Operator-valued Schatten spaces and quantum entropies

Abstract

Operator-valued Schatten spaces were introduced by G. Pisier as a noncommutative counterpart of vector-valued $\ell _p$-spaces. This family of operator spaces forms an interpolation scale which makes it a powerful and convenient tool in a variety of applications. In particular, as the norms coming from this family naturally appear in the definition of certain entropic quantities in quantum information theory (QIT), one may apply Pisier’s theory to establish some features of those quantities. Nevertheless, it could be quite challenging to follow the proofs of the main results of this theory from the existing literature. In this article, we attempt to fill this gap by presenting the underlying concepts and ideas of Pisier’s theory in a self-contained way which we hope to be more accessible, especially for the QIT community at large. Furthermore, we describe some applications of this theory in QIT. In particular, we prove a new uniform continuity bound for the quantum conditional Rényi entropy.

Appendices

Appendix A. Missing proofs from Sects. 3 and 4

Proof of Proposition 3.6

Part (i) is straightforward. We prove (ii) in the special case of ${\mathcal {H}}=\ell _2^d$; generalization to arbitrary ${\mathcal {H}}$ is similar.

We first show that ${\mathcal {S}}_\infty ^d = {\mathcal {C}}^d\otimes _h {\mathcal {R}}^d$. Recall that ${\mathcal {C}}^d$ is the space of column vectors, and ${\mathcal {R}}^d$ is the space of row vectors. Then, for $v^c\in {\mathcal {C}}^d$ and $w^r\in {\mathcal {R}}^d$, their multiplication $v^cw^r$ makes sense and forms a $d\times d$ matrix, i.e., an element of ${\mathcal {S}}_\infty ^d$. Define the linear map $\phi : {\mathcal {C}}^d\otimes _h {\mathcal {R}}^d\rightarrow {\mathcal {S}}_\infty ^d$ by

$$\begin{aligned} \phi (v^c\otimes w^r):= v^cw^r. \end{aligned}$$

We claim that $\phi$ is a complete isometry between ${\mathcal {C}}^d\otimes _h {\mathcal {R}}^d$ and ${\mathcal {S}}_\infty ^d$. To this end, we prove that $\phi =\phi _1$ is an isometry; extending the proof for $\phi _n$, the n-th amplification of $\phi$, is immediate.

Let $Z\in {\mathcal {C}}^d\otimes _h{\mathcal {R}}^d$. Then,

$$\begin{aligned} \Vert Z\Vert _{ {\mathcal {C}}^d\otimes _h{\mathcal {R}}^d}&= \inf \left\{ \Vert X\Vert _{M_{1, m}({\mathcal {C}}^d)} \Vert Y\Vert _{M_{m, 1}({\mathcal {R}}^d)} :\right. \,\\&\quad \left. X\in M_{1, m}({\mathcal {C}}^d) , Y\in M_{m, 1}({\mathcal {R}}^d), Z=X\odot Y \right\} . \end{aligned}$$

Put

$$\begin{aligned} X = [v_1^c \ldots v_m^c]\quad \text { and }\quad Y=\begin{bmatrix} w_1^r\\ \vdots \\ w_m^r \end{bmatrix}. \end{aligned}$$

Then,

$$\begin{aligned} \Vert X\Vert _{M_{1, m}({\mathcal {C}}^d)} = \big \Vert [v_1^c \ldots v_m^c]\big \Vert _{d, m}\quad \text { and }\quad \Vert Y\Vert _{M_{m, 1}({\mathcal {R}}^d)}=\left\| \begin{bmatrix} w_1^r\\ \vdots \\ w_m^r \end{bmatrix}\right\| _{m, d}. \end{aligned}$$

On the other hand, $Z=X\odot Y$ corresponds to $\phi (Z) = \sum _i v^c_i w^r_i$. Therefore, $\Vert Z\Vert _{ {\mathcal {C}}^d\otimes _h{\mathcal {R}}^d}$ equals the infimum of

$$\begin{aligned} \big \Vert [v_1^c \ldots v_m^c]\big \Vert _{d, m} \left\| \begin{bmatrix} w^r_1\\ \vdots \\ w_m^r \end{bmatrix}\right\| _{m, d}, \end{aligned}$$

subject to $\phi (Z) = \sum _i v^c_i w^r_i$. Note that this is nothing but the operator norm of $\phi (Z)$, i.e., $\Vert \phi (Z)\Vert _{{\mathcal {S}}_\infty ^d}$.

Next, we show that ${\mathcal {S}}_1^d = {\mathcal {R}}^d\otimes _h {\mathcal {C}}^d$. As before, thinking of ${\mathcal {R}}^d$ and ${\mathcal {C}}^d$ as row and column vectors, we can define the linear map $\psi : {\mathcal {R}}^d\otimes _h {\mathcal {C}}^d\rightarrow {\mathcal {S}}_1^d$ by

$$\begin{aligned} \psi (v^r\otimes w^c):= w^cv^r. \end{aligned}$$

Again we only show that $\psi$ is an isometry. To see this, let $Z\in {\mathcal {R}}^d\otimes _h {\mathcal {C}}^d$. Then,

$$\begin{aligned} \Vert Z\Vert _{{\mathcal {R}}^d\otimes _h {\mathcal {C}}^d}{} & {} = \inf \Big \{ \Vert X\Vert _{M_{1, m}({\mathcal {R}}^d)} \Vert Y\Vert _{M_{m, 1}({\mathcal {C}}^d)}:\,\\{} & {} \quad X\in M_{1, m}({\mathcal {R}}^d), Y\in M_{m, 1}({\mathcal {C}}^d), Z=X\odot Y\Big \}. \end{aligned}$$

Put

$$\begin{aligned} X = [v_1^r \ldots v_m^r]\quad \text { and }\quad Y=\begin{bmatrix} w_1^c\\ \vdots \\ w_m^c \end{bmatrix}. \end{aligned}$$

Then,

$$\begin{aligned} \Vert X\Vert _{M_{1, m}({\mathcal {R}}^d)} = \big \Vert [v_1^r \ldots v_m^r]\big \Vert _{1, md} = \Big (\sum _i \Vert v^r_i\Vert ^2\Big )^{1/2}, \end{aligned}$$

(A.1)

and

$$\begin{aligned} \Vert Y\Vert _{M_{m, 1}({\mathcal {C}}^d)}=\left\| \begin{bmatrix} w^c_1\\ \vdots \\ w^c_m \end{bmatrix}\right\| _{md, 1}=\Big (\sum _i \Vert w^c_i\Vert ^2\Big )^{1/2}. \end{aligned}$$

(A.2)

On the other hand, $Z=X\odot Y$ corresponds to $\psi (Z) = \sum _i w^c_iv^r_i$. Therefore, $\Vert Z\Vert _{ {\mathcal {R}}^d\otimes _h{\mathcal {C}}^d}$ equals the infimum of

$$\begin{aligned} \Big (\sum _i \Vert v^r_i\Vert ^2\Big )^{1/2}\Big (\sum _i \Vert w^c_i\Vert ^2\Big )^{1/2}, \end{aligned}$$

subject to $\psi (Z) = \sum _i w^c_iv^r_i$. An application of Hölder’s inequality shows that this is equal to $\Vert \psi (Z)\Vert _{{\mathcal {S}}_1^d}$.

Part (iii) follows readily once we note that $M_{1, d}({\mathcal {C}}^d)$ and $M_{d, 1}({\mathcal {R}}^d)$ are naturally identified with the space of $d\times d$ matrices, and by definition, their norms are the same as the norm of ${\mathcal {S}}_\infty ^d$. The isometries $M_{1, d}({\mathcal {R}}^d) = M_{d, 1}({\mathcal {C}}^d) = {\mathcal {S}}_2^d$ are proven essentially in (A.1) and (A.2). $\square$

Proof of Proposition 4.3

(i) First note that by using the reiteration property (Proposition 3.3), we observe that ${\mathcal {R}}\big (1-\tfrac{1}{p_\theta }\big )=\Big [{\mathcal {R}}\big (1-\tfrac{1}{p_0}\big ),{\mathcal {R}}\big (1-\tfrac{1}{p_1}\big )\Big ]_\theta$ and ${\mathcal {R}}\big (\tfrac{1}{p_\theta }\big )=\Big [{\mathcal {R}}\big (\tfrac{1}{p_0}\big ),{\mathcal {R}}\big (\tfrac{1}{p_1}\big )\Big ]_\theta$. Then,

$$\begin{aligned}&{\mathcal {S}}_{p_\theta }[{\mathcal {X}}_\theta ] ={\mathcal {R}}\big (1-\tfrac{1}{p_\theta }\big )\otimes _h \Big ({\mathcal {X}}_\theta \otimes _h{\mathcal {R}} \big (\tfrac{1}{p_\theta }\big )\Big ) \qquad \text {(Theorem~)}\\&\quad ={\mathcal {R}}\big (1-\tfrac{1}{p_\theta }\big )\otimes _h\Big [{\mathcal {X}}_0\otimes _h{\mathcal {R}}\big (\tfrac{1}{p_0}\big ),{\mathcal {X}}_1\otimes _h{\mathcal {R}} \big (\tfrac{1}{p_1}\big )\Big ]_\theta \qquad \text {(Theorem~)}\\&\quad =\Big [{\mathcal {R}}\big (1-\tfrac{1}{p_0}\big )\otimes _h{\mathcal {X}}_0\otimes _h{\mathcal {R}}\big (\tfrac{1}{p_0}\big ),{\mathcal {R}}\big (1-\tfrac{1}{p_1}\big )\otimes _h{\mathcal {X}}_1\otimes _h{\mathcal {R}}\big (\tfrac{1}{p_1}\big )\Big ]_\theta \qquad \text {(Theorem~ )}\\&\quad =\big [{\mathcal {S}}_{p_0}[{\mathcal {X}}_0], \, {\mathcal {S}}_{p_1}[{\mathcal {X}}_1]\big ]_\theta \qquad \text {(Theorem~)}. \end{aligned}$$

With Remark 3.5 in mind, (4.4) is obtained from (4.3) by taking ${\mathcal {X}}_i={\mathcal {S}}_{p_i}$ for $i=0,1$.

(ii) Again, we prove the theorem in the finite-dimensional case, i.e., ${\mathcal {S}}_p^d[{\mathcal {X}}]^*={\mathcal {S}}^d_{p'}[{\mathcal {X}}^*]$. We know that for operator spaces ${\mathcal {Y}}_1, {\mathcal {Y}}_2$ which one of them is finite-dimensional we have $({\mathcal {Y}}_1\otimes _h{\mathcal {Y}}_2)^* = {\mathcal {Y}}_1^*\otimes _h {\mathcal {Y}}_2^*$ completely isometrically [39, Corollary 5.8]. Using this result twice we obtain

$$\begin{aligned} {\mathcal {S}}^d_p[{\mathcal {X}}]^*&= \Big ({\mathcal {R}}^d(1-1/p)\otimes _h {\mathcal {X}}\otimes _h {\mathcal {R}}^d(1/p)\Big )^*\\&= {\mathcal {R}}^d(1-1/p)^*\otimes _h \Big ({\mathcal {X}}\otimes _h {\mathcal {R}}^d(1/p)\Big )^*\\&= {\mathcal {R}}^d(1-1/p)^*\otimes _h {\mathcal {X}}^* \otimes _h {\mathcal {R}}^d(1/p)^*, \end{aligned}$$

completely isometrically. Next, we have

$$\begin{aligned} {\mathcal {R}}^d(1/p)^*{} & {} = [{\mathcal {R}}^d, {\mathcal {C}}^d]_{1/p}^* = \big [\big ({\mathcal {R}}^d\big )^*, \big ({\mathcal {C}}^d\big )^*\big ]_{1/p} \\{} & {} = \big [{\mathcal {C}}^d, {\mathcal {R}}^d\big ]_{1/p} =\big [{\mathcal {R}}^d, {\mathcal {C}}^d\big ]_{1/p'}= {\mathcal {R}}^d(1/p'), \end{aligned}$$

completely isometrically. Here, in the second equality we use the fact that for a compatible couple $({\mathcal {Y}}_1, {\mathcal {Y}}_2)$ of finite-dimensional operator spaces we have $[{\mathcal {Y}}_1, {\mathcal {Y}}_2]_\theta ^* = [{\mathcal {Y}}_1^*, {\mathcal {Y}}_2^*]_\theta$ completely isometrically [39, Theorem 2.7.4]. We also for the third equality use $({\mathcal {C}}^d)^*={\mathcal {R}}^d$ and $({\mathcal {R}}^d)^*={\mathcal {C}}^d$ that are easy to prove [39, p. 41]. Then, we have

$$\begin{aligned} {\mathcal {S}}^d_p[{\mathcal {X}}]^* = {\mathcal {R}}^d(1/p)\otimes _h {\mathcal {X}}^* \otimes _h {\mathcal {R}}^d(1/p') = S^d_{p'}[{\mathcal {X}}^*], \end{aligned}$$

completely isometrically.

(iii) For $p=\infty$ we have

$$\begin{aligned} {\mathcal {S}}_\infty [{\mathcal {S}}_\infty ]&={\mathcal {C}}\otimes _h {\mathcal {S}}_\infty \otimes _h {\mathcal {R}}\\&=({\mathcal {C}}\otimes _h {\mathcal {C}})\otimes _h({\mathcal {R}}\otimes _h {\mathcal {R}})\\&=(\ell _2\otimes _2\ell _2)_c\otimes _h(\ell _2\otimes _2\ell _2)_r \qquad \qquad \text {(Proposition~(i))}\\&={\mathcal {S}}_\infty (\ell _2\otimes _2\ell _2) \qquad \qquad \text {(Proposition~(ii))}. \end{aligned}$$

The case $p=1$ is similar. The general case is now obtained by interpolation:

$$\begin{aligned} {\mathcal {S}}_p[{\mathcal {S}}_p]&=\Big [{\mathcal {S}}_\infty [{\mathcal {S}}_\infty ],{\mathcal {S}}_1[{\mathcal {S}}_1]\Big ]_{\frac{1}{p}} \qquad \qquad \text {(4.4)}\\&=\Big [{\mathcal {S}}_\infty (\ell _2\otimes _2\ell _2),{\mathcal {S}}_1(\ell _2\otimes _2\ell _2)\Big ]_{\frac{1}{p}}\\&={\mathcal {S}}_p(\ell _2\otimes _2\ell _2) \qquad \qquad \text {(3.1)}. \end{aligned}$$

The proof is complete. $\square$

Proof of Remark 4.6

(i) We need to show that for any $\epsilon >0$ there is a factorization $X={\hat{A}}{\hat{Y}}{\hat{A}}^*$ such that

$$\begin{aligned} \Vert X\Vert _{{\mathcal {S}}_p[{\mathcal {S}}_q]} +\epsilon \ge \Vert {\hat{A}}\Vert _{{\mathcal {S}}_{2p}}\Vert {\hat{Y}}\Vert _{{\mathcal {S}}_\infty [{\mathcal {S}}_q]}\Vert {\hat{A}}^*\Vert _{{\mathcal {S}}_{2p}}. \end{aligned}$$

Following Theorem 4.4, assume that $X=AYB$ is such that $\Vert A\Vert _{{\mathcal {S}}_{2p}}= \Vert B\Vert _{{\mathcal {S}}_{2p}}$ and

$$\begin{aligned} \Vert X\Vert _{{\mathcal {S}}_p[{\mathcal {S}}_q]}+\epsilon \ge \Vert A\Vert _{{\mathcal {S}}_{2p}}\Vert Y\Vert _{{\mathcal {S}}_\infty [{\mathcal {S}}_q]}\Vert B\Vert _{{\mathcal {S}}_{2p}}. \end{aligned}$$

Define ${\hat{A}}:=[A ~ B^*]$ and

$$\begin{aligned} {\hat{Y}}= \frac{1}{2}\begin{bmatrix} 0 &{}\quad Y\\ Y^* &{}\quad 0 \end{bmatrix}. \end{aligned}$$

Then we have $X={\hat{A}}{\hat{Y}}{\hat{A}}^*$, and

$$\begin{aligned} \Vert {\hat{A}}\Vert _{{\mathcal {S}}_{2p}}\Vert {\hat{Y}}\Vert _{{\mathcal {S}}_\infty [{\mathcal {S}}_q]}\Vert {\hat{A}}^*\Vert _{{\mathcal {S}}_{2p}}&= \frac{1}{2} \Vert AA^*+B^*B\Vert _{{\mathcal {S}}_{p}} \Vert Y\Vert _{{\mathcal {S}}_\infty [{\mathcal {S}}_q]}\\&\le \frac{1}{2}\big ( \Vert AA^*\Vert _{{\mathcal {S}}_{p}}+\Vert B^*B\Vert _{{\mathcal {S}}_{p}}\big ) \Vert Y\Vert _{{\mathcal {S}}_\infty [{\mathcal {S}}_q]}\\&= \Vert A\Vert _{{\mathcal {S}}_{2p}}\Vert B\Vert _{{\mathcal {S}}_{2p}}\Vert Y\Vert _{{\mathcal {S}}_\infty [{\mathcal {S}}_q]}\\&\le \Vert X\Vert _{{\mathcal {S}}_p[{\mathcal {S}}_q]}+\epsilon , \end{aligned}$$

where in the third line we used $\Vert A\Vert _{{\mathcal {S}}_{2p}}= \Vert B\Vert _{{\mathcal {S}}_{2p}}$.

(ii) Following [11], when X is positive, by Hölder’s inequality we have

$$\begin{aligned} \Vert A X B\Vert _{{\mathcal {S}}_p}&\le \Vert AX^{\frac{1}{2}}\Vert _{{\mathcal {S}}_{2p}}\Vert X^{\frac{1}{2}}B\Vert _{{\mathcal {S}}_{2p}} \\&= \Vert AXA^*\Vert _{{\mathcal {S}}_{p}}^{\frac{1}{2}}\Vert B^*XB\Vert _{{\mathcal {S}}_{p}}^{\frac{1}{2}}\\&\le \max \left\{ \Vert AXA^*\Vert _{{\mathcal {S}}_{p}},\, \Vert B^*XB\Vert _{{\mathcal {S}}_{p}}\right\} . \end{aligned}$$

This shows that the supremum in (4.6) can be restricted to the case of $B=A^*$. Moreover, as argued in the proof of Theorem 4.5 we can apply polar decomposition to assume that A is positive.

To prove the assertion for (4.5) we follow similar steps as in the proof of Theorem 4.5. First, by part (i) of the remark, for any $\epsilon >0$ there exists a factorization $X=AYA^*$ such that

$$\begin{aligned} \Vert A\Vert _{{\mathcal {S}}_{2p}}^2\Vert Y\Vert _{{\mathcal {S}}_\infty [{\mathcal {S}}_q]}\le \Vert X\Vert _{{\mathcal {S}}_p[{\mathcal {S}}_q]}+\epsilon . \end{aligned}$$

Moreover, once again applying the polar decomposition we can assume that $A>0$. Let ${\hat{A}} = A^{\frac{p}{r}}$ and ${\hat{Y}} = A^{\frac{p}{q}}Y A^{\frac{p}{q}}$. Then, we have $X= {\hat{A}} {\hat{Y}}{\hat{A}}^*$ and $\Vert {\hat{A}}\Vert _{{\mathcal {S}}_{2r}}= \Vert {\hat{A}}^*\Vert _{{\mathcal {S}}_{2r}}= \Vert A\Vert _{{\mathcal {S}}_{2p}}^{\frac{p}{r}}$. Moreover, by Theorem 4.4 we have

$$\begin{aligned} \Vert Y\Vert _{{\mathcal {S}}_{\infty }[{\mathcal {S}}_q]}\ge \frac{\Vert A^{\frac{p}{q}}YA^{\frac{p}{q}}\Vert _{{\mathcal {S}}_q[{\mathcal {S}}_q]}}{\Vert A^{\frac{p}{q}}\Vert ^2_{{\mathcal {S}}_{2q}}} = \frac{\Vert {\hat{Y}}\Vert _{{\mathcal {S}}_q[{\mathcal {S}}_q]}}{\Vert A\Vert _{{\mathcal {S}}_{2p}}^{\frac{2p}{q}}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert {\hat{A}}\Vert _{{\mathcal {S}}_{2r}} \Vert {\hat{Y}}\Vert _{{\mathcal {S}}_q[{\mathcal {S}}_q]}\Vert {\hat{A}}^*\Vert _{{\mathcal {S}}_{2r}}&\le \Vert A\Vert _{{\mathcal {S}}_{2p}}^{\frac{2p}{r}}\cdot \Vert A\Vert _{{\mathcal {S}}_{2p}}^{\frac{2p}{q}}\cdot \Vert Y\Vert _{{\mathcal {S}}_{\infty }[{\mathcal {S}}_q]} \\&= \Vert A\Vert _{{\mathcal {S}}_{2p}}^{2}\cdot \Vert Y\Vert _{{\mathcal {S}}_{\infty }[{\mathcal {S}}_q]} \\&\le \Vert X\Vert _{{\mathcal {S}}_p[{\mathcal {S}}_q]}+\epsilon . \end{aligned}$$

The proof is complete. $\square$

Appendix B. Auxiliary lemmas in the proof of Theorem 3.7

1.1 B.1. A selection lemma

Suppose that $\alpha$, $\beta$ and $\gamma$ are norms, respectively, on $M_{d,n}$, $M_{n,d'}$ and $M_{d,d'}$ such that

$$\begin{aligned} \gamma (X)=\inf \left\{ \alpha (A)\cdot \beta (B):X=A B, A\in M_{d,n}, B\in M_{n,d'}\right\} . \end{aligned}$$

Our goal here is to show that almost optimizing pairs (A, B) can be chosen to be measurable functions.

Lemma B.1

Let X(s) be a continuous function on ${\mathbb {R}}$ with values in $M_{d,d'}$. Then for every $\epsilon >0$ there exist measurable functions A(s), B(s) such that for every $s\in {\mathbb {R}}$,

$$\begin{aligned}&X(s)=A(s) B(s),\\&\alpha (A(s))\le 1+\epsilon ,\\&\beta (B(s))\le \gamma (X(s)). \end{aligned}$$

To prove this lemma we make use of a special case of the Kuratowski–Ryll-Nardzewski selection theorem [41]:

Theorem B.2

(Kuratowski–Ryll-Nardzewski selection theorem) Let $cl ({\mathbb {R}}^n)$ be the set of all nonempty closed subsets of ${\mathbb {R}}^n$. Let $\psi :{\mathbb {R}}\rightarrow cl ({\mathbb {R}}^n)$ be a multi-function such that the set $\{s\in {\mathbb {R}}:\psi (s)\cap U\ne \emptyset \}$ is measurable for every closed subset U of ${\mathbb {R}}^n$. Then there is a measurable function $f:{\mathbb {R}}\rightarrow {\mathbb {R}}^n$ such that $f(s)\in \psi (s)$ for all $s\in {\mathbb {R}}$.

Proof of Lemma B.1

Based on the above selection theorem we define $\psi (s)$ to be the set of all almost optimizer pairs (A, B) satisfying

$$\begin{aligned}&X(s)=A B,\\&\alpha (A)\le 1+\epsilon ,\\&\beta (B)\le \gamma (X(s)). \end{aligned}$$

We need to show that $\psi (\cdot )$ satisfies the hypotheses of the above selection theorem. It is straightforward to show that $\psi (s)$ is non-empty and closed for every s. Thus, we need to verify that $\{s\in {\mathbb {R}}:\psi (s)\cap U\ne \emptyset \}$ is measurable. To this end, consider the map

$$\begin{aligned} \Psi :(s,A,B)\mapsto \Big (\gamma (X(s)-A B),\,\alpha (A),\,\gamma (X(s))-\beta (B)\Big ). \end{aligned}$$

Since $\Psi$ is continuous, the set

$$\begin{aligned} {\mathcal {E}}:=\Psi ^{-1}\Big (\{0\}\times [0,1+\epsilon ]\times [0,+\infty )\Big ) \end{aligned}$$

is closed. Now, if U is a closed set of pairs (A, B), then ${\mathcal {E}}\cap ({\mathbb {R}}\times U)$ is closed. Next, observe that

$$\begin{aligned} \{s\in {\mathbb {R}}:\psi (s)\cap U\ne \emptyset \}=\pi _1({\mathcal {E}}\cap ({\mathbb {R}}\times U)), \end{aligned}$$

where $\pi _1$ denotes the projection map onto the first component. Therefore, to see the measurability of $\{s\in {\mathbb {R}}:\psi (s)\cap U\ne \emptyset \}$, it suffices to prove that $\pi _1({\mathcal {E}}\cap ({\mathbb {R}}\times U))$ is measurable.

Note that any Euclidean space can be written as a countable union of compact sets. Then, since the intersection of a closed and a compact set is compact, the closed set ${\mathcal {E}}\cap ({\mathbb {R}}\times U)$ can be written as

$$\begin{aligned} {\mathcal {E}}\cap ({\mathbb {R}}\times U)=\bigcup _{i=1}^\infty K_i \end{aligned}$$

where $K_i$’s are compact. On the other hand, since a continuous image of a compact set is compact, $\pi (K_i)$’s are compact too. This implies that

$$\begin{aligned} \pi _1({\mathcal {E}}\cap ({\mathbb {R}}\times U))=\bigcup _{i=1}^\infty \pi _1(K_i) \end{aligned}$$

is a countable union of compact sets, and is therefore measurable. $\square$

1.2 B.2. A subharmonicity lemma

Let ${\mathbb {D}}$ be the unit disk in the complex plane and $\partial {\mathbb {D}}$ be its boundary. It is well-known that having a sufficiently nice function f on $\partial {\mathbb {D}}$, there is a unique harmonic function on ${\mathbb {D}}$ that matches f on the boundary. This harmonic function can be represented in terms of the Poisson kernel for ${\mathbb {D}}$, that is given by

$$\begin{aligned} P(z,e^{it}):=\Re \left( \frac{e^{it}+z}{e^{it}-z}\right) =\frac{1-|z|^2}{|e^{it}-z|^2}, \end{aligned}$$

for $z=re^{i\theta }\in {\mathbb {D}},e^{it}\in \partial {\mathbb {D}}$. In particular, if $f\in L_1(\partial {\mathbb {D}})$ and

$$\begin{aligned} F(z):=\frac{1}{2\pi }\int _{-\pi }^{\pi } f(e^{it}) P(z,e^{it})\textrm{d}t, \end{aligned}$$

(B.1)

then,

$$\begin{aligned} \lim _{r\rightarrow 1} F(re^{i\theta })=f(e^{i\theta }), \qquad \text {a.e. on} \ \partial {\mathbb {D}}. \end{aligned}$$

Recall that ${\mathbb {S}}:=\{z\in {\mathbb {C}}:0\le \text {Re}(z)\le 1\}$. The Poisson kernel for ${\mathbb {S}}$ may be obtained by carrying P over to ${\mathbb {S}}$ using the conformal map $\phi :{\mathbb {D}}\rightarrow {\mathbb {S}}$ defined by¹²

$$\begin{aligned} \phi (z)=\frac{1}{i\pi }\log \left( i\frac{1+z}{1-z}\right) . \end{aligned}$$

(B.2)

Indeed, for any holomorphic function $f:{\mathbb {S}}\rightarrow {\mathbb {C}}$, the function $f\circ \phi :{\mathbb {D}}\rightarrow {\mathbb {C}}$ is also holomorphic. Thus, using (B.1) we have

$$\begin{aligned} f(z)=\frac{1}{2\pi }\int _{-\pi }^{\pi } f\circ \phi (e^{it}) \cdot P\big (\phi ^{-1}(z),e^{it}\big )\textrm{d}t. \end{aligned}$$

We note that $\phi$ maps the boundary of ${\mathbb {D}}$ onto the boundary of ${\mathbb {S}}$. Then, letting $\phi (e^{it})=b+is$ with $b\in \{0,1\}$ and $s\in {\mathbb {R}}$, and rewriting the integral in terms of s, we find that

$$\begin{aligned} f(z)=\sum _{b=0}^1 \int _{-\infty }^{+\infty } f(b+is)\cdot \frac{P\big (\phi ^{-1}(z),\phi ^{-1}(b+is)\big )}{2\cosh (\pi s)}\textrm{d}s. \end{aligned}$$

Consequently, the Poisson kernel of ${\mathbb {S}}$ is given by the functions $P_b(x+iy,s)$ for $b=0,1$, where

$$\begin{aligned} P_b(x+iy,s)&:=\frac{P\big (\phi ^{-1}(z),\phi ^{-1}(b+is)\big )}{2\cosh (\pi s)}\\&= \frac{\sin (\pi x)}{2\big (\cosh (\pi (y-s))-\cos (\pi (x-b))\big )}. \end{aligned}$$

We note that $P_b(x+iy,s)\ge 0$ and it can be verified that

$$\begin{aligned} \int _{-\infty }^{+\infty } P_0(x+iy,s)\textrm{d}s=1-x, \qquad \quad \int _{-\infty }^{+\infty } P_1(x+iy,s)\textrm{d}s=x. \end{aligned}$$

Lemma B.3

Let $({\mathcal {X}}_0, {\mathcal {X}}_1)$ be a compatible couple and let ${\mathcal {X}}_\theta = [{\mathcal {X}}_0, {\mathcal {X}}_1]_\theta$ for $0<\theta <1$.

• (i) Let $f(z)\in {\mathscr {F}}({\mathcal {X}}_0,{\mathcal {X}}_1)$. Then we have

  $$\begin{aligned} \log \Vert f(\theta )\Vert _{{\mathcal {X}}_\theta }\le \sum _{b=0}^1 \int _{-\infty }^{+\infty } P_b(\theta , s)\log \Vert f(b+is)\Vert _{{\mathcal {X}}_b}\textrm{d}s. \end{aligned}$$

• (ii) If ${\mathcal {X}}_0, {\mathcal {X}}_1$ are finite-dimensional (or more generally, reflexive), then the same inequality holds whenever f(z) is bounded on ${\mathbb {S}}$ and holomorphic in its interior (though not necessarily continuous up to the boundary).

The first part of this lemma is proven in [7, Lemma 4.3.2]. Here, we prove the second part for the sake of completeness.

Proof of (ii)

Let ${\mathcal {X}}^*_\theta$ be the Banach dual of ${\mathcal {X}}_\theta$. Then, there is $\ell \in {\mathcal {X}}^*_\theta$ such that

$$\begin{aligned} \Vert \ell \Vert _{{\mathcal {X}}^*_\theta }=1, \qquad \Vert f(\theta )\Vert _{{\mathcal {X}}_\theta } = |\langle \ell , f(\theta )\rangle |. \end{aligned}$$

Since ${\mathcal {X}}^*_\theta =[{\mathcal {X}}^*_0, {\mathcal {X}}^*_1]_\theta$, for any $\epsilon >0$ there is g(z) such that $g(\theta )=\ell$ and

$$\begin{aligned} \max \left\{ \sup _t \Vert g(it)\Vert _{{\mathcal {X}}^*_0}, \sup _t \Vert g(1+it)\Vert _{{\mathcal {X}}^*_1}\right\} \le \Vert \ell \Vert _{{\mathcal {X}}^*_\theta }+\epsilon =1+\epsilon . \end{aligned}$$

Now define $\psi (z):= \langle g(z), f(z)\rangle$. We note that $\psi (z)$ is holomorphic and bounded. Then, by the subharmonicity of holomorphic functions we have

$$\begin{aligned} \log \Vert f(\theta )\Vert _{{\mathcal {X}}_\theta }&= \log |\psi (\theta )|\\&\le \sum _{b=0}^1 \int _{-\infty }^{+\infty } P_b(\theta , s)\log |\psi (b+is)|\textrm{d}s\\&\le \sum _{b=0}^1 \int _{-\infty }^{+\infty } P_b(\theta , s)\log \big (\Vert g(b+is)\Vert _{{\mathcal {X}}_b^*}\cdot \Vert f(b+is)\Vert _{{\mathcal {X}}_b}\big )\textrm{d}s\\&\le \sum _{b=0}^1 \int _{-\infty }^{+\infty } P_b(\theta , s)\log \big ((1+\epsilon ) \Vert f(b+is)\Vert _{{\mathcal {X}}_b}\big )\textrm{d}s\\&\le \log (1+\epsilon )+ \sum _{b=0}^1 \int _{-\infty }^{+\infty } P_b(\theta , s)\log \Vert f(b+is)\Vert _{{\mathcal {X}}_b}\textrm{d}s. \end{aligned}$$

The desired inequality follows once we note that $\epsilon >0$ is arbitrary. $\square$

Appendix C. Factorization of operator-valued functions

This section is dedicated to the Wiener–Masani theorem on factorization of operator-valued functions [12, 31, 46]. This theorem is used in the proof of Theorem 3.7, and is one of the main technical tools in the development of the theory of operator-valued Schatten spaces. Thus, we devote this section to the statement and proof of this theorem. We follow the approach of [18] for the proof, that is based on characterization of invariant subspaces of the shift operator. To this end, we first need a number of ingredients.

1.1 C.1. Hardy spaces

Following the notation of Sect. 2, let ${\mathbb {D}}$ be the unit disk in the complex plane and let $\partial {\mathbb {D}}$ be its boundary, equipped with the normalized Lebesgue measure. Then, $L_p= L_p(\partial {\mathbb {D}})$ denotes the Banach space of all measurable functions $f:\partial {{\mathbb {D}}}\rightarrow {\mathbb {C}}$ such that the norm

$$\begin{aligned} \Vert f\Vert _p:= {\left\{ \begin{array}{ll} \left( \frac{1}{2\pi }\int _{-\pi }^{\pi } |f(e^{it})|^p \textrm{d}t\right) ^\frac{1}{p}, ~~~~~\qquad \qquad 1 \le p<\infty , \\ \mathop {\mathrm {ess\,sup}}\limits |f|, ~~~~~\qquad \qquad \qquad \qquad \ \ \ p=\infty , \end{array}\right. } \end{aligned}$$

is finite. We note that by Hölder’s inequality, we have $L_q\subseteq L_p$ for $p<q$.

Definition C.1

The Hardy space $H_p$ is the closed subspace of $L_p$ consisting of all functions $f\in L_p$ for which there is a function F holomorphic in the interior of ${\mathbb {D}}$ whose radial limit coincides with f almost everywhere.¹³

There is another description of $H_p$ in terms Fourier coefficients and Fourier series that goes as follows. For each function $f\in L_1$, let

$$\begin{aligned} a_n=\frac{1}{2\pi }\int _{-\pi }^{\pi } f(e^{it})e^{-nit}\textrm{d}t, \end{aligned}$$

and

$$\begin{aligned} f(e^{it})\sim \sum _{n=-\infty }^{\infty }a_n e^{nit}, \end{aligned}$$

be the Fourier coefficients and the Fourier series of f, respectively. Then, $H_p$ is the subset of $L_p$ consisting of all functions f whose Fourier coefficients $a_n$ vanish for $n<0$.

Define the operator ${\mathfrak {S}}$ by $({\mathfrak {S}}f)(e^{it})=e^{it}f(e^{it})$. Consequently, for $n\ge 1$ we have $({\mathfrak {S}}^n f)(e^{it}) = e^{nit}f(e^{it})$. The operator ${\mathfrak {S}}$ is called the shift operator since it shifts the Fourier coefficients.

Definition C.2

A function $f\in H_2$ is said to be outer if the linear span of the set $\{{\mathfrak {S}}^n f:n\ge 0\}$ is dense in $H_2$.

It is well-known that every outer function f can be written in the form

$$\begin{aligned} f(z)=\gamma \exp \left( \frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{e^{it}+z}{e^{it}-z}k(t)\textrm{d}t\right) , \end{aligned}$$

where $\gamma$ is a complex number of modulus 1, and $k(t)=\log |f(e^{it})|$ almost everywhere; see, e.g., [29, 42] for more information on outer functions.

1.2 C.2 Vector-valued Hardy spaces

Let ${\mathcal {H}}$ be a separable (usually finite-dimensional) Hilbert space. An ${\mathcal {H}}$-valued function $F:\partial {{\mathbb {D}}}\rightarrow {\mathcal {H}}$ is said to be measurable if for every $v\in {\mathcal {H}}$ the complex function $e^{it}\mapsto \langle v,F(e^{it})\rangle$ is measurable. Abusing the notation, let $L_p({\mathcal {H}})=L_p(\partial {\mathbb {D}}, {\mathcal {H}})$ be the Banach space of all measurable functions $F:\partial {{\mathbb {D}}}\rightarrow {\mathcal {H}}$ such that the norm

$$\begin{aligned} \Vert F\Vert _p:= {\left\{ \begin{array}{ll} \left( \frac{1}{2\pi }\int _{-\pi }^{\pi } \Vert F(e^{it})\Vert ^p \textrm{d}t\right) ^\frac{1}{p}, ~~~~~\qquad \qquad 1 \le p<\infty , \\ \mathop {\mathrm {ess\,sup}}\limits \Vert F(e^{it})\Vert , ~~~~~\qquad \qquad \qquad \qquad \ \ \ p=\infty \end{array}\right. } \end{aligned}$$

is finite.¹⁴ It is worth noting that the above integral makes sense since the function $e^{it}\mapsto \Vert F(e^{it})\Vert$, as the supremum of measurable functions, is itself measurable. The case $p=2$ is of particular interest since $L_2({\mathcal {H}})$ is a Hilbert space under the inner product

$$\begin{aligned} \langle F,G\rangle =\frac{1}{2\pi }\int _{-\pi }^{\pi }\big \langle F(e^{it}),G(e^{it})\big \rangle \textrm{d}t. \end{aligned}$$

Let $\{e_n:n=1,2,\cdots \}$ is an orthonormal basis for ${\mathcal {H}}$, and for $F\in L_2({\mathcal {H}})$ define the coordinate functions by

$$\begin{aligned} f_n(e^{it}):=\big \langle e_n,F(e^{it})\big \rangle . \end{aligned}$$

(C.1)

Then, we have

$$\begin{aligned} \Vert F\Vert _2^2=\sum _n \Vert f_n\Vert _2^2, \end{aligned}$$

where $\Vert f_n\Vert _2$ is the norm of $f_n$ in $L_2=L_2(\partial {\mathbb {D}}, {\mathbb {C}})$.

For any $F\in L_p({\mathcal {H}})$, one may define the Fourier expansion of F as

$$\begin{aligned} F(e^{it})\sim \sum _{n=-\infty }^{\infty } \hat{a}_n\cdot e^{nit}, \end{aligned}$$

(C.2)

where $\hat{a}_n\in {\mathcal {H}}$ is given by

$$\begin{aligned} {\hat{a}}_n=\frac{1}{2\pi }\int _{-\pi }^{\pi } F(e^{it})e^{-nit}\textrm{d}t. \end{aligned}$$

We note that letting ${\mathcal {H}}={\mathbb {C}}$, we obtain the usual $L_p=L_p(\partial {\mathbb {D}}, {\mathbb {C}})$ space with the usual Fourier transform.

Definition C.3

Let ${\mathcal {H}}$ be a (separable) Hilbert space and $1\le p\le \infty$. The vector-valud Hardy space $H_p({\mathcal {H}})$ is defined to be the closed subspace of $L_p({\mathcal {H}})$ consisting of all functions F for which $\hat{a}_n=0$, for all $n<0$.

Similar to the usual complex-valued Hardy space, $H_p({\mathcal {H}})$ can also be characterized in terms of holomorphic functions. We say that a vector-valued function $\phi : {\mathbb {D}}\rightarrow {\mathcal {H}}$ is holomorphic if for every $v\in {\mathcal {H}}$, the function $z\mapsto \langle v,\phi (z)\rangle$ is holomorphic. Then, $H_p({\mathcal {H}})$ is the space of functions $F\in L_p({\mathcal {H}})$ for which there is a holomorphic function $\phi : {\mathbb {D}}\rightarrow {\mathcal {H}}$ whose radial limit coincides with F almost everywhere, i.e.,

$$\begin{aligned} F(e^{it}) = \lim _{r\rightarrow 1^+} \phi (re^{it}), \qquad \text {a.e.} \end{aligned}$$

1.3 C.3 Invariant subspaces

The shift operator ${\mathfrak {S}}$ can also be defined on the space of vector-valued functions: $({\mathfrak {S}}F)(e^{it})=e^{it}F(e^{it})$.

Definition C.4

A closed subspace ${\mathcal {M}}$ of $L_2({\mathcal {H}})$ is said to be invariant if ${\mathfrak {S}}({\mathcal {M}})\subseteq {\mathcal {M}}$, where ${\mathfrak {S}}({\mathcal {M}}):=\{{\mathfrak {S}}F:\, F\in {\mathcal {M}}\}$. We say that ${\mathcal {M}}$ is doubly invariant if ${\mathfrak {S}}({\mathcal {M}})\subseteq {\mathcal {M}}$ and ${\mathfrak {S}}^{-1}({\mathcal {M}})\subseteq {\mathcal {M}}$. Also, ${\mathcal {M}}$ is said to be simply invariant if it is invariant but not doubly invariant.

We may also consider operator-valued (matrix-valued) functions. We say that such a function $U(e^{it})$ is measurable if the function $e^{it}\mapsto \langle v,U(e^{it}) w\rangle$ is measurable for all vectors v, w.

Lemma C.5

Let ${\mathcal {M}}$ be a simply invariant subspace of $L_2({\mathcal {H}})$ that contains no nontrivial doubly invariant subspace. Then there exist an auxiliary Hilbert space ${\mathcal {K}}$ and a measurable operator-valued function U such that

• (i) $U(e^{it}):{\mathcal {K}}\rightarrow {\mathcal {H}}$ is an isometry for any $e^{it}\in \partial {{\mathbb {D}}}$.

• (ii) ${\mathcal {M}}$ consists of functions $e^{it}\mapsto U(e^{it})F(e^{it})$ for $F\in H_2({\mathcal {K}})$.

We note that vector-valued Hardy spaces $H_2({\mathcal {K}})$ are simply invariant subspaces and do not contain any nontrivial doubly invariant subspace. The above lemma says that all such subspaces are essentially vector-valued Hardy spaces.

Proof

For each n, define ${\mathcal {M}}_n:={\mathfrak {S}}^n({\mathcal {M}})$. As ${\mathcal {M}}$ is invariant, we have

$$\begin{aligned} {\mathcal {M}}={\mathcal {M}}_0\supseteq {\mathcal {M}}_1\supseteq {\mathcal {M}}_2\supseteq \cdots . \end{aligned}$$

We note that

$$\begin{aligned} {\mathcal {M}}_\infty =\bigcap \limits _{n\ge 0}{\mathcal {M}}_n \end{aligned}$$

is doubly invariant, so by assumption ${\mathcal {M}}_\infty =0$.

For any $n\ge 0$, let ${\mathcal {N}}_n$ be the orthogonal complement of ${\mathcal {M}}_{n+1}$ in ${\mathcal {M}}_n$ (with respect to the inner product of $L_2({\mathcal {H}})$). Thus, we have ${\mathcal {M}}_n={\mathcal {N}}_n\oplus {\mathcal {M}}_{n+1}$. Moreover, since ${\mathcal {M}}_\infty =0$, we have

$$\begin{aligned} {\mathcal {M}}={\mathcal {N}}_0\oplus {\mathcal {N}}_1\oplus {\mathcal {N}}_2\oplus \cdots . \end{aligned}$$

(C.3)

Let ${\mathcal {B}}_0=\{E_1,E_2,\cdots \}$ be an orthonormal basis for ${\mathcal {N}}_0$, and note that ${\mathcal {B}}_n:={\mathfrak {S}}^n({\mathcal {B}}_0)=\{{\mathfrak {S}}^n E_k:k\ge 1\}$, for $n\ge 1$, is a subset of ${\mathcal {M}}_n\subseteq {\mathcal {M}}_1$. Thus, ${\mathcal {B}}_0$ is orthogonal to ${\mathcal {B}}_n$ for all $n\ge 1$. This means that for every j, k and $n\ge 1$ we have

$$\begin{aligned} \langle E_j,{\mathfrak {S}}^n E_k\rangle =\frac{1}{2\pi }\int _{-\pi }^{\pi } e^{nit}\big \langle E_j(e^{it}),E_k(e^{it})\big \rangle \textrm{d}t=0. \end{aligned}$$

Moreover, as ${\mathcal {B}}_0$ is orthonormal, the same equation holds when $n=0$ and $j\ne k$. More generally, we observe that

$$\begin{aligned} \langle {\mathfrak {S}}^n E_j,{\mathfrak {S}}^m E_k\rangle =\frac{1}{2\pi }\int _{-\pi }^{\pi } e^{(m-n)it}\big \langle E_j(e^{it}),E_k(e^{it})\big \rangle dt=\delta _{jk}\cdot \delta _{mn}. \end{aligned}$$

This equation shows that all the nontrivial Fourier coefficients of the functions $e^{it}\mapsto \big \langle E_j(e^{it}),E_k(e^{it})\big \rangle$ are zero, and hence it is almost everywhere identical to the constant function $\delta _{jk}$.

We claim that for each n, the orthonormal set ${\mathcal {B}}_n$ is a basis for ${\mathcal {N}}_n$. To this end, we need to show that ${\mathcal {B}}_n\subseteq {\mathcal {N}}_n=\mathop {\textrm{span}}\limits {\mathcal {B}}_n$. Letting $F\in {\mathcal {M}}_{n+1}$, by definition there is $G\in {\mathcal {M}}$ such that $F={\mathfrak {S}}^{n+1} G$. On the other hand, for ${\mathfrak {S}}^n E_j\in {\mathcal {B}}_n$ we have

$$\begin{aligned} \langle {\mathfrak {S}}^n E_j,F\rangle =\langle {\mathfrak {S}}^n E_j,{\mathfrak {S}}^{n+1}G\rangle =\langle E_j,{\mathfrak {S}}G\rangle =0, \end{aligned}$$

where we used the fact that ${\mathfrak {S}}G\in {\mathcal {M}}_1$ and that $E_j\in {\mathcal {N}}_0$. Thus, ${\mathfrak {S}}^nE_j\in {\mathcal {M}}_n$ and is orthogonal to ${\mathcal {M}}_{n+1}$, which means that ${\mathfrak {S}}^n E_j\in {\mathcal {N}}_n$. Therefore, ${\mathcal {B}}_n\subseteq {\mathcal {N}}_n$.

Next, let $F\in {\mathcal {M}}_n$. Then, there is $G\in {\mathcal {M}}$ such that $F={\mathfrak {S}}^n G$. Since $G\in {\mathcal {M}}={\mathcal {N}}_0\oplus {\mathcal {M}}_1$, there are $G_0\in {\mathcal {N}}_0$ and $G_1\in {\mathcal {M}}$ such that $G=G_0+{\mathfrak {S}}G_1$. Therefore

$$\begin{aligned} F={\mathfrak {S}}^n G_0+{\mathfrak {S}}^{n+1}G_1. \end{aligned}$$

By definition we have ${\mathfrak {S}}^{n+1}G_1\in {\mathcal {M}}_{n+1}$. Also, since $G_0\in {\mathcal {N}}_0=\mathop {\textrm{span}}\limits {\mathcal {B}}_0$, we have ${\mathfrak {S}}^n G_0\in \mathop {\textrm{span}}\limits {\mathcal {B}}_n$. As a result, $F\in \mathop {\textrm{span}}\limits {\mathcal {B}}_n\oplus {\mathcal {M}}_{n+1}$ for any $F\in {\mathcal {M}}_n$. Thus, using ${\mathcal {B}}_n\subseteq {\mathcal {N}}_n$ proven above, and the definition of ${\mathcal {N}}_n$, we find that ${\mathcal {N}}_n=\mathop {\textrm{span}}\limits {\mathcal {B}}_n$.

Putting (C.3) and ${\mathcal {N}}_n=\mathop {\textrm{span}}\limits {\mathcal {B}}_n$ together, we find that every $F\in {\mathcal {M}}$ can be expanded as

$$\begin{aligned} F=\sum _j \sum _{n=0}^\infty a_{nj}{\mathfrak {S}}^n E_j, \ \ \ \quad \sum _{n,j} |a_{nj}|^2<\infty . \end{aligned}$$

(C.4)

Now, let ${\mathcal {K}}$ be a Hilbert space isomorphic to ${\mathcal {N}}_0$ with orthonormal basis $\{e_1,e_2,\cdots \}$. Every function in $L_2({\mathcal {K}})$ can be written in the form $\sum f_j e_j$ with $f_j\in L_2$, $j=1, 2,\ldots$, being coordinate functions (see (C.1)). Define $\hat{U}:L_2({\mathcal {K}})\rightarrow L_2({\mathcal {H}})$ by

$$\begin{aligned} \hat{U}\Big (\sum f_j e_j\Big )=\sum f_j E_j, \end{aligned}$$

meaning that $\hat{U}\Big (\sum f_j e_j\Big )(e^{it})=\sum f_j(e^{it}) E_j(e^{it})$. Then, $\hat{U}$ is a well-defined linear operator. Moreover, since $\langle E_j(e^{it}),E_k(e^{it})\rangle =\delta _{jk}$, almost everywhere, $\hat{U}$ is an isometry.

For every $e^{it}$ we define a linear map $U(e^{it}):{\mathcal {K}}\rightarrow {\mathcal {H}}$ as follows. For every $v\in {\mathcal {K}}$ let ${\hat{v}}\in L_2({\mathcal {K}})$ be the constant function $\hat{v}(e^{it})=v$. Then, define

$$\begin{aligned} U(e^{it})v:=(\hat{U}\hat{v})(e^{it}). \end{aligned}$$

It is easy to verify that $e^{it}\mapsto U(e^{it})$ is measurable. We claim that $U(e^{it})$ is an isometry for almost every t. To show this, fix $v, w\in {\mathcal {K}}$. Using the fact that $\hat{U}$ is an isometry, for every n we obtain

$$\begin{aligned} \big \langle \hat{U}\hat{v},\hat{U}({\mathfrak {S}}^n\hat{w})\big \rangle&=\langle \hat{v},{\mathfrak {S}}^n\hat{w}\rangle \nonumber \\&=\frac{1}{2\pi }\int _{-\pi }^{\pi } e^{nit}\langle v, w\rangle \textrm{d}t \nonumber \\&=\frac{1}{2\pi }\langle v, w\rangle \int _{-\pi }^{\pi } e^{nit}\textrm{d}t\nonumber \\&=\delta _{n0}\langle v, w\rangle . \end{aligned}$$

(C.5)

On the other hand, the definition of $\hat{U}$ implies that $\hat{U}({\mathfrak {S}}^n \hat{w})={\mathfrak {S}}^n\hat{U}(\hat{w})$. Therefore,

$$\begin{aligned} \langle \hat{U}\hat{v},\hat{U}({\mathfrak {S}}^n\hat{w})\rangle&=\langle \hat{U}\hat{v}, {\mathfrak {S}}^n\hat{U}(\hat{w})\rangle \\&=\frac{1}{2\pi }\int _{-\pi }^{\pi } e^{nit}\langle \hat{U}\hat{v}(e^{it}),\hat{U}\hat{w}(e^{it})\rangle \textrm{d}t. \end{aligned}$$

Consequently, comparing to the previous equation, all the nontrivial Fourier coefficients of the function $e^{it}\mapsto \langle \hat{U}\hat{v}(e^{it}),\hat{U}\hat{w}(e^{it})\rangle =\langle U(e^{it})v, U(e^{it})w\rangle$ vanish. This means that this function is constant almost everywhere. Thus, we have, almost everywhere,

$$\begin{aligned} \langle U(e^{it})v, U(e^{it})w\rangle&=\frac{1}{2\pi }\int _{-\pi }^{\pi }\langle U(e^{is})v, U(e^{is})w\rangle \textrm{d}s\\&=\frac{1}{2\pi }\int _{-\pi }^{\pi }\langle \hat{U}\hat{v}(e^{is}), \hat{U}\hat{w}(e^{is})\rangle \textrm{d}s\\&=\langle \hat{U}\hat{v}, \hat{U}\hat{w}\rangle \\&=\langle v, w\rangle , \end{aligned}$$

where in the last line we use (C.5). This means that $U(e^{it})$ is an isometry almost everywhere (hence one can modify $U(e^{it})$ on a set of measure zero so that $U(e^{it})$ is an isometry for any $e^{it}$).

We note that $H_2({\mathcal {K}}) = \left\{ \sum f_je_j:\, f_j\in H_2, \Vert f_j\Vert _2^2<\infty \right\}$, and every $f_j\in H_2$ can be written as $f_j(e^{it})=\sum _{n\ge 0} a_{nj}e^{nit}$. Therefore, (C.4) yields

$$\begin{aligned} \hat{U}H_2({\mathcal {K}})&=\left\{ \sum _j f_j E_j:f_j\in H_2({\mathbb {C}}),\sum \Vert f_j\Vert _2^2<\infty \right\} \\&=\left\{ \sum _j \sum _{n\ge 0}a_{nj}{\mathfrak {S}}^nE_j:\sum |a_{nj}|^2<\infty \right\} \\&={\mathcal {M}}. \end{aligned}$$

Finally, from the definition of U it follows that $U(e^{it})e_j=(\hat{U}e_j)(e^{it})=E_j(e^{it})$. Hence, for every $F=\sum _j f_je_j\in H_2({\mathcal {K}})$ we obtain

$$\begin{aligned} (\hat{U}F)(e^{it})&=\Big (\sum f_j E_j\Big )(e^{it})\\&=\sum f_j(e^{it}) E_j(e^{it})\\&=\sum f_j(e^{it}) U(e^{it})e_j\\&=U(e^{it})\Big (\sum f_j(e^{it})e_j\Big )\\&=U(e^{it})F(e^{it}). \end{aligned}$$

That is, ${\mathcal {M}}$ consists of functions of the form $e^{it}\mapsto U(e^{it})F(e^{it})$ for $F\in H_2({\mathcal {K}})$. $\square$

Remark C.6

In the above lemma, if ${\mathcal {H}}$ is finite-dimensional, then so is ${\mathcal {K}}$ since $U(e^{it}):{\mathcal {K}}\rightarrow {\mathcal {H}}$ is an isometry almost everywhere.

1.4 C.4 Wiener–Masani factorization theorem

In the following we assume that ${\mathcal {H}}$ is finite-dimensional with orthonormal basis $\{e_1, \ldots , e_d\}$. For every $e^{it}\in \partial {\mathbb {D}}$ we assume that $T(e^{it}):{\mathcal {H}}\rightarrow {\mathcal {H}}$ is a linear map such that $e^{it}\mapsto T(e^{it})$ is an operator-valued measurable function. We further assume that entries of T (in the basis $\{e_1, \ldots , e_d\}$) are integrable, i.e., belong to $L_1$.

Suppose that $T(e^{it})$ is positive, and uniformly bounded almost everywhere, i.e., there exists a constant $c>0$ such that

$$\begin{aligned} 0\le T(e^{it})\le cI, \qquad \text { almost everywhere}. \end{aligned}$$

Let $Q(e^{it})=\sqrt{T(e^{it})}$. We note that with the above assumptions Q can be written as the limit of measurable functions (consider the truncated Taylor expansion of $x\mapsto \sqrt{x}$ at point c), and is measurable itself. We also note that Q can be considered as an element of $B(L_2({\mathcal {H}}))$. Indeed, for every $F\in L_2({\mathcal {H}})$ we have $(QF)(e^{it})=Q(e^{it})F(e^{it})$ and

$$\begin{aligned} \Vert QF\Vert _2^2&=\frac{1}{2\pi }\int _{-\pi }^{\pi }\big \langle Q(e^{it})F(e^{it}),Q(e^{it})F(e^{it})\big \rangle \textrm{d}t\\&=\frac{1}{2\pi }\int _{-\pi }^{\pi }\langle F(e^{it}),T(e^{it})F(e^{it})\rangle \textrm{d}t\\&\le \frac{1}{2\pi }\int _{-\pi }^{\pi }c\langle F(e^{it}),F(e^{it})\rangle \textrm{d}t\\&=c\Vert F\Vert _2^2. \end{aligned}$$

Define

$$\begin{aligned} {\mathcal {M}}(T)=\overline{\mathop {\textrm{span}}\limits \{{\mathfrak {S}}^n (Qe_j):n\ge 0,\, j=1,\ldots , d\}}, \end{aligned}$$

where $Qe_j\in L_2({\mathcal {H}})$ is given by $(Qe_j)(e^{it}) = Q(e^{it})e_j$. Observe that by definition, ${\mathcal {M}}(T)$ is a closed and invariant subspace of $L_2({\mathcal {H}})$. In fact, ${\mathcal {M}}(T)$ is the smallest invariant subspace of $L_2({\mathcal {H}})$ generated by columns of Q.

Lemma C.7

Let $T(e^{it})$ be a measurable operator-valued function as above, and $\lambda (e^{it})$ be a positive measurable function such that

$$\begin{aligned} \log \lambda (e^{it})\in L_1, \end{aligned}$$

(C.6)

and

$$\begin{aligned} 0<\lambda (e^{it})I\le T(e^{it})\le cI, \qquad \hbox { a.e. on}\ \partial {{\mathbb {D}}}, \end{aligned}$$

for some $c>0$. Then ${\mathcal {M}}(T)$ does not contain any nontrivial doubly invariant subspace.

Proof

Observe that by (C.6) there is $\psi \in H_2$ such that $|\psi (e^{it})|=\sqrt{\lambda (e^{it})}$. To construct such a function we use the Poisson kernel of ${\mathbb {D}}$ to define

$$\begin{aligned} g(z):= \frac{1}{2\pi } \int _{-\pi }^\pi \Re \frac{e^{it}+z}{e^{it}-z} \log \sqrt{\lambda (e^{it})} \textrm{d}t. \end{aligned}$$

We note that $g(e^{it})=\log \sqrt{\lambda (e^{it})}$ almost everywhere, and that g(z) is harmonic. Then, considering its harmonic conjugate, there is a holomorphic function ${\hat{g}}(z)$ with $\Re {\hat{g}}(z) = g(z)$. Now, letting $\psi (z) = e^{{\hat{g}}(z)}$, we have almost everywhere

$$\begin{aligned} |\psi (e^{it})| = |e^{{\hat{g}}(it)}| = e^{\Re {\hat{g}}(it)} = e^{g(it)} = \sqrt{\lambda (e^{it})}. \end{aligned}$$

We say that $P\in H_2({\mathcal {H}})$ is a polynomial if its Fourier expansion (C.2) consists of finitely many $n\ge 0$ terms, i.e., P is a polynomial in variable $e^{it}$ with coefficients in ${\mathcal {H}}$. For such a polynomial $P\in H_2({\mathcal {H}})$ we have $\psi P\in H_2({\mathcal {H}})$ where $(\psi P)(z) = \psi (z)P(z)$. Define

$$\begin{aligned} {\mathcal {N}}=\overline{\mathop {\textrm{span}}\limits \{\psi P: P \text { polynomial}\}}. \end{aligned}$$

Observe that ${\mathcal {N}}\subseteq H_2({\mathcal {H}})$ is an invariant subspace.

Next, consider the subspace ${\mathcal {W}}=\mathop {\textrm{span}}\limits \{Q P:\, P \text { polynomial}\}\subseteq L_2({\mathcal {H}})$ where as before $Q=\sqrt{T}$. We note that by definition $\overline{{\mathcal {W}}}= {\mathcal {M}}(T)$. Define U from ${\mathcal {W}}$ to ${\mathcal {N}}$ by $U(Q P)=\psi P$. Since $\sqrt{\lambda (e^{it})} I\le Q(e^{it})$ almost everywhere, we have

$$\begin{aligned} \Vert \psi P\Vert ^2_2&= \frac{1}{2\pi }\int _{-\pi }^\pi \big \Vert \psi (e^{it}) P(e^{it}) \big \Vert ^2\\&= \frac{1}{2\pi }\int _{-\pi }^\pi \lambda (e^{it}) \big \Vert P(e^{it}) \big \Vert ^2\\&\le \frac{1}{2\pi }\int _{-\pi }^\pi \big \Vert (QP)(e^{it}) \big \Vert ^2\\&= \Vert QP\Vert ^2_2. \end{aligned}$$

This means that $\Vert U\Vert \le 1$, and U can be extended to $U:{\mathcal {M}}(T)\rightarrow {\mathcal {N}}$ while $\Vert U\Vert \le 1$.

We claim that U is one-to-one. To prove this, suppose that $F\in {\mathcal {M}}(T)$ and $UF=0$. Since $F\in {\mathcal {M}}(T)$, there is a sequence of polynomials $\{P_n:\, n\ge 1\}\subset H_2({\mathcal {H}})$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }Q P_n=F \ \ \ \qquad \text {(in } L^2({\mathcal {H}}) {\text{) }}, \end{aligned}$$

and since $QF=0$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\psi P_n=0 \ \ \ \qquad \text {(in } L^2({\mathcal {H}}) \text { )}. \end{aligned}$$

Define $R_m=\{e^{it}:\lambda (e^{it})\ge 1/m\}$, and let $J_m$ be the characteristic function of $R_m$. We compute

$$\begin{aligned} 0&\le \frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{1}{m} J_m(e^{it})\big \Vert P_n(e^{it})\big \Vert ^2\,\textrm{d}t\\&\le \frac{1}{2\pi }\int _{-\pi }^{\pi } \lambda (e^{it})J_m(e^{it})\big \Vert P_n(e^{it})\big \Vert ^2\,\textrm{d}t\\&\le \frac{1}{2\pi }\int _{-\pi }^{\pi } \lambda (e^{it})\big \Vert P_n(e^{it})\big \Vert ^2\,\textrm{d}t\\&=\Vert \psi P_n\Vert _2^2. \end{aligned}$$

Therefore

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{2\pi }\int _{-\pi }^{\pi } J_m(e^{it})\big \Vert P_n(e^{it})\big \Vert _2^2\,\textrm{d}t=0, \ \ \ \quad \forall m. \end{aligned}$$

(C.7)

On the other hand, we have

$$\begin{aligned}&\big \Vert F-QP_n\big \Vert _2^2\\&\quad =\frac{1}{2\pi }\int _{-\pi }^{\pi } \big \Vert F(e^{it})-Q(e^{it})P_n(e^{it})\big \Vert ^2\,\textrm{d}t\\&\qquad \ge \frac{1}{2\pi }\int _{-\pi }^{\pi } J_m(e^{it})\big \Vert F(e^{it})-Q(e^{it})P_n(e^{it})\big \Vert ^2\,\textrm{d}t\\&\quad \ge \frac{1}{2\pi }\int _{-\pi }^{\pi } J_m(e^{it})\Big (\Vert F(e^{it})\big \Vert -\Vert Q(e^{it})P_n(e^{it})\big \Vert \Big )^2\,\textrm{d}t\\&\quad =\frac{1}{2\pi }\int _{-\pi }^{\pi } J_m(e^{it})\Big (\big \Vert F(e^{it})\big \Vert ^2+\big \Vert Q(e^{it})P_n(e^{it})\big \Vert ^2\\&\qquad -2\big \Vert F(e^{it})\big \Vert \cdot \big \Vert Q(e^{it})P_n(e^{it})\big \Vert \Big )\textrm{d}t\\&\quad \ge \frac{1}{2\pi }\int _{-\pi }^{\pi } J_m(e^{it})\big \Vert F(e^{it})\big \Vert ^2\,\textrm{d}t\\&\qquad -\frac{1}{\pi }\left( \int _{-\pi }^{\pi } \big \Vert F(e^{it})\big \Vert ^2\,\textrm{d}t \int _{-\pi }^{\pi } J_m(e^{it})\big \Vert Q(e^{it})P_n(e^{it})\big \Vert ^2\,\textrm{d}t\right) ^{\frac{1}{2}}\\&\quad \ge \frac{1}{2\pi }\int _{-\pi }^{\pi } J_m(e^{it})\big \Vert F(e^{it})\big \Vert ^2\,\textrm{d}t-2\Vert F\Vert _2\cdot \frac{c}{2\pi }\int _{-\pi }^{\pi } J_m(e^{it})\big \Vert P_n(e^{it})\big \Vert ^2\,\textrm{d}t, \end{aligned}$$

where the penultimate inequality follows by ignoring a non-negative term and the Cauchy-Schwarz inequality, and the last inequality follows from $Q(e^{it})\le \sqrt{c}I$. Taking the limit of both sides as $n\rightarrow \infty$, using (C.7) and the non-negativity of the remaining term, we obtain

$$\begin{aligned} \Vert J_m F\Vert _2^2=\frac{1}{2\pi }\int _{-\pi }^{\pi } J_m(e^{it})\big \Vert F(e^{it})\big \Vert ^2\,\textrm{d}t=0, \ \ \ \quad \forall m. \end{aligned}$$

Now, the monotone convergence theorem implies $\Vert F\Vert _2^2=\lim \limits _{m\rightarrow \infty } \Vert J_mF\Vert _2^2=0$. This means that $F=0$. Thus, $U:{\mathcal {M}}(T)\rightarrow {\mathcal {N}}$ is one-to-one, as claimed.

Finally, suppose that ${\mathcal {M}}'\subseteq {\mathcal {M}}(T)$ is a nontrivial doubly invariant subspace. Then $\overline{U{\mathcal {M}}'}\subseteq {\mathcal {N}}\subseteq H_2({\mathcal {H}})$ is a doubly invariant subspace. But $H_2({\mathcal {H}})$ does not contain nontrivial doubly invariant subspaces. This implies that $\overline{U{\mathcal {M}}'}=0$. Hence, ${\mathcal {M}}'=0$ since U is one-to-one. $\square$

Similar to the scalar case (see Definition C.2), an operator-valued function A(z) is called outer if $\{A(z)P(z):\, P(z)\in H_2({\mathcal {H}}) \text { polynomial}\}$ is dense in $H_2({\mathcal {H}})$. Equivalently, A(z) is outer if

$$\begin{aligned} \overline{\mathop {\textrm{span}}\limits \{{\mathfrak {S}}^n (Ae_j):\, n\ge 0,\, j=1,\ldots , d\}} = H_2({\mathcal {H}}). \end{aligned}$$

Lemma C.8

[29] Let A(z) be an outer operator-valued function in $H_\infty ({\mathcal {H}})$. If $A(e^{it})$ is invertible almost everywhere on $\partial {{\mathbb {D}}}$, and

$$\begin{aligned} \log \Vert A^{-1}(e^{it})\Vert \in L_1, \end{aligned}$$

(C.8)

then A(z) is invertible for all z in the interior of ${\mathbb {D}}$.

Proof

Since $\Vert A^{-1}(e^{it})\Vert ^{-1} \le \Vert A(e^{it})\Vert$ almost everywhere on $\partial {{\mathbb {D}}}$ and $A(z) \in H_\infty ({\mathcal {H}})$, it is evident that the function ${\tilde{\lambda }}(e^{it}):=\Vert A^{-1}(e^{it})\Vert ^{-1}$ lies in $L_\infty$. On the other hand, it follows from (C.8) that $\log {\tilde{\lambda }}(e^{it})\in L_1$. These facts imply that the outer function $\lambda (z)$, defined by

$$\begin{aligned} \lambda (z):=\exp \left( \frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{e^{it}+z}{e^{it}-z}\log {\tilde{\lambda }}(e^{it})\textrm{d}t\right) , \end{aligned}$$

lies in $H_\infty$, and

$$\begin{aligned} \big \Vert \lambda (e^{it}) A^{-1}(e^{it})\big \Vert =1, \qquad \hbox { a.e.\ on}\ \partial {{\mathbb {D}}}. \end{aligned}$$

(C.9)

Let h be an arbitrary but fixed vector in ${\mathcal {H}}$, and define $\psi _h(e^{it})=\overline{\lambda (e^{it})} A^{-1}(e^{it})^*h$. Then, $\psi _h \in L_2({\mathcal {H}})$ since $\Vert \psi _h\Vert _{L_2({\mathcal {H}})} \le \Vert h\Vert _{\mathcal {H}}$ by (C.9).

Now let $\Psi _h$ be the bounded linear functional on $L_2({\mathcal {H}})$ corresponding to $e^{-it}\psi _h(e^{it}) \in L_2({\mathcal {H}})$, i.e.,

$$\begin{aligned} \Psi _h(F)&=\big \langle e^{-it}\psi _h(e^{it}), F(e^{it}) \big \rangle _{L_2({\mathcal {H}})}\\&=\frac{1}{2\pi }\int _{-\pi }^{\pi } e^{it}\big \langle \psi _h(e^{it}), F(e^{it}) \big \rangle _{{\mathcal {H}}}\textrm{d}t. \end{aligned}$$

We are going to show that $\Psi _h$ vanishes on $H_2({\mathcal {H}})$. To this end, note first that the last expression above is indeed a Fourier coefficient, with negative index, of the function $\big \langle \psi _h(e^{it}), F(e^{it}) \big \rangle _{{\mathcal {H}}} \in H_2$. As a result, for every function $u(z) \in H_2({\mathcal {H}})$ we have

$$\begin{aligned} \Psi _h(F)=\frac{1}{2\pi }\int _{-\pi }^{\pi } e^{it}\big \langle \overline{\lambda (e^{it})}h, u(e^{it})\big \rangle _{{\mathcal {H}}}\textrm{d}t=0, \end{aligned}$$

where $F(z)=A(z)u(z)$, since $\big \langle \overline{\lambda (e^{it})}h, u(e^{it}) \big \rangle _{{\mathcal {H}}} \in H_2$. Thus, $\Psi _h$ vanishes on $A(z) H_2({\mathcal {H}})$, which in turn implies that $\Psi _h$ vanishes on the entire $H_2({\mathcal {H}})$ as by assumption A(z) is outer.

Let g be an arbitrary (but fixed) vector in ${\mathcal {H}}$, and $z \notin {\mathbb {D}}$. Then, the vector-valued function $F^{g}_z(e^{it}):=(e^{it}-z)^{-1}g$ is readily seen to be in $H_2({\mathcal {H}})$. Hence, by the preceding paragraph $\Psi _h(F^{g}_z)=0$, or,

$$\begin{aligned} \frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{\lambda (e^{it})\big \langle h, A^{-1}(e^{it})g \big \rangle _{{\mathcal {H}}}}{e^{it}-z}e^{it}\textrm{d}t=0. \end{aligned}$$

Consequently, letting $B(e^{it})=A^{-1}(e^{it})$, the function $\lambda (e^{it})\big \langle h, A^{-1}(e^{it})g \big \rangle _{{\mathcal {H}}}$ extends to a function $\lambda (z)\langle h, B(z)g \rangle _{{\mathcal {H}}}$ in $H_\infty$ via Poisson integral. Finally, since $\lambda (z)$ is outer, we have $\lambda (z)\ne 0$ for all z in the interior of ${\mathbb {D}}$, and hence the function $\langle h, B(z)g \rangle _{{\mathcal {H}}}$ is holomorphic as well. It is easily verified that indeed $B(z)=A^{-1}(z)$ for all z in the interior of ${\mathbb {D}}$. $\square$

We can now state and prove the Wiener–Masani theorem.

Theorem C.9

(Wiener–Masani) Let $T(e^{it})$ be a measurable operator-valued function as above, and $\lambda (e^{it})$ be a positive measurable function such that

$$\begin{aligned} \log \lambda (e^{it})\in L_1, \end{aligned}$$

(C.10)

and

$$\begin{aligned} 0<\lambda (e^{it})I\le T(e^{it})\le cI, \qquad \hbox { a.e. on}\ \partial {{\mathbb {D}}}, \end{aligned}$$

(C.11)

for some $c>0$. Then there exists an outer operator-valued function $A(z)\in H_\infty ({\mathcal {H}})$ such that A(z) is invertible for all z in the interior of ${\mathbb {D}}$, and

$$\begin{aligned} T(e^{it})=A^*(e^{it})A(e^{it}), \qquad \hbox { a.e. on}\ \partial {{\mathbb {D}}}. \end{aligned}$$

Proof

By Lemma C.7, ${\mathcal {M}}(T)$ does not contain any doubly invariant subspace. Therefore, by Lemma C.5, there exist a Hilbert space ${\mathcal {K}}$ and a function $U(e^{it}):{\mathcal {K}}\rightarrow {\mathcal {H}}$, that is an isometry almost everywhere, such that ${\mathcal {M}}(T)=UH_2({\mathcal {K}})$.

Recall that $Qe_j\in {\mathcal {M}}(T)$ for every basis element $e_j\in {\mathcal {H}}$. Then, for each j, there is $A_j\in H_2({\mathcal {K}})$ such that $Qe_j=UA_j$. Define $A(e^{it}): {\mathcal {H}}\rightarrow {\mathcal {K}}$ by $A(e^{it})e_j=A_j(e^{it})$. Observe that A is a holomorphic operator-valued function since $A_j\in H_2({\mathcal {K}})$. Moreover, using the fact that U is an isometry almost everywhere, we have

$$\begin{aligned} A^*(e^{it}) A(e^{it})&=A^*(e^{it}) U^*(e^{it}) U(e^{it})A(e^{it}) \nonumber \\&=Q^*(e^{it}) Q(e^{it}) \nonumber \\&=T(e^{it}), \qquad \hbox { a.e.\ on}\ \partial {{\mathbb {D}}}. \end{aligned}$$

(C.12)

Now we show that $A(e^{it})$ is outer. Let $F\in H_2({\mathcal {K}})$ be arbitrary. Then, $UF\in UH_2({\mathcal {K}})= {\mathcal {M}}(T)$, so for every $n\ge 1$ there is a polynomial $P_n$ such that $\Vert QP_n- UF\Vert _2\le 1/n$. Next, using the fact that U is an isometry almost everywhere we have

$$\begin{aligned} \Vert AP_n- F\Vert _2= \Vert UAP_n- UF\Vert _2=\Vert QP_n- UF\Vert _2\le 1/n. \end{aligned}$$

This shows that $\{AP: P \text { polynomial}\}$ is dense in ${\mathcal {H}}_2({\mathcal {K}})$ and A is outer.

It follows from (C.12) and our assumptions on $T(e^{it})$ that $A(e^{it})$ is invertible almost everywhere on $\partial {{\mathbb {D}}}$, and $\Vert A^{-1}(e^{it})\Vert ^2=\Vert T^{-1}(e^{it})\Vert$. On the other hand, by (C.11) we have

$$\begin{aligned} 2\log \Vert A^{-1}(e^{it})\Vert&=\log \Vert T^{-1}(e^{it})\Vert \\&\le \log \lambda (e^{it})^{-1}, \qquad \hbox { a.e. on}\ \partial {{\mathbb {D}}}. \end{aligned}$$

This, together with (C.10), yields that $\log \Vert A^{-1}(e^{it})\Vert \in L_1$. Finally, Lemma C.8 is applied to conclude that A(z) is invertible for all z in the interior of ${\mathbb {D}}$. $\square$

The Wiener–Masani theorem can also be stated for operator-valued functions defined on the boundary of the strip ${\mathbb {S}}$. To this end, having measurable operator-valued functions $T_b(s)\in B({\mathcal {H}})$, for $b=0,1$, we may use the conformal map (B.2) to transform them into measurable operator-valued function on $\partial {\mathbb {D}}$. Then, using the Wiener–Masani theorem, finding an appropriate holomorphic operator-valued function A, and pulling it back as a function on ${\mathbb {S}}$ through the conformal map, the desired factorization of $T_b$, $b=0,1$, is derived.

Corollary C.10

Let $T_b(b+is)$, $b=0,1$, be two measurable operator-valued functions, and $\lambda _b(b+is)$ be two positive measurable functions such that¹⁵

$$\begin{aligned} \log \lambda _b(b+is)\in L_1\left( \frac{\textrm{d}s}{\cosh (\pi s)}\right) , \end{aligned}$$

and

$$\begin{aligned} 0< \lambda _b(b+is)I\le T_b(b+is)\le cI, \qquad \hbox { a.e.\ on}\ \partial {{\mathbb {S}}}, \end{aligned}$$

for some $c>0$. Then there exists a bounded holomorphic operator-valued function A(z) on ${\mathbb {S}}$ such that A(z) is invertible for all z in the interior of ${\mathbb {S}}$, and

$$\begin{aligned} T_b(b+is)=A^*(b+is)A(b+is), \qquad \hbox { a.e.\ on}\ \partial {{\mathbb {S}}}. \end{aligned}$$