## 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.