Fuglede's theorem

In mathematics, Fuglede's theorem is a result in operator theory, named after Bent Fuglede.

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:

<math display="block">TN^* = (NT)^* = (TN)^* = N^*T.</math>

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

<math display="block">N = \sum\nolimits_i \lambda_i P_i </math>

where P_i are pairwise orthogonal projections. One expects 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

<math display="block">N^* = \sum\nolimits_i {\bar \lambda_i} P_i.</math>

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

<math display="block">N = \int_{\sigma(N)} \lambda d P(\lambda). </math>

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

<math display="block">\rho = \sum\nolimits_i {\bar \lambda} P_{\Omega_i}.</math>

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 <math>P_{\Omega_i}</math>, 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.

Putnam's generalization

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) 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.

Then M*T = TN*.

First proof (Marvin Rosenblum):

By induction, the hypothesis implies that M^kT = TN^k for all k.

Thus for any λ in <math>\Complex</math>,

<math display="block">e^{\bar\lambda M}T = T e^{\bar\lambda N}.</math>

Consider the function

<math display="block">F(\lambda) = e^{\lambda M^*} T e^{-\lambda N^*}.</math>

This is equal to

<math display="block">e^{\lambda M^*} \left[e^{-\bar\lambda M}T e^{\bar\lambda N}\right] e^{-\lambda N^*} = U(\lambda) T V(\lambda)^{-1},</math>

where <math>U(\lambda) = e^{\lambda M^* - \bar\lambda M}</math> because <math>M</math> is normal, and similarly <math>V(\lambda) = e^{\lambda N^* - \bar\lambda N}</math>. However we have

<math display="block">U(\lambda)^* = e^{\bar\lambda M - \lambda M^*} = U(\lambda)^{-1}</math>

so U is unitary, and hence has norm 1 for all λ; the same is true for V(λ), so

<math display="block">\|F(\lambda)\| \le \|T\|\ \forall \lambda.</math>

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.