### Topics in Operator Theory, Lecture 2: Dilations of contractions

In this lecture we will study the first chapter in the theory of dilations of contractions. To proceed in our study of operator spaces and operator algebras, the material we will cover is not strictly needed. However, this is where I want to begin, for several reasons:

1. The objects and theorems here motivate (and have motivated historically) the development of the general theory, and help understand it better and appreciate it more.
2. We will reach very quickly a nontrivial application of operator theory to function theory, which is quite different from what you all saw in your studies, probably.
3. I am stalling, so that the students who need to fill in some prerequisites (like commutative C*-algebras and the spectral theorem, will have time to do so).
4. I love this stuff!

Okay, enough explaining, let us begin.

#### 1. Overview

Definition: An operator $T \in B(H)$ is said to be

1. selfadjoint if $T = T^*$,
2. normal if $T T^* = T^* T$,
3. unitary if $T T^* = T^* T = I$,
4. isometric (or an isometry) if $T^* T = I$ (equivalently, if $\|T h\| = \|h\|$ for all $h \in H$,
5. coisometric (or a coisometry) if $T T^* = I$ (i.e., if $T^*$ is an isometry),
6. contractive (or a contraction) if $\|T\| \leq 1$.

Normal (and hence, selfadjoint and unitary) operators are well understood. In this lecture our goal will be to understand first isometries (and hence also coisometries) and then contractions a little bit better. Before we begin with this,  I will now repeat some material from the first lecture in my von Neumann algebras notes to remind ourselves of the nice structure theorem for normal operators.

The spectral theorem is the basic structure theorem for normal operators. It tells us how a general normal operator looks like. Recall that if $T$ is a normal operator acting on a finite dimensional space $H = \mathbb{C}^n$, then $T$ is unitarily equivalent to a diagonal operator, that is, there exists a unitary operator $U \in \mathcal{U}(H)$ such that

$U T U^* = \begin{pmatrix}\lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix}$,

where $\sigma(T) = \{\lambda_1, \ldots, \lambda_n\}$ (some points in $\sigma(T)$ are possibly repeated).

Moreover, if $T$ is a compact normal operator on a Hilbert space, then $T$ unitarily equivalent to a diagonal operator (an infinite diagonal matrix, acting by multiplication on $\ell^2$), the diagonal of which corresponds to the eigenvalues of $T$, which form a sequence converging to $0$:

$U T U^* = \begin{pmatrix}\lambda_1 & & \\ & \lambda_2 & \\ & & \ddots \end{pmatrix}$.

If $T$ is unitarily equivalent to a diagonal operator where the diagonal elements form a bounded sequence of numbers (not necessarily converging to $0$), then $T$ is a bounded normal operator (which is not necessarily compact). However, a general bounded normal operator need not be unitarily equivalent to a diagonal operator.

Example: The operator $T : L^2[0,1] \to L^2[0,1]$ given by $T f(x) = xf(x)$ is a selfadjoint bounded operator, and it is an easy exercise to see that this operator has no eigenvalues (so it cannot be unitarily equivalent to a diagonal operator). However, the operator in this example is rather well understood, and it is “sort of” diagonal. The general case is not significantly more complicated than this.

To understand general normal operators, one needs to recall the notions of measure space and of $L^p$ spaces. Let $(X, \mu)$ be a measure space and consider the Hilbert space $L^2 = L^2(X,\mu)$. Every $f \in L^\infty = L^\infty(X,\mu)$ defines a (normal) bounded operator $M_f : h \mapsto fh$ on $L^2$.

Exercise A: In case you never have, prove the following facts (or look them up; Kadison-Ringrose have a nice treatment relevant to our setting). Let $(X, \mu)$ be a $\sigma$-finite measure space and $f \in L^\infty(X, \mu)$.

1. $\|M_f\| = \|f\|_\infty$ (where $\|f\|_\infty$ is the essential supremum of $f$, which is defined to be $\inf \{t \geq 0 : \mu\{x:|f(x)|> t\} = 0\}$).
2. $M_f^* = M_{\overline{f}}$.
3. If $g : X \to \mathbb{C}$ and $h \mapsto gh$ defines a bounded operator on $L^2(X, \mu)$, then $g$ is essentially bounded: $\|g\|_\infty < \infty$.
4. If $f,g \in L^\infty(X, \mu)$, then $M_f M_g = M_{fg}$ and $M_f + M_g = M_{f+g}$.
5. $M_f$ is selfadjoint if and only if $f$ is real valued almost everywhere.

The algebra $L^\infty$ is an abstract C*-algebra with the usual algebraic operations, the $*$-operation $f^* = \overline{f}$, and norm $\|f\| = \|f\|_\infty = ess-sup_{x \in X}|f(x)|$. The map $\pi : L^\infty \to B(L^2)$

$\pi(f) = M_f$

is a $*$-representation (i.e., and algebraic homomorphism that preserves the adjoint $\pi(f^*) = \pi(f)^*$), which is isometric ($\|\pi(f)\| = \|f\|$), so omitting $\pi$ we can think of $L^\infty$ as a C*-subalgebra of $B(L^2)$. Since $(M_f)^* = M_{\overline{f}}$, the operator $f \sim M_f$ is always normal. It is selfadjoint if and only $f$ is a.e. real valued, and it is unitary if and only if $|f|=1$ a.e., which happens if and only if it is isometric. The operator $M_f$, where $f \in L^\infty$, is called a multiplication operator. Multiplication operators form a rich collection of examples of normal operators. The spectral theorem says that this collection exhausts all selfadjoint operators: every normal operator is unitarily equivalent to a multiplication operator.

Theorem 1 (the spectral theorem): Let $T$ be a normal operator on a Hilbert space $H$. Then $T$ is unitarily equivalent to a multiplication operator, that is, there exists a measure space $(X, \mu)$, a unitary operator $U : L^2(X,\mu) \to H$, and a complex valued $f \in L^\infty(X,\mu)$, such that

$U^* T U = M_f$

When $H$ is separable, $X$ can be taken to be a locally compact Hausdorff space, and $\mu$ a regular Borel probability measure.

The spectral theorem allows, in principle, to “solve all problems about normal operators”. Okay, that’s maybe an exaggeration; the spectral theorem reduces any problem about normal operators to problems about multiplication operators. As an example, we prove

Proposition 2 (von Neumann’s inequality for normal operators): Let $T$ be a normal contraction. Then for any polynomial $p \in \mathbb{C}[z]$

$\|p(T)\| \leq \sup_{|z|\leq 1} |p(z)|\,\,$     $\,\, \Big( = \sup_{|z|=1}|p(z)| \Big)$ .

Proof: Since unitary equivalence preserves everything, we may assume that $T = M_f$. Now, $f$ is a contraction so $|f| \leq 1$ a.e. So we find

$p(T) = p(M_f) = M_{p \circ f}$,

and since $|f| \leq 1$ a.e., we have $|p \circ f| \leq \sup_{|z|\leq 1} |p(z)|$ a.e., so

$\|p(T)\| = \|p \circ f\|_\infty \leq \sup_{|z|\leq 1} |p(z)|$,

as required. (The second equality simply follows from the maximum modulus principle, which, by the way, can be proved using the methods we will use in this lecture).

#### 2. Isometries

Unitary operators are isometric, that’s one basic example. An isometry that is not a unitary is called a proper isometry.

Example (the unilateral shift): Let $G$ be a Hilbert space, suppose $\dim G = d$. Let $H$ be the Hilbert space consisting of square summable sequences with values in $G$:

$H = \ell^2(\mathbb{N}, G) = \left\{x = (x_n)_{n=1}^\infty : \|x\| := \sum_n \|x_n\|^2 < \infty \right \}$.

$H$ has an inner product

$\langle x, y \rangle = \sum \langle x_n, y_n \rangle$.

On $H$ we define the unilateral shift (of multiplicity $d$$S \in B(H)$ to be given by

$S(x_0, x_1, x_2, \ldots) = (0, x_0, x_1, x_2, \ldots)$.

A simple computation reveals $S^*(x_0, x_1, x_2, \ldots) = (x_1, x_2, x_3, \ldots)$ and so

$S S^* = P_{Im S} \neq I_H = S^*S$.

Definition: Let $V \in B(H)$ be an isometry. A subspace $L \subseteq H$ is called a wandering subspace for $V$ if

$\forall m,m\in \mathbb{N} . m \neq n \Rightarrow V^m L \perp V^n L$.

(Equivalently $V^n L \perp L$ for all $n$). We define

(*)   $M_+(L) = \bigoplus_{n=0}^\infty V^n L$.

We have that $VM_+(L) = \bigoplus_{n=1}^{\infty} V^n L = M_+(L) \ominus L$, and this means that $L = M_+(L) \ominus V M_+(L)$.

Example: Let $F \subseteq G$ be a subspace. Fix $k \in \mathbb{N}$, and let $L \subseteq \ell^2(\mathbb{N}, G)$ be given by

$L = \{(x_n) : x_k \in F$ and $x_n = 0$ for all $n \neq k \}$.

Then $L$ is a wandering subspace for $S$. In particular,

$G \oplus 0 \oplus 0 \oplus \cdots$

is a wandering subspace.

Definition: An isometry $V \in B(H)$ is said to be a unilateral shift (of multiplicity $d$) if it has a wandering subspace $L \subseteq H$ (with $\dim L = d$) for $V$ such that $H = M_+(L)$.

Note that $L$ is determined uniquely by (*) as

(**)   $L = H \ominus VH = Im V^\perp$.

Alright: we have defined THE unilateral shift and the notion of A unilateral shift, what gives?

Exercise A: Let $V$ be a unilateral shift. Show that

1. $V$ is unitarily equivalent to THE unilateral shift $S$ on $\ell^2(\mathbb{N}, L)$, where $L$ is given by (**).
2. A unilateral shift $V'$ is unitarily equivalent to $V$ if and only if $L \cong L'$.
3. $\forall h \in H . \lim_n (V^*)^n h = 0$ (in other words, $SOT-\lim_n V^{*n} = 0$).

Definition: Let $T \in B(H)$. A subspace $K \subseteq H$ is said to be

1. invariant if $TK \subseteq K$,
2. coinvariant if $T^* K \subseteq K$ (equivalent, if $K^\perp$ is invariant for $T$).
3. reducing if it is invariant for both $T$ and $T^*$ (equivalently, if $K$ and $K^\perp$ are both invariant for $T$).

Sometimes, to clarify the operator with respect to which the subspace should satisfy the properties, the terminology invariant for $T$ (etc.) is used.

Theorem 3 (Wold decomposition): Let $V \in B(H)$ be an isometry. Then there exists a decomposition of $H$ to two reducing subspaces $H = H_0 \oplus H_1$ such that $V\big|_{H_0}$ is unitary on $H_0$ and $V\big|_{H_1}$ is a unilateral shift. Moreover, this decomposition is determined uniquely by $H_0 = \cap V^n H$ and $H_1 = M_+(L)$, where $L = H \ominus VH$

Proof: If we define $L = H \ominus VH$, then $L$ is wandering for $V$ because for all $n$ we have by definition $V^n L \subseteq V^n H \subseteq VH \perp L$.

Let us define $H_1 = M_+(L) = \bigoplus_{n=0}^\infty V^n L$, and $H_0 = H \ominus H_1$. Then $H_1$ is clearly invariant, and $V\big|_{H_1}$ is a unilateral shift.

We will soon show that $H_0 = \cap_{n=0}^\infty V^n H$. Assuming that for the moment, we have that $VH_0 = \cap_{n=1}^\infty V^n H = \cap_{n=0}^\infty V^n H = H_0$, so $H_0$ is invariant, and thus $H_0, H_1$ are reducing. Moreover, $V\big|_{H_0}$ is a surjective isometry, so it is a unitary (showing that a surjective isometry is a unitary is a basic exercise).

To show that $H_0 = \cap_{n=0}^\infty V^n H$ we take $h \in H$ and note that $h \in H_0 = H \ominus H_1$ if and only if $h \perp L \oplus VL \oplus \ldots \oplus V^{m-1} L$ for every $m$. But

$L \oplus VL \oplus \ldots \oplus V^{m-1} L = (H \ominus VH) \oplus (VH \ominus V^2H) \oplus \ldots \oplus (V^{m-1} H \ominus V^m H) = H \ominus V^m H$

where we cancelled out all the terms in the above “telescopic sum”. Now, $h \perp H \ominus V^m H$ for all $m$ is the same as $h \in V^m H$ for all $m$, and we have established that $H_0 = \cap_{m} V^m H$.

We leave the uniqueness as an exercise.

Definition: Let $A \in B(H)$, and suppose that $K$ is a Hilbert space containing $H$. An operator $B \in B(K)$ is said to be an extension of $A$ if $A = B\big|_H$.

Theorem 4 (unitary extension of an isometry): Every isometry has a unitary extension.

Proof: Let $V$ be an isometry and let $V = V_0 \oplus V_1$ be its Wold decomposition on $H_0 \oplus H_1$. By Exercise A, up to unitary equivalence, $V_1 = S$, the unilateral shift of some multiplicity $d$ on $\ell^2(\mathbb{N},G)$. Letting $U$ be the bilateral shift on $\ell^2(\mathbb{Z}, G)$, that is, $U$ is defined by

$U(\ldots, x_{-2}, x_{-1},\underbrace{x_0}, x_1, x_2, \ldots) = (\ldots x_{-3}, x_{-2},\underbrace{x_{-1}}, x_0, x_1, \ldots)$.

Clearly, $U$ is an extension of $V_1$, and so $V_0 \oplus U$ is an extension of $V = V_0 \oplus V_1$.

Corollary: Von Neumann’s inequality holds for isometries.

Proof: Let $V$ be an isometry and let $U$ be a unitary extension. From $U\big|_H = V$ follows $U^n\big|_H = V^n$ and so $\|p(V)\| = \|p(U)\big|_H\| \leq \|p(U)\|\leq \sup_{|z|=1}|p(z)|$.

#### 3. Contractions

In the previous section we showed that every isometry has a unitary extension, and this immediately led to the application that isometries satisfy von Neumann’s inequality. Of course, we cannot hope to show that any operator other than an isometry has a unitary extension, so we need another trick.

Definition: Let $A \in B(H)$, and suppose that $K$ is a Hilbert space containing $H$. An operator $B \in B(K)$ is said to be an dilation of $A$ if $A^n = P_H B^n \big|_H$ for all $n=1, 2, \ldots$.

Sometimes, the term dilation refers to the relation $A = P_H B\big|_H$, and then the above notion is called a power dilationThere are also situations when $A^n = P_H B^n \big|_H$ is only assumed for all $n=1, 2, \ldots, N$ for some $N$, and then $B$ is said to be an N-dilation of $A$. It is usually not hard to understand from context what notion is the one intended.

Examples: If $B$ is an extension of $A$, then $B$ is a dilation of $A$. Note that in this case

$B = \begin{pmatrix} A & * \\ 0 & * \end{pmatrix}$.

Likewise, if $B$ is a co-extension of $A$ (meaning that $B^*$ is a dilation of $A^*$) then it is a dilation, and in this case we have the picture:

$B = \begin{pmatrix} A & 0 \\ * & * \end{pmatrix}$,

or (simply changing basis)

$B = \begin{pmatrix} * & * \\ 0 & A \end{pmatrix}$.

Both of the above situations are a special case of the following picture

(#)    $B = \begin{pmatrix} * & * & * \\ 0 & A & * \\ 0 & 0 & * \end{pmatrix}$

in which $B$ is easily seen to be a dilation of $A$:

$B^n = \begin{pmatrix} * & * & * \\ 0 & A^n & * \\ 0 & 0 & * \end{pmatrix}$.

It turns out that (#) is the most general form of a dilation.

Proposition 5 (Sarason’s Lemma): Let $H \subseteq K$ be a Hilbert spaces, and suppose that $B \in B(K)$ is a (power) dilation of $A := P_H B\big|_H$. Then there exist two subspaces $M \subseteq N \subseteq K$ invariant for $B$ such that $H = N \ominus M$

Remark: The case where $M = 0$ corresponds to the case where $H$ is invariant and $B$ is an extension of $A$; the case where $N = K$ correspond to the case where $H$ is co-invariant and $B$ is a co-extension of $A$.

Remark: In the situation of the above proposition, we say that $H$ is a semi-invariant subspace for $B$. More generally, if $\mathcal{A} \subseteq B(K)$ is an operator algebra, then a subspace $H \subseteq K$ is said to be semi-invariant for $\mathcal{A}$  if the map $T \mapsto P_H T \big|_H$ is a homomorphism. Clearly, $H$ is semi-invariant for $B$ if and only it is semi-invariant for $\mathcal{A} := alg(I,B)$.

Proof of Sarason’s Lemma: (maybe I should leave this as an exercise)

Let us write $\mathcal{A} := alg(I,B)$ as in the above remark, and let $N := [\mathcal{A} H]$ (this notation means the closed linear subspace spanned by $ah$ where $a \in \mathcal{A}$ and $h \in H$). Then $N$ is clearly invariant, and $H \subseteq N$. For this to work out we really have no choice but to define $M = N \ominus H$. The only thing that remains to do is to prove that $M$ is invariant for $\mathcal{A}$.

If $a, b \in \mathcal{A}$ then the definition of semi-invariant means that $P_H ab \big|_H = P_H a P_H b \big|_H$, and so for every $h \in H$ we have that $P_H a b h = P_H a P_H b h$. By taking sums and limits, we find that $P_H a n = P_H a P_H n$ for all $n \in N$.

To show that $M$ is invariant, let $m \in M$, $a \in \mathcal{A}$, and consider $a m$. Since $am \in N$, to show that it is in $M$ we need to show that it is orthogonal to $H$; equivalently, we need to show that $P_H a m = 0$. But by the previous paragraph $P_H a m = P_H a P_H m = 0$, because $m \in M = N \ominus H$. This completes the proof.

Theorem 6 (Sz.-Nagy’s isometric dilation theorem): Let $T$ be a contraction on a Hilbert space $H$. Then there exists a Hilbert space $K \supseteq H$ and an isometry $V \in B(K)$ such that

1. $V$ is a co-extension of $T$
2. $K$ is the smallest subspace of $K$ invariant for $V$ that contains $H$

Moreover, the pair $(V,K)$ is unique, in the sense that if $(V',K')$ is another such pair satisfying the above requirements, then there exists a unitary $W : K \to K'$ such that $Wh = h$ for all $h \in H$, and such that $W V = V' W$

Remark: The isometry $V$ (or more precisely, the pair $(V,K)$) is known as the minimal isometric dilation of $T$. The theorem can be reformulated (and sometimes is) by saying that every contraction has a minimal coisometric extension.

Proof: Define the defect operator $D_T = (I - T^* T)^{1/2}$. Note that $\|T\| \leq 1$ is equivalent to $T^*T \leq I$, so the square root makes sense. The operator $D_T$ measures how much $T$ fails to be an isometry. We now define $\mathcal{D}_T = [D_T H]$ (the defect space) and

$K = H \oplus \mathcal{D}_T \oplus \mathcal{D}_T \oplus \mathcal{D}_T \oplus \ldots$.

On $K$ we define an operator $V$ by

$V(h, d_1, d_2, \ldots) = (Th, D_Th, d_1, d_2, \ldots)$

or, in matrix form

$V = \begin{pmatrix} T & & & \\ D_T & & & \\ & I & & \\ & & I & \\ & & & \ddots \end{pmatrix}$

(all the empty slots are understood to be zero).

We leave it as an exercise for the reader to complete the proof.

Exercise B: Prove that $V$ is an isometry, a co-extension of $T$, and that it satisfies the minimality requirement. While you are at it, please also show that $V$ is unique.

The uniqueness of the minimal dilation can be strengthened as follows.

Exercise C: Prove that if $(U, L)$ is an isometric co-extension of $T$, the there exist two $U$ reducing subspaces $K_1,K_2 \subseteq L$, such that $H \subseteq K_1$ and $L = K_1 \oplus K_2$, and such that $U_1 := U\big|_{K_1}$ is unitary equivalent to the minimal isometric dilation of $T$. Thus, $U$ breaks up as that $U = U_1 \oplus U_2$, where $U_2$ is essentially irrelevant.

Note that if we didn’t insist on creating a minimal isometric dilation, we could have taken

$K = H \oplus H \oplus H \oplus \cdots = \ell^2(\mathbb{N}, H)$

and $V$ to be given by the same formula:

$V = \begin{pmatrix} T & & & \\ D_T & & & \\ & I & & \\ & & I & \\ & & & \ddots \end{pmatrix}$

Exercise D: If $A_1, \ldots, A_d$ are $d$ contractions, then there exists a Hilbert space $K$ and $d$ isometries $V_1, \ldots, V_d$ on $K$ such that

$A_{i_1} A_{i_2} \cdots A_{i_N} = P_H V_{i_1} V_{i_2} \cdots V_{i_N} \big|_H$

for all $N$ and all $i_1, \ldots, i_N \in \mathbb{N}$.

Theorem 7 (Sz.-Nagy’s unitary dilation theorem): Let $T$ be a contraction on a Hilbert space $H$. Then there exists a Hilbert space $L \supseteq H$ and a unitary $U \in B(L)$ such that

1. $U$ is a dilation of $T$
2. $H$ is the smallest subspace of $L$ reducing for $U$ that contains $H$

Moreover, the pair $(U,L)$ is unique, in the sense that if $(U',L')$ is another such pair satisfying the above requirements, then there exists a unitary $W : L \to L'$ such that $Wh = h$ for all $h \in H$, and such that $WU = U' W$

Remark: The unitary $U$ (or more precisely, the pair $(U, L)$) is called the minimal unitary dilation of $T$.

Proof: Let $(V,K)$ be the minimal isometric dilation of $T$, and let $(U,L)$ be the unitary extension of $V$ constructed in Theorem 4. Then $U$ is a dilation of $T$ (“my dilation’s dilation is a dilation of mine”). If $(U, L)$ is not minimal by some chance, then restrict $U$ to $L' = \bigvee_{n=-\infty}^\infty U^n H$ (the notation should be self-explanatory), and then $(U\big|_{L'}, L')$ is a minimal unitary dilation. Uniqueness is left as an exercise.

Corollary (von Neumann’s inequality): Let $T$ be a contraction on a Hilbert space. For every polynomial $p \in \mathbb{C}[z]$

$\|p(T)\| \leq \sup_{|z|=1}|p(z)|$.

Proof:

$\|p(T)\| = \|P_H p(U) \big|_H\| \leq \|p(U)\| \leq \sup_{|z|=1}|p(z)|$.