The Spectral Theorem and Functional Calculus for Compact Normal Operators

If $\left\{ P_{n}\right\}$ is a sequence of pairwise orthogonal nonzero projections and $P = \sum P_{n}$, show that $\left\| P - \sum_{j=1}^{n} P_{j}\right\| = 1$ for all $n$.

If $\mathcal{H}$ is separable, show that the definitions of a diagonalizable operator in (4.6) and (7.3) are equivalent.

Proposition 4.6 (context): Let $\mathcal{H}$ be a separable Hilbert space with basis $\left\{ e_{n}\right\}$; let $\left\{ \alpha_{n}\right\} \subseteq \mathbb {F}$ with $M = \sup \left\{ \left|\alpha_{n}\right|: n \geq 1\right\} < \infty$. If $Ae_{n} = \alpha_{n} e_{n}$ for all $n$, then $A$ extends by linearity to a bounded operator on $\mathcal{H}$ with $\left\| A\right\| = M$. The operator $A$ is compact if and only if $\alpha_{n} \rightarrow 0$ as $n \rightarrow \infty$. (Such an operator $A$ is said to be diagonalizable.)

(7.3) (context): An operator $A$ on $\mathcal{H}$ is diagonalizable if there is a partition of the identity on $\mathcal{H}$, $\left\{ P_{i}: i \in I\right\}$, and a family of scalars $\left\{ \alpha_{i}: i \in I\right\}$ such that $\sup_{i} \left|\alpha_{i}\right| < \infty$ and $Ah = \alpha_{i} h$ whenever $h \in \operatorname {ran} P_{i}$.

If $A = \sum \alpha_{i} P_{i}$ as in (7.3), show that $A$ is compact if and only if:

Prove Proposition 7.4: An operator $A$ on $\mathcal{H}$ is diagonalizable if and only if there is an orthonormal basis for $\mathcal{H}$ consisting of eigenvectors for $A$.

If $A = \oplus_{i} \alpha_{i} P_{i}$, show that $A^{*} = \oplus_{i} \overline{\alpha_{i}} P_{i}$, $A$ is normal, and $\left\| A\right\| = \sup \left\{ \left|\alpha_{i}\right|: i \in I\right\}$.

Give the remaining details in the proof of (7.6).

Theorem 7.6 (Spectral Theorem for Compact Normal Operators, context): If $T$ is a compact normal operator on the complex Hilbert space $\mathcal{H}$, then $T$ has only a countable number of distinct eigenvalues. If $\left\{ \lambda_{1}, \lambda_{2}, \ldots \right\}$ are the distinct nonzero eigenvalues of $T$, and $P_{n}$ is the projection of $\mathcal{H}$ onto $\ker (T-\lambda_{n})$, then $P_{n} P_{m} = P_{m} P_{n} = 0$ if $n \neq m$ and

where this series converges to $T$ in the metric defined by the norm on $\mathcal{B}(\mathcal{H})$.

If $A \in \mathcal{B}(\mathcal{H})$ and $AT=TA$ for every compact operator $T$, show that $A$ is a multiple of the identity operator.

Suppose $T$ is a compact normal operator on a $\mathbb {C}$-Hilbert space such that $\dim \ker (T-\lambda ) \leq 1$ for all $\lambda$ in $\mathbb {C}$. Show that if $A \in \mathcal{B}(\mathcal{H})$ and $AT=TA$, then $A = \phi (T)$ for some $\phi$ in $l^{\infty }(\mathbb {C})$.

Definition 7.10 (context): Denote by $l^{\infty }(\mathbb {C})$ all the bounded functions $\phi : \mathbb {C} \rightarrow \mathbb {C}$. If $T$ is a compact normal operator satisfying $T = \sum_{n=1}^{\infty } \lambda_{n} P_{n}$ (7.7), define $\phi (T): \mathcal{H} \rightarrow \mathcal{H}$ by

where $P_{0}$ is the projection of $\mathcal{H}$ onto $\ker T$.

Prove a converse to Exercise 8: if $T$ is a compact normal operator such that $\left\{ A \in \mathcal{B}(\mathcal{H}): AT=TA\right\} = \left\{ \phi (T): \phi \in l^{\infty }(\mathbb {C})\right\}$, then $\dim \ker (T-\lambda ) \leq 1$ for all $\lambda$ in $\mathbb {C}$.

Let $T$ be a compact normal operator and show that $\dim \ker (T-\lambda ) \leq 1$ for all $\lambda$ in $\mathbb {C}$ if and only if there is a vector $h$ in $\mathcal{H}$ such that $\left\{ p(T)h: p \text{ is a polynomial in one variable}\right\}$ is dense in $\mathcal{H}$. (Such a vector $h$ is called a cyclic vector for $T$.)

If $\lambda \in \mathbb {C}$, let $\delta_{\lambda }$ be the unit point mass at $\lambda$; that is, $\delta_{\lambda }$ is the measure on $\mathbb {C}$ such that $\delta_{\lambda }(\Delta ) = 1$ if $\lambda \in \Delta$ and $\delta_{\lambda }(\Delta ) = 0$ if $\lambda \notin \Delta$. If $\left\{ \lambda_{1}, \lambda_{2}, \ldots \right\}$ is a bounded sequence of distinct complex numbers and $\left\{ \alpha_{n}\right\}$ is a sequence of real numbers with $\alpha_{n} > 0$ and $\sum_{n} \alpha_{n} < \infty$, let $\mu = \sum_{n=1}^{\infty } \alpha_{n} \delta_{\lambda_{n}}$; so $\mu$ is a finite measure. If $\phi \in l^{\infty }(\mathbb {C})$, let $M_{\phi }$ be the multiplication operator on $L^{2}(\mu )$. Define $T: L^{2}(\mu ) \rightarrow L^{2}(\mu )$ by $(Tf)(\lambda_{n}) = \lambda_{n} f(\lambda_{n})$. Prove:

Using the notation of Theorem 7.11, give necessary and sufficient conditions on $T$ and $\phi$ that $\phi (T)$ be compact. (Hint: consider separately the cases where $\ker T$ is finite or infinite dimensional.)

Definition 7.10 (context): Denote by $l^{\infty }(\mathbb {C})$ all the bounded functions $\phi : \mathbb {C} \rightarrow \mathbb {C}$. If $T$ is a compact normal operator satisfying $T = \sum_{n=1}^{\infty } \lambda_{n} P_{n}$ (7.7), define $\phi (T): \mathcal{H} \rightarrow \mathcal{H}$ by

where $P_{0}$ is the projection of $\mathcal{H}$ onto $\ker T$.

Prove the uniqueness part of Theorem 7.16: If $T$ is a positive compact operator, then there is a unique positive compact operator $A$ such that $A^{2} = T$.

If $T \in \mathcal{B}(\mathcal{H})$, show that $T^{*}T \geq 0$.

Let $T$ be a compact normal operator and show that there is a compact positive operator $A$ and a unitary operator $U$ such that $T = UA = AU$. discuss the uniqueness of $A$ and $U$.

Let $T \in \mathcal{B}_{0}(\mathcal{H})$ and let $A$ be the unique positive square root of $T^{*}T$ [(7.16) and Exercise 14].

Prove the following uniqueness statement for the functional calculus (7.11). If $T$ is a compact normal operator on a $\mathbb {C}$-Hilbert space $\mathcal{H}$ and $\tau : l^{\infty }(\mathbb {C}) \rightarrow \mathcal{B}(\mathcal{H})$ is a multiplicative linear map such that $\left\| \tau (\phi )\right\| = \sup \left\{ \left|\phi (\lambda )\right|: \lambda \in \sigma_{p}(T)\right\}$, $\tau (1) = 1$, and $\tau (\psi ) = T$ whenever $\psi (z) = z$ on $\sigma_{p}(T) \cup \left\{ 0\right\}$, then $\tau (\phi ) = \phi (T)$ for every $\phi$ in $l^{\infty }(\mathbb {C})$.