In mathematics, Fuglede's theorem is a result in operator theory, named after Bent Fuglede.
Theorem (Fuglede) Let T and N be bounded operators on a complex Hilbert space with N being normal.
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 Pi are pairwise orthogonal projections.
One expects that TN = NT if and only if TPi = PiT.
Indeed, it can be proved to be true by elementary arguments (e.g. it can be shown that all Pi are representable as polynomials of N and for this reason, if T commutes with N, it has to commute with Pi...).
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 PΩ to each Borel subset of σ(N).
Differently from the finite dimensional case, it is by no means obvious that TN = NT implies TPΩ = PΩT.
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 N*, giving rise to a vicious circle!
That is the relevance of Fuglede's theorem: The latter hypothesis is not really necessary.
The following contains Fuglede's result as a special case.
The proof by Rosenblum pictured below is just that presented by Fuglede for his theorem when assuming N=M.
Theorem (Calvin Richard Putnam)[1] Let T, M, N be linear operators on a complex Hilbert space, and suppose that M and N are normal, T is bounded and MT = TN.
First proof (Marvin Rosenblum): By induction, the hypothesis implies that MkT = TNk for all k. Thus for any λ in
So F is a bounded analytic vector-valued function, and is thus constant, and equal to F(0) = T. Considering the first-order terms in the expansion for small λ, we must have M*T = TN*.
The original paper of Fuglede appeared in 1950; it was extended to the form given above by Putnam in 1951.
[1] The short proof given above was first published by Rosenblum in 1958; it is very elegant, but is less general than the original proof which also considered the case of unbounded operators.
Comparing entries then gives the desired result.
From Putnam's generalization, one can deduce the following: Corollary If two normal operators M and N are similar, then they are unitarily equivalent.
Proof: Suppose MS = SN where S is a bounded invertible operator.
Putnam's result implies M*S = SN*, i.e.
Let S*=VR, with V a unitary (since S is invertible) and R the positive square root of SS*.
As R is a limit of polynomials on SS*, the above implies that R commutes with M. It is also invertible.
Corollary If M and N are normal operators, and MN = NM, then MN is also normal.
Proof: The argument invokes only Fuglede's theorem.
The theorem can be rephrased as a statement about elements of C*-algebras.
Theorem (Fuglede-Putnam-Rosenblum) Let x, y be two normal elements of a C*-algebra A and z such that xz = zy.