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 ''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
$N^* = \sum_i P_i.$

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 ''P''_Ω to each Borel subset of ''σ''(''N''). ''N'' can be expressed as
$N = \int_ \lambda d P(\lambda).$

Differently from the finite dimensional case, it is by no means obvious that ''TN = NT'' implies ''TP''_Ω = ''P''_Ω''T''. Thus, it is not so obvious that ''T'' also commutes with any simple function of the form
$\rho = \sum_i P_.$

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 $P_$, 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''^''k''''T'' = ''TN''^''k'' for all ''k''.
Thus for any λ in $\Complex$,
$e^T = T e^.$

Consider the function
$F(\lambda) = e^ T e^.$
This is equal to
e^ \left ^T e^\righte^ = U(\lambda) T V(\lambda)^,
where $U(\lambda) = e^$ because $M$ is normal, and similarly $V(\lambda) = e^$. However we have
$U(\lambda)^* = e^ = U(\lambda)^$
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

$T' = \begin 0 & 0 \\ T & 0 \end \quad \text \quad N' = \begin N & 0 \\ 0 & M \end.$

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.
$S^ M^* S = N^*.$

Take the adjoint of the above equation and we have
$S^* M (S^)^* = N.$

So
$S^* M (S^)^* = S^ M S \quad \Rightarrow \quad SS^* M (SS^*)^ = M.$

Let ''S*=VR'', with ''V'' a unitary (since ''S'' is invertible) and ''R'' the positive square root of ''SS*''. As ''R'' is a limit of polynomials on ''SS*'', the above implies that ''R'' commutes with ''M''. It is also invertible. Then
$N = S^*M (S^*)^=VRMR^V^*=VMV^*.$

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
$(MN) (MN)^* = MN (NM)^* = MN M^* N^*.$

By Fuglede, the above becomes
$= M M^* N N^* = M^* M N^*N.$

But ''M'' and ''N'' are normal, so
$= M^* N^* MN = (MN)^* MN.$

''C*''-algebras

The theorem can be rephrased as a statement about elements of C*-algebras.

Theorem (Fuglede-Putnam-Rosenblum) 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*''.

References

* Fuglede, Bent
A Commutativity Theorem for Normal Operators — PNAS
* .
*

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Operator theory
Theorems in functional analysis