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:

TN^* = (NT)^* = (TN)^* = N^*T.\,

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

N = \sum_i \lambda_i P_i \,

where Pi are pairwise orthogonal projections. One aspects that TN = NT if and only if TPi = PiT. Indeed it can be proved to be true by elementary arguments. Therefore T must also commute with

N^* = \sum_i {\bar \lambda_i} P_i.

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). N can be expressed as

N = \int_{\sigma(N)} \lambda d P(\lambda). \,

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

\rho = \sum_i {\bar \lambda} P_{\Omega_i}.

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 P_{\Omega_i}, 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 and MT = TN. Then M*T = TN*.

First proof (Marvin Rosenblum): By induction, the hypothesis implies that MkT = TNk for all k. Thus for any λ in \mathbb{C},

e^{\bar\lambda M}T = T e^{\bar\lambda N}.

Consider the function

F(\lambda) = e^{\lambda M^*} T e^{-\lambda N^*}.

This is equal to

e^{\lambda M^*} \left[e^{-\bar\lambda M}T e^{\bar\lambda N}\right] e^{-\lambda N^*} = U(\lambda) T V(\lambda)^{-1},

where U(\lambda) = e^{\lambda M^* - \bar\lambda M} and V(\lambda) = e^{\lambda N^* - \bar\lambda N}. However we have

U(\lambda)^* = e^{\bar\lambda M - \lambda M^*} = U(\lambda)^{-1}

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

\|F(\lambda)\| \le \|T\|\ \forall \lambda.

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. 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. Another simple proof of Putnam's theorem is as follows:

Second proof: Consider the matrices

T' = \begin{bmatrix} 0 & 0\\ T & 0 \end{bmatrix} \quad \mbox{and} \quad N' = \begin{bmatrix} N & 0 \\ 0 & M \end{bmatrix}.

The operator N' is normal and, by assumption, T' N' = N' T' . By Fuglede's theorem, one has

T' (N')^* = (N')^*T'. \,

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.

S^{-1} M^* S = N^*. \,

Take the adjoint of the above equation and we have

S^* M (S^{-1})^* = N. \,

So

S^* M (S^{-1})^* = S^{-1} M S \quad \Rightarrow \quad SS^* M (SS^*)^{-1} = M.

Therefore, on Ran(M), SS* is the identity operator. SS* can be extended to Ran(M)⊥ = Ker(M). Therefore, by normality of M, SS* = I, the identity operator. Similarly, S*S = I. This shows that S is unitary.

Corollary If M and N are normal operators, and MN = NM, then MN is also normal.

Proof: The argument invokes only Fuglede's theoerm. One can directly compute

(MN) (MN)^* = MN (NM)^* = MN M^* N^*. \,

By Fuglede, the above becomes

= M M^* N N^* = M^* M N^*N. \,

But M and N are normal, so

= M^* N^* MN = (MN)^* MN. \,

References