbert space preserve a proper nonzero closed subspace? Linearity means that applying the rule to a sum or a scalar multiple gives the corresponding sum or multiple of its outputs. “Bounded” means a single finite constant limits the amplification of every vector's length. “Separable” means a countable set of approximations is dense. A Hilbert space is complete in the norm from its inner product: a sequence whose vectors eventually lie arbitrarily close to one another has a limit in the space.

One model is $\ell^2$: infinite complex coordinate lists for which the sum of squared coordinate magnitudes is finite. The completeness condition ensures that limits measured in this norm are still allowed vectors. A vector subspace is closed under addition and scalar multiplication. A closed subspace also contains limits of its own convergent sequences.

For contrast, the finite rule $T(a,b)=(2a,3b)$ preserves the line of vectors $(a,0)$. The line through $(1,1)$ is not preserved, since $T(1,1)=(2,3)$ is not a multiple of $(1,1)$. In the actual infinite-dimensional target, $H$ denotes the specified Hilbert space and $T$ any bounded linear operator on it. One seeks a closed subspace $M$ such that $$M\ne\{0\},\quad M\ne H,$$ $$T(M)\subseteq M.$$ This asks for a stable subspace, not necessarily an eigenvector or a decomposition into two mutually orthogonal invariant pieces. Counterexamples in other kinds of vector spaces do not settle this specified complex Hilbert-space problem.[1] [3]

The infinite-dimensional complex Hilbert-space case is unresolved.[1]

The 2026 withdrawal supplies no solution. The later primary introduction still states the general problem as open.[1] [2]

If a transformation keeps a smaller space invariant, starting in that space keeps every repeated transformation inside it. This gives a stable part on which to study the rule separately. The problem asks whether such a part is unavoidable, even for an arbitrary bounded transformation of the specified infinite-dimensional space.[1] [3]

A closed invariant subspace supplies a smaller Hilbert space on which the same operator acts, retaining limit operations and repeated transformations. That makes invariant subspaces a basic tool for understanding operator structure. A universal existence theorem would show that no bounded operator in this setting is wholly without such a proper stable part; a counterexample would reveal how infinite-dimensional action can evade that kind of reduction.[1] [3]