Unbounded Normal Operators and the Spectral Theorem

Let $(X, \Omega )$ be a measurable space, $\mathcal{H}$ a Hilbert space, and $E$ a spectral measure for $(X, \Omega , \mathcal{H})$. Let $\Phi (X, \Omega )$ be the algebra of all $\Omega$-measurable functions $\phi : X \to \mathbb {C}$ and define $\rho : \Phi (X, \Omega ) \to \mathcal{C}(\mathcal{H})$ by $\rho (\phi ) = \int \phi \, dE$. Prove that for $\phi , \psi$ in $\Phi (X, \Omega )$:

Show that if $A$ is a symmetric operator that is normal, then $A$ is self-adjoint.

Let $E$ be a spectral measure for $(X, \Omega , \mathcal{H})$, let $\phi : X \to \mathbb {C}$ be an $\Omega$-measurable function, and let

where $E_{h,k}(\Delta ) = \left\langle E(\Delta )h, k \right\rangle$. (Theorem 4.7 shows that for $h$ in $\mathcal{D}_\phi$ and $f$ in $\mathcal{H}$, $\left\langle \left(\int \phi \, dE\right) h, f \right\rangle = \int \phi \, dE_{h,f}$.)

Show that for $h$ in $\mathcal{D}_\phi$, $\left\| \left(\int \phi \, dE\right) h\right\|^2 = \int \left|\phi \right|^2 \, dE_{h,h}$.

Using the notation of Theorem 4.10 (see X.4.1 above: $\rho (\phi ) = \int \phi \, dE$ for $E$ a spectral measure and $\phi$ an $\Omega$-measurable function), what is $\sigma \left(\int \phi \, dE\right)$?

Recall the construction from the proof of the Spectral Theorem (4.11): given a normal operator $N$, one lets $B = (1 + N^*N)^{-1} = \int_0^1 t \, dP(t)$ be the spectral decomposition of $B$, puts $P_n = P\left(\left[1/(n+1), 1/n\right]\right)$ and $\mathcal{H}_n = P_n \mathcal{H}$ for $n \geq 1$, so that $\mathcal{H}_n$ reduces $N$ and $N_n \equiv N \rvert \mathcal{H}_n$ is a bounded normal operator with $\sigma (N_n) \subseteq \Delta_n \equiv \left\{ z \in \mathbb {C} : (n-1)^{1/2} \leq \left|z\right| \leq n^{1/2}\right\}$; letting $N_n = \int z \, dE_n(z)$ be the spectral decomposition of $N_n$, the spectral measure of $N$ is then defined by $E(\Delta ) = \sum_{n=1}^{\infty } E_n(\Delta \cap \Delta_n)$.

If $\Delta_n$ and $E_n$ are as in this construction, show that $E_n(\Delta_{n+1}) = E_n(\Delta_{n-1}) = 0$.

Use the Spectral Theorem to show that if $0 < a \leq b < \infty$, $\Delta = \left\{ z \in \mathbb {C} : a \leq \left|z\right| \leq b\right\}$, and $N = \int z \, dE(z)$ is the spectral decomposition of the normal operator $N$, then

State and prove a polar decomposition for operators in $\mathcal{C}(\mathcal{H}, \mathcal{K})$.

If $A$ is self-adjoint, prove that $\exp (iA)$ is unitary.

If $N$, $M$ are normal operators and $A$ is a bounded operator such that $AN \subseteq MA$, then $AN^* \subseteq M^*A$.

Recall that if $N$ is a normal operator on $\mathcal{H}$, a vector $e_0$ is a star-cyclic vector for $N$ if for all non-negative integers $k$ and $l$, $e_0 \in \operatorname {dom}(N^{*k}N^l)$ and $\mathcal{H} = \bigvee \left\{ N^{*k}N^l e_0 : k, l \geq 0\right\}$.

Prove: if $N$ is a normal operator on $\mathcal{H}$ with a star-cyclic vector $e_0$, then there is a finite measure $\mu$ on $\mathbb {C}$ such that every polynomial in $z$ and $\overline{z}$ belongs to $L^2(\mu )$ and there is an isomorphism $W : \mathcal{H} \to L^2(\mu )$ such that $We_0 = 1$ and $WNW^{-1} = N_\mu$.

Let $\mu$ be a finite measure on $\mathbb {C}$ such that every polynomial in $z$ and $\overline{z}$ belongs to $L^2(\mu )$ and the collection of such polynomials is dense in $L^2(\mu )$. Let $\mathcal{D}_\mu = \left\{ f \in L^2(\mu ) : zf \in L^2(\mu )\right\}$ and define $N_\mu f = zf$ for $f$ in $\mathcal{D}_\mu$. (Recall that $N_\mu$ is a normal operator and $1$ is a star-cyclic vector for $N_\mu$.)

If $\mu_1, \mu_2$ are finite measures on $\mathbb {C}$ satisfying the conditions above and $N_{\mu_1}, N_{\mu_2}$ are defined accordingly, show that $N_{\mu_1} \cong N_{\mu_2}$ iff $[\mu_1] = [\mu_2]$.

Recall Theorem 4.19: if $N$ is a normal operator on the separable Hilbert space $\mathcal{H}$, then there is a $\sigma$-finite measure space $(X, \Omega , \mu )$ and an $\Omega$-measurable function $\phi$ such that $N$ is unitarily equivalent to $M_\phi$ on $L^2(\mu )$.

The proof is only sketched in the text as follows. Write $N$ as the (unbounded) direct sum of bounded normal operators $\left\{ N_n\right\}$. For each $n$ there is a $\sigma$-finite measure space $(X_n, \Omega_n, \mu_n)$ and a bounded $\Omega_n$-measurable function $\phi_n$ such that $N_n \cong M_{\phi_n}$. Let $X$ be the disjoint union of $\left\{ X_n\right\}$ and let $\Omega = \left\{ \Delta \subseteq X : \Delta \cap X_n \in \Omega_n \text{ for every } n\right\}$. If $\Delta \in \Omega$, let $\mu (\Delta ) = \sum_{n=1}^{\infty } \mu_n(\Delta \cap X_n)$. Let $\phi : X \to \mathbb {C}$ be defined by $\phi (x) = \phi_n(x)$ if $x \in X_n$. Then $\phi$ is $\Omega$-measurable and $N \cong M_\phi$ on $L^2(X, \Omega , \mu )$.

Fill in the details of this proof.