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/Part 7: Operator Algebras/Chapter 41: Operator Systems/Section 41.8: Schur Products

Proposition 41.8.2 (Trigonometric Moment Sequences).label Let $\mathbb{T}= \partial B_{\complex}(0, 1)$, and $C(\mathbb{T}; \complex)$ act on $L^{2}(\mathbb{T}; \complex)$ via multiplication. For each $n \in \natp$ and $P = \sum_{-n \le j \le n}\lambda_{j}z^{j} \in \complex[z, \ol{z}]$, then the following are equivalent:

  1. (1)

    $P \ge 0$.

  2. (2)

    For each $q \in \complex[z, \ol{z}]$, $\dpn{Pq, q}{L^2(\mathbb{T}; \complex)}\ge 0$.

  3. (3)

    For each $N \in \natp$[1], the Toeplitz matrix $\sum_{i, j = 1}^{N}E_{ij}\lambda_{i - j}$ is positive.

For any complex Hilbert space $H$ and $\bracsn{A_n|n \in \integer}\subset H$, the following are equivalent:

  1. (T1)

    For each $n \in \natp$, the Toeplitz matrix $\sum_{i, j = 1}^{n} E_{ij}\otimes A_{i - j}$ is positive.

  2. (T2)

    The mapping $\phi: C(\mathbb{T}; \complex) \to B(H)$ defined by extending $\phi(z^{n}) = A_{n}$ is completely positive.

Proof. (1) $\Leftrightarrow$ (2): By the Stone-Weierstrass Theorem, $\complex[z, \ol z]$ is dense in $C(\mathcal{T}; \complex)$. By Proposition 25.1.7, $C(\mathcal{T}; \complex)$ is dense in $L^{2}(\mathcal{T}; \complex)$.

(2) $\Leftrightarrow$ (3): Let $q = \sum_{-N \le j \le N}\mu_{j} z^{j}$ and $\xi = (\mu_{-N}, \cdots, \mu_{N}) \in \complex^{2N+1}$, then

\begin{align*}\dpn{Pq, q}{L^2(\mathbb{T}; \complex)}&= \sum_{-N \le j, k, l \le N}\dpn{\lambda_j z^j \cdot \mu_k z^k, \mu_l z^l}{L^2(\mathbb{T}; \complex)}\\&= \sum_{-N \le k, l \le N}\dpn{\lambda_{l-k}\mu_k z^l, \mu_l z^l}{L^2(\mathbb{T}; \complex)}\\&= \sum_{-N \le k, l \le N}\lambda_{l - k}u_{k}\ol{\mu_l}= \sum_{k, l = 1}^{2N+1}\lambda_{l - k}\xi_{k}\ol{\xi_{l}}\\&= \angles{\paren{\sum_{i, j = 1}^{2N+1} E_{ij}\lambda_{i - j}}\xi, \xi}\\\end{align*}

Therefore $\dpn{Pq, q}{L^2(\mathbb{T}; \complex)}\ge 0$ for all $q \in \complex[z, \ol z]$ if and only if the Toeplitz matrix $\sum_{i, j = 1}^{N}E_{ij}\lambda_{i - j}$ is positive for all $N \in \natp$.

(T1) $\Rightarrow$ (T2): Using Stinespring’s Theorem, it is sufficient to show that $\phi$ is positive. Using Lemma 41.2.4 and Proposition 41.2.7, it is sufficient to show that $\phi(P) \ge 0$ for all $P \in \complex[z, \ol z]$ with $P \ge 0$.

For each $P = \sum_{-n \le j \le n}\lambda_{j}z^{j} \in \complex[z, \ol z]$ with $P \ge 0$ and $m \in \natp$, the matrix

\[P_{m} = \sum_{i, j = 1}^{m} \lambda_{i - j}E_{ij}\otimes A_{i - j}\in M_{m}(B(H))\]

is positive. For any $\xi \in H$, let $\xi_{m} = (\xi, \cdots, \xi) \in H^{m}$, then

\begin{align*}0&\le \dpn{P_m\xi_m, \xi_m}{H^m}= \sum_{i, j = 1}^{m} \lambda_{i - j}\dpn{A_{i - j}\xi, \xi}{H}= \sum_{k = -n}^{n} (m - |k|) \lambda_{k} \dpn{A_{k}\xi, \xi}{H}\\ 0&\le \limv{m}\sum_{k = -n}^{n} \frac{(m - |k|)}{m}\lambda_{k} \dpn{A_{k}\xi, \xi}{H}= \sum_{k = -n}^{n} \dpn{\lambda_k A_k\xi, \xi}{H}= \dpn{\phi(P)\xi, \xi}{H}\end{align*}

so $\phi(P)$ is positive.

(T2) $\Rightarrow$ (T1): For each $n \in \natp$,

\begin{align*}0&\le \braks{\sum_{j = 1}^n E_{1j} \otimes z^j}^{*}\braks{\sum_{j = 1}^n E_{1j} \otimes z^j}= \sum_{j, k = 1}^{n} E_{jk}\otimes z^{j-k}\\ 0&\le \phi\braks{\sum_{j, k = 1}^n E_{jk} \otimes z^{j-k}}= \sum_{j, k = 1}^{n} E_{jk}\otimes A_{j - k}\end{align*}

$\square$

  1. Padding the coefficients with zeroes if necessary.keyboard_return

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