Fuglede\'s theorem - Citizendia

In mathematics, Fuglede's theorem is a result in operator theory, named after Bent Fuglede. Mathematics is the body of Knowledge and Academic discipline that studies such concepts as Quantity, Structure, Space and In Mathematics, operator theory is the branch of Functional analysis which deals with Bounded linear operators and their properties Bent Fuglede (born October 8, 1925) is a Danish Mathematician, and now retired Professor ( Emeritus) from University

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.

Colloquially, the theorem claims that commutativity between operators is transitive under the given assumptions. The claim does not hold in general if N is not normal. A simple counterexample is provided by letting N be the unilateral shift and T = N. In Mathematics, and in particular Functional analysis, the shift operators are examples of Linear operators important for their simplicity and natural occurrence Also, when T is self-adjoint, the claim is trivial regardless of whether N is normal:

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

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 P_i are pairwise orthogonal projections. In Mathematics, particularly Linear algebra and Functional analysis, the spectral theorem is any of a number of results about Linear operators TN = NT if and only if TP_i = P_iT. Therefore T must also commute with

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

In general, 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). In Mathematics, particularly Functional analysis a projection-valued measure is a function defined on certain subsets of a fixed set and whose values are self-adjoint N can be expressed as

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

As in the finite dimensinal case, TN = NT implies TP_Ω = P_ΩT. Thus T commutes with any simple function of the form

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

A limiting argument then shows that T commutes with

$N^* = \int_{\sigma(N)} {\bar \lambda} d P(\lambda). \,$

Putnam's generalization

The following contains Fuglede's result as a special case.

Theorem (Putnam) Let T, M, N be linear operators on a complex Hilbert space, and suppose that M and N are normal and MT = TN. In Mathematics, a linear map (also called a linear transformation, or linear operator) is a function between two Vector spaces that This article assumes some familiarity with Analytic geometry and the concept of a limit. In Mathematics, especially Functional analysis, a normal operator on a Hilbert space H (or more generally in a C* algebra) is a Then M*T = TN*.

First proof (Rosenblum ): By induction, the hypothesis implies that M^kT = TN^k 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

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

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

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

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

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