Spectral Theory

Let $T \in B(H)$, $H$ a separable Hilbert space.
Consider:

1. norm-closed algebra generated by $T$ and $I$
2. $C^*(F)$ - norm-closed algebra generated by $T$, $T^*$, and $I$
3. $W^*(T)$ - weak operator topology closed algebra generated by $T, T^*$, and $I$.

$B(H)$ has 6 or more interesting and useful topologies:

1. Norm topology - $A_i \rightarrow A$ if $\sup_{x \in \text{ball }H} \Vert(A_i - A)(x)\Vert \rightarrow 0$ as $i \rightarrow \infty$.
2. Strong operator topology: $A_i \rightarrow A$ if $\forall x$, $A_i x \rightarrow Ax$
3. Weak operator topology: $A_i \rightarrow A$ if $\forall x,y$, $\left\langle{ A_ix, y }\right\rangle \rightarrow \left\langle{ Ax, y }\right\rangle$

Each of these is successively strictly weaker. In fact, $B(H)$ is the dual of the trace class operators:

Definition 7.0.1   Pick a basis $x_i$ for $H$. Define tr$: B_1(H) \rightarrow \mathbb{C}$,

   tr$$(A) = \sum_i \left\langle{ Ax_i, x_i }\right\rangle$$

Then, $(B_1(H), \Vert A\Vert _1 =$   tr$((A^* A)^\frac12))^* = B(H)$, so $B(H)$ has a wk $^*$-topology.

Definition 7.0.2   The spectrum of $T$ is

$$\sigma(T) = \{ \lambda \in \mathbb{C}: T - \lambda I$$    is not invertible $$\}$$

Definition 7.0.3   By a functional calculus for $T \in B(H)$, we mean an algebra homomorphism from a space of functions $\mathcal{F}$ on $\sigma(T)$ into $B(H)$

Let $1 \in \mathcal{F}$ denote the identically $1$ function. $z \in \mathcal{F}$ denotes the function that sends $z$ to $z$. We denote, for $f \in \mathcal{F}$, the image of $f$ under this functional calculus by $f(T)$. We want

1. $1(T) = I_H$
2. $z(T) = T$
   This way, $p(T)$ is correct for polynomial $p$, i.e.
   $$p(T) = \sum a_i T^i$$
   if $p(Z) = \sum a_i z^i$.
3. continuity properties
4. the range of the functional calculus is sometimes $C^*(T)$, $W^*(T)$.

There are three standard functional calculuses:

1. Analytic (Riesz) functional calculus. Space of functions is $f:N \rightarrow \mathbb{C}$, $N$ an open neighborhood of $\sigma(T)$. $f$ analytic on $N$.
   Continuity property: If $f(z) = \sum_{i=0}^\infty a_i z^i$, then $f(T) = \sum_{i=0}^\infty a_i T^i$, provided the radius of convergence is sufficiently large.

2. Continuous functional calculus. Space of functions: $C(\sigma(T))$. Operators: normal operators on $B(H)$. Then,
   $$f \mapsto f(T)$$
   is a $*$-monomorphism (injective), isometric from $\Vert\cdot\Vert _\infty$ to operator norm, range in $C^*(T)$, and extends the analytic functional calculus.
   

3. Borel functional calculus. Let $T$ be a normal operator on a separable $H$. There is a measure $\mu$ on $\sigma(T)$ and space of functions is $L^\infty(\mu)$. The function calculus maps $L^\infty(\mu)$ into $W^*(T)$. Finally, it is an isometric $*$-isomorphism and it is a $($wk $^*, WOT)$-homeomorphism.

Friday, April 7, 2006:

Application of the functional calculus - isometric dilation.
Let $T \in B(H)$, $H$ a Hilbert space. We call $B$ a dilation of $T$ if $B \in B(K)$, $H \leq K$ and $P_H B^n P_H = T^n$.

We call two dilations of $T, B \in B(K)$ and $C \in B(L)$ isomorphic if there is a unitary $U:K \rightarrow L$ such that

1. $UB = CU$
2. $U\vert _H = I_H$.

Theorem 7.0.4   If $T \in B(H)$, $\Vert T\Vert \leq 1$, then $T$ has an isometric dilation $V \in B(K)$ which is minimal in the sense that

$$\overline{\text{span}}\left( \bigcup_{n} \{ V^n H : n \geq 0 \} \right) = K$$

and $V$ is unique, up to isomorphism.

Proof. Let

$$H_\infty = \bigoplus_{n=0}^\infty H$$

i.e. sequences,

$$(h_0, h_1, h_2, \ldots)$$

such that

$$\sum_{i} \Vert h_i\Vert^2 < \infty$$

Embed $H$ in $H_\infty$ by

$$h \mapsto (h,0, 0, \ldots )$$

As $\Vert T\Vert \leq 1$, $\left\langle{ Th, Th }\right\rangle \leq \left\langle{ h,h }\right\rangle$ for all $h \in H$. Then,

$$\left\langle{ h,h }\right\rangle - \left\langle{ Th, Th }\right\rangle \geq 0$$

so

$$\left\langle{ (I - T^* T) h, h }\right\rangle \geq 0$$

I.e., $I - T^* T \geq 0$. Thus,

$$\sigma( I - T^* T) \subset [0, +\infty).$$

Let

$$D_T = \sqrt{I - T^* T}$$

in the functional calculus. Notice $D_I^2 = I - T^* T$. Then,

$$\Vert D_T h \Vert^2 = \left\langle{ D_T h, D_T h }\right\rangle =... ...e - \left\langle{ T^* T h, h }\right\rangle = \Vert h\Vert^2 - \Vert Th\Vert^2$$

$$\Vert h\Vert^2 = \Vert Th\Vert^2 + \Vert D_T h \Vert^2$$

$D_T$ is called the defect operator.

Now, define $V:H_\infty \rightarrow H_\infty$ by

$$V(h_0, h_1, h_2, \ldots ) = (Th_0, D_T h_0, h_1, h_2, \ldots )$$

Then,

$$\Vert V(h_0, h_1, \ldots)\Vert^2 = \Vert Th\Vert^2 + \Vert D_T h_0\Vert^2 + \Vert h_1\Vert^2 + \cdots$$

$$= \Vert h_0\Vert^2 + \Vert h_1\Vert^2 + \cdots = \Vert(h_0, h_1, \cdots)\Vert^2$$

So, $V$ is an isometry. Notice,

$$V^n(h_0, h_1, \ldots ) = (T^n h_0, D_T T^{n-1} h_0, D_T^2 T^{n-2} h_0, \cdots, h_1, h_2 )$$

So,

$$P_H V^n( h_0, 0, \ldots) = T^n h_0$$

for all $h_0 \in H$. So, $V$ is a dilation. Let

$$K = \overline{\text{span}} \left( \bigcup_{n \geq 0 } V^n H \right)$$

Clearly, $K \in$   Lat $V$. Then, $V\vert _K$ is a minimal isometric dilation.

Uniqueness:
For any isometric dilation,

$$\left\langle{ V^n h, V^m k }\right\rangle = \begin{cases}\left\la... ...ht\rangle = \left\langle{ h, T^{m-n} k }\right\rangle , & m \geq n \end{cases}$$

So, $\left\langle{ V^m h, V^n k }\right\rangle , h, k \in H$ does not depend on the choice of $V$. Let $V_1, V_2$ be two minimal isometric dilations of $T$ on $K_1, K_2$, respectively. For $i=1, i=2$,

$$L_i :=$$   span$$\{ V_i^n h : h \in H, n \geq 0 \}$$

$L_i$ is a (not necessarily closed) subspace.

Define: $U: L_2 \rightarrow L_1$ by

$$U(V_2^n h ) V_1^n h$$

and extend by linearity. By the fact $\left\langle{ V^m h, V^n k }\right\rangle$ does not depend on the choice of $V$, $U$ is isometric (hence linear and well-defined). By the minimality of $\bar L_1 = K_i$ and so we can extend $U$ to a unitary $U:K_2 \rightarrow K_1$. So,

$$U(h) = U(V_2^\circ h) = V_1^\circ h = h$$

so $U\vert _H = I\vert _H$. Then,

$$U(V_2 k ) = U( V_2( \sum_{k=1}^m V_2^n h_K)) = U( \sum_{k=1}^m V_2^{n+1} h_k)$$

$$= \sum_{k=1}^m V_1^{n+1} h_k = \sum_{k=1}^m V_1^{n+1} h_k = V_1(\sum_{k=1}^m V_1^n h_k ) = V_1 Uk$$

Thus, $U(V_2k) = V_1 Uk$ for all $k \in L_2$. As $U, V_1, V_2$ are continuous and $\bar L_2 = K_2$,

$$U(V_2 k ) = V_1 U k$$

for all $k \in K_2$. Thus, $U V_2 = V_1 U$ and so $V_1, V_2$ are isomorphic. $\qedsymbol$

Source: Focias and Sz. Nagy, ``Harmonic Analysis of Operators on Hilbert Space", Section I.5.