Given a Hilbert space $\mathcal{H}$.

Consider normal operators: $$N:\mathcal{D}N\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$

Regard their algebra: $$\mathcal{A}(N):=\{\eta(N):\eta\in\mathcal{B}(\sigma(N)):\|\eta\|_\infty<\infty\}$$

They are commutative: $$\eta(N)\vartheta(N)=(\eta\vartheta)(N)=(\vartheta\eta)(N)=\vartheta(N)\eta(N)$$

They are uniformly closed: $$M:=\lim_n\eta_n(N)=(\lim_n\eta_n)(N)=:\eta(N)$$

But also weakly or strongly closed?

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    $\begingroup$ What if you look at $(Mf)(x)=xf(x)$ on $L^{2}[0,1]$ and start with Borel functions $F$ and $F(M)=M_{F}$. Do you still end up with Borel functions $F$? $\endgroup$ – Disintegrating By Parts Jul 17 '15 at 17:49
  • $\begingroup$ @TrialAndError: Hehe right just take a nonmeasurable set and all finite subsets of it. :) $\endgroup$ – C-Star-W-Star Jul 17 '15 at 18:12

In general, they are not weakly closed (not even strongly closed). To see this, consider the Hilbert space $\ell^2([0,1])$ and the operator $N=M_{\rm id}$. Then the (bounded) Borel functional calculus will yield all operators $M_f$ (multiplication with $f$, where $f$ is Borel measurable and bounded.

But now consider the set $A:=\{I\subset [0,1]\,\mid\,I \text{ finite}\}$, ordered by inclusion. This is a directed set. Now let $f:[0,1]\to \Bbb{R}$ be an arbitrary non-measurable (bounded) function. Let $f_I := f \cdot \chi_I$, where $\chi_I$ is the indicator function of $I$.

It is not hard to see (since each "sequence" in $\ell^2([0,1])$ has countable support) that $M_{f_I} x \to M_f x$ for all $x \in \ell^2$, but $M_f \notin A(N)$.

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  • $\begingroup$ Not sure but shouldn't it read $N=M_\mathrm{id}$ instead of $N=\mathrm{id}$? Otherwise it would be simply $\mathcal{A}(N)=\mathbb{C}$? $\endgroup$ – C-Star-W-Star Jul 17 '15 at 18:24
  • $\begingroup$ @Freeze_S: Oh, yes. Good catch! Will edit! Oh, you already did. Thanks! $\endgroup$ – PhoemueX Jul 17 '15 at 20:42

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