Theorem 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 takingT=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 Pi are pairwiseorthogonal projections. One expects 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 In general, when the Hilbert space is not finite-dimensional, the normal operatorN gives rise to a projection-valued measureP on its spectrum, σ, which assigns a projection PΩ to each Borel subset of σ. N can be expressed as 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.
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 Let T, M, N be linear operators on a complex Hilbert space, and suppose that M and N are normal, M is bounded and MT = TN. Then M*T = TN*. First proof : By induction, the hypothesis implies that MkT = TNk for all k. Thus for any λ in, Consider the function This is equal to where because is normal, and similarly. However we have so U is unitary, and hence has norm 1 for all λ; the same is true for V, so So F is a bounded analytic vector-valued function, and is thus constant, and equal to F = 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 The operator N' is normal and, by assumption, T' N' = N' T' . By Fuglede's theorem, one has Comparing entries then gives the desired result. From Putnam's generalization, one can deduce the following: Corollary If two normal operatorsM 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. Take the adjoint of the above equation and we have So Let S*=VR, with V a unitary and R the positive square root ofSS*. As R is a limit of polynomials on SS*, the above implies that R commutes with M. It is also invertible. Then Corollary If M and N are normal operators, and MN = NM, then MN is also normal. Proof: The argument invokes only Fuglede's theorem. One can directly compute By Fuglede, the above becomes But M and N are normal, so
''C*''-algebras
The theorem can be rephrased as a statement about elements of C*-algebras. Theorem Let x, y be two normal elements of a C*-algebra A and z such that xz = zy. Then it follows that x* z = z y*.
