The Functional Calculus for Normal Operators: The Conclusion of the Saga

What is a scalar-valued spectral measure for a diagonalizable normal operator?

Let $N_1$ and $N_2$ be normal operators with scalar-valued spectral measures $\mu_1$ and $\mu_2$. What is a scalar spectral measure for $N_1 \oplus N_2$?

Let $\{ e_n\}$ be an orthonormal basis for $\mathcal{H}$ and put $\mu (\Delta ) = \sum_{n=1}^\infty 2^{-n} \left\| E(\Delta ) e_n\right\|^2$. Show that $\mu$ is a scalar-valued spectral measure for $N$.

Give an example of a normal operator on a nonseparable space which has no scalar-valued spectral measure.

Prove that the map $\rho$ in the following theorem (Theorem 8.10: if $N$ is a normal operator on the separable Hilbert space $\mathcal{H}$ and $\mu$ is a scalar-valued spectral measure for $N$, then there is a well-defined map $\rho : L^\infty (\mu ) \to W^*(N)$ given by the formula $\rho (\phi ) = \phi (N)$ such that (a) $\rho$ is a $*$-isomorphism and an isometry; (b) $\rho : (L^\infty (\mu ), \text{weak*}) \to (W^*(N), \text{WOT})$ is a homeomorphism) is an isometry without using (VIII.4.8).

Prove the following proposition (Proposition 8.12). Let $N$ be a normal operator on a separable space with scalar-valued spectral measure $\mu$, and let $\phi \in L^\infty (\mu )$ (so by the Spectral Mapping Theorem, $\sigma (\phi (N))$ is the $\mu$-essential range of $\phi$). If $N = \int z \, dE$, then $\mu \circ \phi^{-1}$ is a scalar-valued spectral measure for $\phi (N)$ and $E \circ \phi^{-1}$ is its spectral measure.

Show that if $\mu$ and $\nu$ are compactly supported measures on $\mathbb {C}$, the following statements are equivalent:

1. $N_\mu \oplus N_\nu$ is $*$-cyclic;

2. $W^*(N_\mu \oplus N_\nu ) = W^*(N_\mu ) \oplus W^*(N_\nu )$;

3. $\mu \perp \nu$.

If $M$ and $N$ are normal operators with scalar-valued spectral measures $\mu$ and $\nu$, respectively, show that the following are equivalent:

1. $W^*(M \oplus N) = W^*(M) \oplus W^*(N)$;

2. $\{ M \oplus N\} ' = \{ M\} ' \oplus \{ N\} '$;

3. there is no operator $A$ such that $MA = AN$ other than $A = 0$;

4. $\mu \perp \nu$.

If $M$ and $N$ are normal operators, show that $C^*(M \oplus N) = C^*(M) \oplus C^*(N)$ if and only if $\sigma (M) \cap \sigma (N) = \varnothing$.

Give an example of two normal operators $M$ and $N$ such that $W^*(M \oplus N) = W^*(M) \oplus W^*(N)$ but $C^*(M \oplus N) \neq C^*(M) \oplus C^*(N)$. In fact, find $M$ and $N$ such that $W^*(M \oplus N)$ splits, but $\sigma (M) = \sigma (N)$.

If $U$ is the bilateral shift and $V$ is any unitary operator, show that $W^*(U \oplus V) = W^*(U) \oplus W^*(V)$ if and only if $V$ has a spectral measure that is singular to arc length on $\partial \mathbb {D}$. (See Exercise 3.7.)

If $\mathcal{A}$ is an abelian von Neumann algebra on a separable space, show that there is a compactly supported measure $\mu$ on $\mathbb {R}$ such that $\mathcal{A}$ is $*$-isomorphic to $L^\infty (\mu )$. (Hint: Use Exercise 7.7.)

(This exercise assumes a knowledge of Exercise 2.21.) Let $N = \int z \, dE(z)$ be a normal operator with scalar-valued spectral measure $\mu$ and define $\alpha : \mathcal{B}_1(\mathcal{H}) \to L^1(\mu )$ by $\alpha (T)(\Delta ) = \operatorname {tr}(TE(\Delta ))$. Show that $\alpha$ is a surjective contraction. What is $\alpha^*$? [$L^1(\mu )$ is identified, via the Radon--Nikodym Theorem, with the set of complex-valued measures that are absolutely continuous with respect to $\mu$.]

Let $\pi : \mathcal{B}(\mathcal{H}) \to \mathcal{B}(\mathcal{H})/\mathcal{B}_0(\mathcal{H})$ be the natural map and let $A \in \mathcal{B}(\mathcal{H})$.

(L.G. Brown) If $\pi : \mathcal{B}(\mathcal{H}) \to \mathcal{B}(\mathcal{H})/\mathcal{B}_0(\mathcal{H})$ is the natural map and $A$ and $B$ are hermitian operators such that $\pi (A) \leq 0 \leq \pi (B)$, then there is a hermitian compact operator $K$ such that $A \leq K \leq B$.