Fuglede’s Theorem in operator theory

The following theorem of Fuglede is a classical result in the theory of normal operators. In the proof, one can appreciate how classical analysis are used in solving problems in operator theory through functional calculus.

 

Let \mathcal{H} be a complex Hilbert space and \mathcal{B(H)} be the C*-algebra of all bounded operators on \mathcal{H}.

Theorem (Fuglede). Let N, X \in \mathcal{B(H)} be such that N is normal and NX = XN. Then we have N*X = XN*.

Proof. Firstly we have X = \exp(-\overline{z}N)X\exp(\overline{z}N) for any complex number z, by approximating the exponentials with polynomials and using the fact that X commutes with every polynomial of N. Now define

f(z) = \exp(zN^*)X\exp(-zN^*)

and we obtain a (\mathcal{B(H)}-valued) entire function. Since N and N* commutes, we have

f(z) = \exp(zN^*)\exp(-\overline{z}N)X\exp(\overline{z}N)\exp(-zN^*) = \exp(zN^* - \overline{z}N)X\exp(\overline{z}N - zN^*).

Note that zN^* - \overline{z}N is anti-hermitian (an operator A is said to be anti-hermitian if A* = -A), so zN^* - \overline{z}N = iB for some self-adjoint B \in \mathcal{B(H)}. It follows that f(z) = \exp(iB)X\exp(-iB) and thus f is bounded. By Liouville’s Theorem, f is a constant function. Therefore for any z,

0 = f'(z) = N^*\exp(zN^*)X\exp(-zN^*) - \exp(zN^*)X\exp(-zN^*)N^*.

Put z = 0, we get the result.

The following is a generalization of Fuglede’s Theorem. A way of proof is to imitate the above proof by considering a different bounded entire function. But we can directly apply the Fuglede’s Theorem and use a “2 x 2 matrix trick”.

Corollary (Putnam). Let M, N, X \in \mathcal{B(H)} be such that M and N are normal and MX = XN. Then we have M*X = XN*.

Proof. Observe that \begin{pmatrix}M & 0 \\ 0 & N \end{pmatrix} is normal and

\begin{pmatrix}M & 0 \\ 0 & N \end{pmatrix} \begin{pmatrix} 0 & X \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & MX \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & XN \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & X \\ 0 & 0 \end{pmatrix} \begin{pmatrix} M & 0 \\ 0 & N \end{pmatrix},

i.e. \begin{pmatrix}M & 0 \\ 0 & N\end{pmatrix} and \begin{pmatrix}0 & X \\ 0 & 0 \end{pmatrix} commute. By Fuglede’s Theorem,  \begin{pmatrix} M^* & 0 \\ 0 & N^* \end{pmatrix} and \begin{pmatrix} 0 & X \\ 0 & 0 \end{pmatrix} commute. On the one hand,

\begin{pmatrix}M^* & 0 \\ 0 & N^* \end{pmatrix} \begin{pmatrix} 0 & X \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & M^*X \\ 0 & 0 \end{pmatrix}.

On the other hand,

\begin{pmatrix} 0 & X \\ 0 & 0 \end{pmatrix} \begin{pmatrix} M^* & 0 \\ 0 & N^* \end{pmatrix} = \begin{pmatrix} 0 & XN^* \\ 0 & 0 \end{pmatrix}.

We are done.

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About Ken Leung

I'm a ordinary boy who likes Science and Mathematics.
This entry was posted in Analysis, Functional analysis, Operator Theory. Bookmark the permalink.

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