Proof of the Spectral Theorem for Normal Operators I am reading Invariant Subspaces by H. Radjavi and P. Rosenthal. I need help understanding the proof of the spectral theorem for normal operators, as stated below.

Theorem. If $A$ is a normal operator on a separable space, then there exists a finite measure space $(X,\mu)$ and $h\in L^\infty(X,\mu)$ such that $A$ is unitarily equivalent to the operator
$M_h$ on $L^2(X,\mu)$.

I will reproduce the proof here, and ask questions simultaneously. I understand the proof in the case where $A\in \mathcal B(\mathcal H)$ has a cyclic vector $f$, i.e. $\bigvee_{m,n = 0}^\infty \{A^n(A^*)^m f\} = \mathcal H$, where $\bigvee$ denotes the closed linear span. So, let us move to the general case directly.

Proof. Since for any $f\in \mathcal H$, $\bigvee_{m,n = 0}^\infty \{A^n(A^*)^m f\}$ reduces $A$, Zorn's lemma implies that there exists a collection $\{M_n\}_{n=1}^\infty$ of pairwise-orthogonal subspaces of $\mathcal H$ such that $\mathcal H = \bigoplus_{n=1}^\infty M_n$, each $M_n$ reduces $A$, and $A\vert_{M_n}$ has a cyclic vector for each $n$ (the fact that the collection is countable follows from separability of $\mathcal H$).


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*How exactly is Zorn's lemma being used here, to produce the required collection of subspaces? It would be helpful if someone could provide the details.


By the cyclic case, for each $n$, there is a finite measure $\mu_n$ on $X_n:= \sigma(A\vert_{M_n})$ such that $A\vert_{M_n}$ is unitarily equivalent to $M_z$ on $L^2(X_n,\mu_n)$. We can assume $\mu_n(X_n) \le 2^{-n}$. Let $h_n(z) = z$ for all $z\in X_n$. Then, $A$ is unitarily equivalent to $\bigoplus_{n=1}^\infty M_{h_n}$ on the space $\bigoplus_{n=1}^\infty L^2(X_n,\mu_n)$. We must show how to regard $\bigoplus_{n=1}^\infty L^2(X_n,\mu_n)$ as $L^2(X,\mu)$ in such a way that $\bigoplus_{n=1}^\infty M_{h_n}$ is unitarily equivalent to $M_h$ on $L^2(X,\mu)$ for some $h$. Relabel the elements of the sets $\{X_n\}$ to make the sets pairwise disjoint, and then let $X = \bigcup_{n=1}^\infty X_n$. Define the measurable subsets of $X$ as the subsets which are countable unions of measurable subsets of the $\{X_n\}$, and if $S = \bigcup_{n=1}^\infty S_n$ is a measurable set with $S_n \subset X_n$ for every $n$, define $\mu(S) := \sum_{n=1}^\infty \mu_n(S_n)$. Then, $(X,\mu)$ is a finite measure space. If we define $h$ on $X$ by $h(x) = h_n(x)$ for all $x\in X_n$, then $h\in L^\infty(X,\mu)$ (in fact $\|h\|_\infty = \|A\|$).



*How is $\|h\|_\infty = \|A\|$? I believe this has something to do with $\|T\| = r(T)$ (the spectral radius) for normal operators $T\in \mathcal B(\mathcal H)$.


It is easily verified that $M_h$ on $L^2(X,\mu)$ is unitarily equivalent to $\bigoplus_{n=1}^\infty M_{h_n}$, and hence to $A$.



*Could someone please help me with explicit details of the above verification? I'm confused primarily because $\bigoplus_{n=1}^\infty M_{h_n}$ is not an operator on $L^2(X,\mu)$, but on $\bigoplus_{n=1}^\infty L^2(X_n,\mu_n)$. I understand that we are identifying $L^2(X,\mu)$ and $\bigoplus_{n=1}^\infty L^2(X_n,\mu_n)$ using the recipe above; but I am not clear about how to proceed.

Thank you very much! Please let me know if any of the notations is not clear.
 A: *

*Consider the set $\{\oplus_{i\in I} M_i | M_i \text{ is cyclic and closed} \}$. Define the partial order $\oplus_{i\in I} M_i\le\oplus_{j\in J} M_j$ if for each $i\in I$, there is $j_i$ such that $M_i=M_{j_i}$. By Zorn's lemma, there is a maximal element $N=\oplus_{i\in I} M_i$. We can show that $N=\mathcal H$, because if not then pick $f\in N^{\perp}$ and $N\oplus \bigvee_{m,n = 0}^\infty \{A^n(A^*)^m f\}$ extends $N$. (Well, if we define $\oplus_{i\in I} M_i\le \oplus_{j\in J} M_j$ as the notation orignally means, i.e. the former is a linear subspace of the latter, the whole argument works as well.)


*Normality hence spectrum is not important for this step. Just use the fact that given $A: \oplus_{i\in I} M_i\rightarrow \oplus_{i\in I} M_i$, then $\|A\|=\sup_i \|A|_{M_i}\|$.


*This is a subtle point: In the theory of Hilbert spaces, we define $\oplus_{i\in I} M_i$ as the completion of the algebraic direct sum. More explicitly, $$\oplus_{i\in I} M_i =\{(f_i)|f_i\in M_i, \sum_{i\in I}\|f_i\|^2<\infty\}$$ With this in mind, it's easy to define $\oplus M_{h_n}$ as an operator on $L^2(X, \mu)$.
