An alternate proof of Fuglede's theorem

Question

To prove Fuglede's Theorem for normal operators on a separable Hilbert space, why does it suffice to show that $E(S_1)T E(S_2)=0$ for all disjoint Borel sets $S_1$ and $S_2$, where $E$ is the spectral measure associated to the given normal operator $N$?

Further, why is this true for $S_1$ and $S_2$ at a positive distance from one another, and how can an approximation by compact sets be used to prove it in the general case?

Answer 1

(several years later, a comment made this answer resurface so I'm updating it so that it keeps up with the comments. I still don't know how to finish the proof through the outline suggested by Douglas)

The question comes from this exercise in Ron Douglas' Banach Algebra Techniques in Operator Theory, Exercise 4.38:

Given an alternate proof of Fuglede's theorem for normal operators on a separable Hilbert space as follows: Show that it is enough to prove that $E(S_1)TE(S_2)=0$ for $S_1$ and $S_2$ disjoint Borel sets; show this first for Borel sets at positive distance from each other and the approximate form within by compact sets in the general case.

The hypothesis is that $TN=NT$. By the Spectral Theorem, $$ N=\int_{\sigma(N)}\,\lambda\,dE(\lambda) $$ for a certain spectral measure $E$, and $$ N^*=\int_{\sigma(N)}\,\overline\lambda\,dE(\lambda). $$ By definition, we have that $N$ is a norm-limit of operators of the form $\sum_j\lambda_jE(S_j)$ (i.e. "simple functions"), where $S_1,\ldots,S_n$ is a partition of $\sigma(N)$, and $$ E(S_j)=\int_{\sigma(N)}\,1_{S_j}(\lambda)\,dE(\lambda). $$ This last integral, in turn, is a wot limit of integrals of polynomials.

Now, since $TN=NT$, $TN^2=N^2T$, and $TN^k=N^kT$ for all $k\in\mathbb N$. So $Tp(N)=p(N)T$ for all polynomials $p$. Then $TE(S_j)=E(S_j)T$ (see the second edit) for Borel sets as above, and so $T$ commutes with all operators of the form $\sum_j\overline\lambda_j\,E(S_j)$. As $N^*$ is a norm limit of such operators, $T$ commutes with $N^*$.

Edit: I'll leave the answer above since it is now mentioned in your question. Regarding the condition you care about, you can use it the following way: Let $S_1,\ldots,S_n$ be a partition of the spectrum of $N$. Then $$ TE(S_j)=ITE(s_j)=\sum_kE(S_k)TE(S_j)=E(S_j)TE(s_j). $$ So $TE(S_j)=E(S_j)TE(S_j)$. Doing the same process on the right side we get $E(S_j)T=E(S_j)TE(S_j)$. So $TE(S_j)=E(S_j)T$. That is, $T$ commutes with the spectral projections of $N$, and thus it commutes with $N^*$.