Fuglede's Theorem - The Result

The Result

Theorem (Fuglede) Let T and N be bounded operators on a complex Hilbert space with N being normal. If TN = NT, then TN* = N*T, where N* denotes the adjoint of N.

Normality of N is necessary, as is seen by taking T=N. When T is self-adjoint, the claim is trivial regardless of whether N is normal:

Tentative Proof: If the underlying Hilbert space is finite-dimensional, the spectral theorem says that N is of the form

where P_i are pairwise orthogonal projections. One aspects that TN = NT if and only if TP_i = P_iT. Indeed it can be proved to be true by elementary arguments (e.g. it can be shown that all P_i are representable as polynomials of N and for this reason, if T commutes with N, it has to commute with P_i...). Therefore T must also commute with

In general, when the Hilbert space is not finite-dimensional, the normal operator N gives rise to a projection-valued measure P on its spectrum, σ(N), which assigns a projection P_Ω to each Borel subset of σ(N). N can be expressed as

Differently from the finite dimensional case, it is by no means obvious that TN = NT implies TP_Ω = P_ΩT. Thus, it is not so obvious that T also commutes with any simple function of the form

Indeed, following the construction of the spectral decomposition for a bounded, normal, not self-adjoint, operator T, one sees that to verify that T commutes with, the most straightforward way is to assume that T commutes with both N and, giving rise to a vicious circle!

That is the relevance of Fuglede's theorem: The latter hypothesis is not really necessary.