Fuglede's Theorem - Putnam's Generalization

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 ,

Consider the function

This is equal to

,

where and . However we have

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

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

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 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.

Take the adjoint of the above equation and we have

So

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

By Fuglede, the above becomes

But M and N are normal, so