From d240588d5c13acc4fc5fc60e2ae18669fadb34e6 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sun, 18 Oct 2026 16:36:10 -0400 Subject: [PATCH] First draft of two additional exercises. --- src/op/system/index.tex | 2 + src/op/system/numerical-radius.tex | 168 +++++++++++++++++++++++++++++ src/op/system/os.tex | 114 ++++---------------- src/op/system/schur.tex | 71 ++++++++++++ 4 files changed, 259 insertions(+), 96 deletions(-) create mode 100644 src/op/system/numerical-radius.tex create mode 100644 src/op/system/schur.tex diff --git a/src/op/system/index.tex b/src/op/system/index.tex index b2610e9..b2aa376 100644 --- a/src/op/system/index.tex +++ b/src/op/system/index.tex @@ -7,3 +7,5 @@ \input{./completely.tex} \input{./complete-upgrade.tex} \input{./module.tex} +\input{./numerical-radius.tex} +\input{./schur.tex} diff --git a/src/op/system/numerical-radius.tex b/src/op/system/numerical-radius.tex new file mode 100644 index 0000000..65e2513 --- /dev/null +++ b/src/op/system/numerical-radius.tex @@ -0,0 +1,168 @@ +\section{The Numerical Radius} +\label{section:numerical-radius} + + +\begin{definition}[Numerical Radius] +\label{definition:bh-numerical-radius} + Let $H$ be a complex Hilbert space and $T \in B(H)$, then + \[ + w(T) = \sup_{\substack{\xi \in H \\ \norm{\xi}_H = 1}}|\dpn{T\xi, \xi}{H}| + \] + + is the \textbf{numerical radius} of $T$. +\end{definition} + +\begin{proposition} +\label{proposition:bh-numerical-radius-gymnastics} + Let $H$ be a complex Hilbert space, then: + \begin{enumerate} + \item The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$. + \item For any $T \in B(H)$, $w(T) \le 1$ if and only if $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$. + \item For each $T \in B(H)_{sa}$, $w(T) = \norm{T}_{B(H)}$. + \item For each $T \in B(H)$, $w(T) \le \norm{T}_{B(H)} \le 2w(T)$. Both inequalities are sharp in general. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): For any $S, T \in B(H)$ and $\xi \in H$ with $\norm{\xi}_H = 1$, + \[ + |\dpn{(S + T)\xi, \xi}{H}| \le |\dpn{S\xi, \xi}{H}| + |\dpn{T\xi, \xi}{H}| \le w(S) + w(T) + \] + + As the above holds for all $\xi \in H$, $w(S + T) \le w(S) + w(T)$. Should $w(T) = 0$, then $\dpn{T\xi, \xi}{H} = 0$ for all $\xi \in H$. By \hyperref[polarisation]{proposition:polarisation-complex}, for any $\xi, \eta \in H$, + \[ + \dpn{T\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} = 0 + \] + + so $T = 0$ and $w$ is a norm on $B(H)$. + + (2): For any $\xi \in H$ with $\norm{\xi}_H = 1$ and $\lambda \in \partial B_\complex(0, 1)$, + \begin{align*} + \dpn{[2 + (\lambda T) + (\lambda T)^*]\xi, \xi}{H} &= 2 + \dpn{\lambda T \xi, \xi}{H} + \dpn{\xi, \lambda T \xi}{H} \\ + &= 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H} + \end{align*} + + If $w(T) \le 1$, then $2\text{Re}(\dpn{\lambda T\xi, \xi}{H}) \ge -2$, and $2 + (\lambda T) + (\lambda T)^* \ge 0$. + + If $2 + (\lambda T) + (\lambda T)^*$ for all $\lambda \in \partial B_\complex(0, 1)$, then for any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \ne 0$, and $\lambda = -\ol{\sgn(\dpn{T\xi, \xi}{H})}$, + \begin{align*} + \dpn{\lambda T\xi, \xi}{H} &= \lambda \dpn{T\xi, \xi}{H} = -\ol{\sgn(\dpn{T\xi, \xi}{H})} \dpn{T\xi, \xi}{H} = -|\dpn{T\xi, \xi}{H}| \\ + 0 &\le 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H} = 2 - 2|\dpn{T\xi, \xi}{H}| + \end{align*} + + so $|\dpn{T\xi, \xi}{H}| \le 1$, and $w(T) \le 1$. + + (3): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. On the other hand, if $w(T) \le 1$, then for each $\xi, \eta \in H$, + \begin{align*} + \dpn{T\xi, \eta}{H} &= \frac{1}{4}\sum_{k = 0}^3 i^k\dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} \\ + &= \frac{1}{4}(\dpn{T(\xi + \eta), \xi + \eta}{H} - \dpn{T(\xi - \eta), \xi - \eta}{H}) \\ + &\le \frac{w(T)}{4}(\norm{\xi + \eta}_H^2 + \norm{\xi - \eta}_H^2) \\ + &= \frac{1}{2}(\norm{\xi}_H^2 + \norm{\eta}_H^2) + \end{align*} + + by \hyperref[polarisation]{proposition:polarisation-complex} and the parallelogram law. In particular, + \begin{align*} + \norm{T\xi}_{H}^2 &\le \frac{1}{2}(\norm{\xi}_H^2 + \norm{T\xi}_H^2) \\ + \norm{T\xi}_H^2 &\le \norm{\xi}_H^2 + \end{align*} + + (4): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. The sharp case is given by $I_H$. + + On the other hand, if $w(T) \le 1$, then $2 + T + T^* = 2 + 2\text{Re}(T) \ge 0$ and $2 - T - T^* = 2 - 2\text{Re}(T) \ge 0$ imply that $\norm{\text{Re}(T)}_{B(H)} \le 1$. Similarly, $\norm{\text{Im}(T)}_{B(H)} \le 1$ as well, so $\norm{T}_{B(H)} \le \norm{\text{Re}(T)}_{B(H)} + \norm{\text{Im}(T)}_{B(H)} \le 2$. Therefore $\norm{T}_{B(H)} \le 2w(T)$. + + For any $\xi, \eta \in \complex$ with $\norm{(\xi, \eta)}_{\complex^2} = 1$, by \hyperref[Young's inequality]{lemma:young-inequality}, + \[ + \abs{\angles{\begin{bmatrix} + 0 &2 \\ + 0 & 0 + \end{bmatrix}\begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}, \begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}}_{\complex^2}} = 2|\eta \ol{\xi}| \le |\xi|^2 + |\eta|^2 \le 1 + \] + + so the above matrix has numerical radius $1$. However, its operator norm is $2$. Thus the second inequality is also sharp. +\end{proof} + +\begin{proposition} +\label{proposition:bh-numerical-radius-positive} + Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator space, $H$ be a complex Hilbert space, and $\phi: S \to B(H)$ be a unital positive map, then for each $a \in S$, $w(\phi(a)) \le \norm{a}_A$. +\end{proposition} +\begin{proof} + Let $a \in S$ with $\norm{a}_A \le 1$, then for each $\lambda \in \complex$, $2 + (\lambda a) + (\lambda a)^* \ge 0$. Since $\phi$ is positive, $2 + (\lambda \phi(a)) + (\lambda \phi(a))^* \ge 0$ as well. Therefore $w(\phi(a)) \le 1$ by \autoref{proposition:bh-numerical-radius-gymnastics}. +\end{proof} + + +\begin{proposition} +\label{proposition:cyclic-forward-shift-radius} + Let $H$ be a complex Hilbert space, $T \in B(H)$, $R = \sum_{j = 2}^nE_{(j-1)j} \otimes T \in M_n(B(H))$, $\bracsn{e_j}_0^{n-1} \subset \complex^n$ be the canonical basis for $\complex^n$, and + \[ + S_n: \complex^n \to \complex^n \quad S_ne_j = e_{j+1 \mod n} + \] + + then: + \begin{enumerate} + \item Let $\omega = e^{2\pi i/n}$ be the primitive $n$-th root of unity, then $S$ is unitarily equivalent to $\text{diag}(\omega^0, \cdots, \omega^{n-1})$. + \item $w(T) = w(T \otimes S_n)$. + \item $w(R) \le w(T \otimes S_n)$. + \item $\text{Re}\dpn{R\xi, \xi}{H^n} \le 1$ for all $\xi \in H^n$ with $\norm{\xi}_{H^n} = 1$ if and only if $w(R) \le 1$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): For each $0 \le k \le n - 1$, let $f_k = \sum_{j = 0}^{n-1}\omega^{-jk}e_j$, then + \begin{align*} + S_nf_k &= \sum_{j = 0}^{n-1}\omega^{-jk}e_{j+1 \mod n} = \sum_{j = 0}^{n-1}\omega^{-(j-1)k}e_{j} \\ + &= \omega^{k}\sum_{j = 0}^{n-1}\omega^{-jk}e_{j+1} = \omega^kf_k + \end{align*} + + For each $0 \le k < l \le n-1$, + \[ + \dpn{f_k, f_l}{\complex^n} = \sum_{j = 0}^{n-1} \omega^{-jk} \cdot \ol{\omega^{-jl}} = \sum_{j = 0}^{n-1} \omega^{-j(k-l)} = 0 + \] + + so $\bracsn{f_k}_0^{n-1} \subset \complex^n$ is an orthogonal eigenbasis for $S_n$, with eigenvalues $\bracsn{\omega^k}_0^{n-1}$. + + (2): Let $\bracsn{f_j}_{0}^{n-1} \subset \complex^n$ be an orthonormal eigenbasis for $S_n$, with eigenvalues $\bracsn{\omega^k}_0^{n-1}$. For any $\bracsn{\xi_j}_0^{n-1} \subset H$, let $\xi = \sum_{j = 0}^{n-1}\xi_j \otimes f_j$, then + \begin{align*} + \dpn{(T \otimes S_n)\xi, \xi}{H^n} &= \sum_{j, k = 0}^{n-1}\dpn{(T \otimes S_n)(\xi_j \otimes f_j), \xi_k \otimes f_k}{H^n} \\ + &= \sum_{j, k = 0}^{n-1}\dpn{T\xi_j, \xi_k}{H} \dpn{Sf_j, f_k}{H} = \sum_{j = 0}^{n-1}\omega^j\dpn{T\xi_j, \xi_j}{H} \\ + |\dpn{(T \otimes S_n)\xi, \xi}{H^n}| &\le \sum_{j = 0}^{n-1}|\omega^j| \cdot w(T) \cdot \norm{\xi_j}_{H}^2 = w(T)\norm{\xi}_{H^n}^2 + \end{align*} + + so $w(T \otimes S) \le w(T)$. On the other hand, for any $\xi \in H$, + \[ + \dpn{T\xi, \xi}{H} = \dpn{(T \otimes S_n)(\xi \otimes f_0), \xi \otimes f_0}{H} \le w(T \otimes S_n)\norm{\xi}_{H}^2 + \] + + so $w(T) \le w(T \otimes S)$. + + (3): Let $\xi = (\xi_1, \cdots, \xi_{n}) \in H^n$, then + \begin{align*} + \dpn{R\xi, \xi}{H^n} &= \sum_{j = 2}^{n}\dpn{(E_{(j-1)j}\otimes T)\xi, \xi}{H^n} = \sum_{j = 1}^{n-1} \dpn{T\xi_j, \xi_{j+1}}{H} \\ + &= \dpn{T0, \xi_1}{H} + \dpn{T\xi_n, 0}{H} + \sum_{j = 1}^{n-1} \dpn{T\xi_j, \xi_{j+1}}{H} \\ + &= \dpn{(T \otimes S_{n+1})(\xi_1, \cdots, \xi_n, 0), (\xi_1, \cdots, \xi_n, 0)}{H^{n+1}} + \end{align*} + + so $w(R) \le w(T \otimes S_{n+1}) = w(T) = w(T \otimes S_n)$ by (2). + + (4): Suppose that $\text{Re}\dpn{R\xi, \xi}{H^n} \le 1$ for all $\xi \in H^n$ with $\norm{\xi}_{H^n} \le 1$. + + Let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ with $\norm{\xi}_{H^n} = 1$, and assume without loss of generality that $\dpn{R\xi, \xi}{H^n} \ne 0$. Let $\omega = \sgn(\dpn{R\xi, \xi}{H^n})$ and $\xi_\omega = (\omega^1\xi_1, \cdots, \omega_n\xi_n)$, then + \begin{align*} + \dpn{R\xi_\omega, \xi_\omega}{H^n} &= \sum_{j = 1}^{n-1} \dpn{\omega^j T\xi_j, \omega^{j+1}\xi_{j+1}}{H} = \sum_{j = 1}^{n-1} \ol{\omega}\dpn{T\xi_j, \xi_{j+1}}{H} \\ + &= \ol{\sgn(\dpn{R\xi, \xi}{H^n})} \dpn{R\xi, \xi}{H^n} = |\dpn{R\xi, \xi}{H^n}| + \end{align*} + + As $\norm{\xi_\omega}_{H^n} = \norm{\xi}_{H^n} = 1$, + \[ + |\dpn{R\xi, \xi}{H^n}| = \dpn{R\xi_\omega, R\xi_\omega}{H^n} = \text{Re}\dpn{R\xi_\omega, \xi_\omega}{H^n} \le 1 + \] + + Therefore $w(T) \le 1$ as well. +\end{proof} + + + + diff --git a/src/op/system/os.tex b/src/op/system/os.tex index 771d55c..90bf729 100644 --- a/src/op/system/os.tex +++ b/src/op/system/os.tex @@ -21,6 +21,24 @@ Let $A$ be a $C^*$-algebra and $S \subset A$ be an operator system, then $S_{+} = \bracsn{x \in S_{sa}|x \ge 0}$ is the \textbf{positive cone} of $S$. \end{definition} +\begin{lemma} +\label{lemma:operator-system-positive-approximate} + Let $A$ be a $C^*$-algebra, $S \subset A$ be an operator system, then $(\ol{S})_+ = \ol{S_+}$. +\end{lemma} +\begin{proof} + Let $x \in (\ol{S})_+$, then there exists $\seq{x_n} \subset S$ such that $x_n \to x$ as $n \to \infty$. In which case, + \[ + x = \frac{x + x^*}{2} = \limv{n}\frac{x_n + x_n^*}{2} \in \ol{S_{sa}} + \] + + Assume without loss of generality that $\seq{x_n} \subset S_{sa}$, so that $\sigma_A(x_n) \subset \real$ for all $n \in \natp$. By \hyperref[upper semicontinuity of the spectrum]{proposition:spectrum-continuous}, $\lim_{n \to \infty}\inf(\sigma_A(x_n)) \ge 0$. Therefore + \[ + x = \limv{n}x_n = \limv{n}[x_n + \inf(\sigma_A(x_n)) \wedge 0] \in \ol{S_+} + \] + + by \autoref{corollary:spectrum-characterisation-iff}. +\end{proof} + \begin{proposition} \label{proposition:operator-system-positive-span} @@ -131,99 +149,3 @@ (3): By \autoref{theorem:positive-commutative-norm}, any such extension would have norm $1$. \end{proof} -\begin{definition}[Numerical Radius] -\label{definition:bh-numerical-radius} - Let $H$ be a complex Hilbert space and $T \in B(H)$, then - \[ - w(T) = \sup_{\substack{\xi \in H \\ \norm{\xi}_H = 1}}|\dpn{T\xi, \xi}{H}| - \] - - is the \textbf{numerical radius} of $T$. -\end{definition} - -\begin{proposition} -\label{proposition:bh-numerical-radius-gymnastics} - Let $H$ be a complex Hilbert space, then: - \begin{enumerate} - \item The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$. - \item For any $T \in B(H)$, $w(T) \le 1$ if and only if $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$. - \item For each $T \in B(H)_{sa}$, $w(T) = \norm{T}_{B(H)}$. - \item For each $T \in B(H)$, $w(T) \le \norm{T}_{B(H)} \le 2w(T)$. Both inequalities are sharp in general. - \end{enumerate} -\end{proposition} -\begin{proof} - (1): For any $S, T \in B(H)$ and $\xi \in H$ with $\norm{\xi}_H = 1$, - \[ - |\dpn{(S + T)\xi, \xi}{H}| \le |\dpn{S\xi, \xi}{H}| + |\dpn{T\xi, \xi}{H}| \le w(S) + w(T) - \] - - As the above holds for all $\xi \in H$, $w(S + T) \le w(S) + w(T)$. Should $w(T) = 0$, then $\dpn{T\xi, \xi}{H} = 0$ for all $\xi \in H$. By \hyperref[polarisation]{proposition:polarisation-complex}, for any $\xi, \eta \in H$, - \[ - \dpn{T\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} = 0 - \] - - so $T = 0$ and $w$ is a norm on $B(H)$. - - (2): For any $\xi \in H$ with $\norm{\xi}_H = 1$ and $\lambda \in \partial B_\complex(0, 1)$, - \begin{align*} - \dpn{[2 + (\lambda T) + (\lambda T)^*]\xi, \xi}{H} &= 2 + \dpn{\lambda T \xi, \xi}{H} + \dpn{\xi, \lambda T \xi}{H} \\ - &= 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H} - \end{align*} - - If $w(T) \le 1$, then $2\text{Re}(\dpn{\lambda T\xi, \xi}{H}) \ge -2$, and $2 + (\lambda T) + (\lambda T)^* \ge 0$. - - If $2 + (\lambda T) + (\lambda T)^*$ for all $\lambda \in \partial B_\complex(0, 1)$, then for any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \ne 0$, and $\lambda = -\ol{\sgn(\dpn{T\xi, \xi}{H})}$, - \begin{align*} - \dpn{\lambda T\xi, \xi}{H} &= \lambda \dpn{T\xi, \xi}{H} = -\ol{\sgn(\dpn{T\xi, \xi}{H})} \dpn{T\xi, \xi}{H} = -|\dpn{T\xi, \xi}{H}| \\ - 0 &\le 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H} = 2 - 2|\dpn{T\xi, \xi}{H}| - \end{align*} - - so $|\dpn{T\xi, \xi}{H}| \le 1$, and $w(T) \le 1$. - - (3): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. On the other hand, if $w(T) \le 1$, then for each $\xi, \eta \in H$, - \begin{align*} - \dpn{T\xi, \eta}{H} &= \frac{1}{4}\sum_{k = 0}^3 i^k\dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} \\ - &= \frac{1}{4}(\dpn{T(\xi + \eta), \xi + \eta}{H} - \dpn{T(\xi - \eta), \xi - \eta}{H}) \\ - &\le \frac{w(T)}{4}(\norm{\xi + \eta}_H^2 + \norm{\xi - \eta}_H^2) \\ - &= \frac{1}{2}(\norm{\xi}_H^2 + \norm{\eta}_H^2) - \end{align*} - - by \hyperref[polarisation]{proposition:polarisation-complex} and the parallelogram law. In particular, - \begin{align*} - \norm{T\xi}_{H}^2 &\le \frac{1}{2}(\norm{\xi}_H^2 + \norm{T\xi}_H^2) \\ - \norm{T\xi}_H^2 &\le \norm{\xi}_H^2 - \end{align*} - - (4): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. The sharp case is given by $I_H$. - - On the other hand, if $w(T) \le 1$, then $2 + T + T^* = 2 + 2\text{Re}(T) \ge 0$ and $2 - T - T^* = 2 - 2\text{Re}(T) \ge 0$ imply that $\norm{\text{Re}(T)}_{B(H)} \le 1$. Similarly, $\norm{\text{Im}(T)}_{B(H)} \le 1$ as well, so $\norm{T}_{B(H)} \le \norm{\text{Re}(T)}_{B(H)} + \norm{\text{Im}(T)}_{B(H)} \le 2$. Therefore $\norm{T}_{B(H)} \le 2w(T)$. - - For any $\xi, \eta \in \complex$ with $\norm{(\xi, \eta)}_{\complex^2} = 1$, by \hyperref[Young's inequality]{lemma:young-inequality}, - \[ - \abs{\angles{\begin{bmatrix} - 0 &2 \\ - 0 & 0 - \end{bmatrix}\begin{bmatrix} - \xi \\ - \eta - \end{bmatrix}, \begin{bmatrix} - \xi \\ - \eta - \end{bmatrix}}_{\complex^2}} = 2|\eta \ol{\xi}| \le |\xi|^2 + |\eta|^2 \le 1 - \] - - so the above matrix has numerical radius $1$. However, its operator norm is $2$. Thus the second inequality is also sharp. -\end{proof} - -\begin{proposition} -\label{proposition:bh-numerical-radius-positive} - Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator space, $H$ be a complex Hilbert space, and $\phi: S \to B(H)$ be a unital positive map, then for each $a \in S$, $w(\phi(a)) \le \norm{a}_A$. -\end{proposition} -\begin{proof} - Let $a \in S$ with $\norm{a}_A \le 1$, then for each $\lambda \in \complex$, $2 + (\lambda a) + (\lambda a)^* \ge 0$. Since $\phi$ is positive, $2 + (\lambda \phi(a)) + (\lambda \phi(a))^* \ge 0$ as well. Therefore $w(\phi(a)) \le 1$ by \autoref{proposition:bh-numerical-radius-gymnastics}. -\end{proof} - - - - - diff --git a/src/op/system/schur.tex b/src/op/system/schur.tex new file mode 100644 index 0000000..81a2f28 --- /dev/null +++ b/src/op/system/schur.tex @@ -0,0 +1,71 @@ +\section{Schur Products} +\label{section:schur-product} + + +\begin{lemma} +\label{lemma:schur-product-positive} + Let $H$ be a complex Hilbert space, $A = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(B(H))$ be positive, and $\Lambda = \sum_{i, j = 1}^n \lambda_{ij}E_{ij} \in M_n(B(H))$ be positive, then $S_A(\Lambda) = \sum_{i, j = 1}^n \lambda_{ij}E_{ij} \otimes a_{ij}$ is also positive. +\end{lemma} +\begin{proof} + Using the \hyperref[Spectral Theorem]{definition:linfty-functional-calculus}, assume without loss of generality that there exists $\bracsn{\lambda_j}_1^n \subset [0, \infty)$ such that $\Lambda = \text{diag}(\lambda_1, \cdots, \lambda_n)$. In which case, + \[ + S_A(\lambda) = \sum_{i = 1}^n \lambda_i E_{ii} \otimes a_{ii} + \] + + is a sum of positive operators, and hence positive. +\end{proof} + + +\begin{proposition}[Trigonometric Moment Sequences] +\label{proposition:trig-moment-sequence-operator} + 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 $P = \sum_{-n \le j \le n}\lambda_jz^j \in \complex[z, \ol{z}]$, then the following are equivalent: + \begin{enumerate} + \item $P \ge 0$. + \item For each $q \in \complex[z, \ol{z}]$, $\dpn{Pq, q}{L^2(\mathbb{T}; \complex)} \ge 0$. + \item For each $N \in \natp$\footnote{Padding the coefficients with zeroes if necessary.}, the Toeplitz matrix $\sum_{i, j = 1}^NE_{ij}\lambda_{i - j}$ is positive. + \end{enumerate} + + For any complex Hilbert space $H$ and $\bracsn{A_n|n \in \integer} \subset H$, the following are equivalent: + \begin{enumerate}[label=(T\arabic*)] + \item For each $n \in \natp$, the Toeplitz matrix $\sum_{i, j = 1}^n E_{ij} \otimes A_{i - j}$ is positive. + \item The mapping $\phi: C(\mathbb{T}; \complex) \to B(H)$ defined by extending $\phi(z^n) = A_n$ is completely positive. + \end{enumerate} + + +\end{proposition} +\begin{proof} + (1) $\Leftrightarrow$ (2): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $\complex[z, \ol z]$ is dense in $C(\mathcal{T}; \complex)$. By \autoref{proposition:radon-cc-dense}, $C(\mathcal{T}; \complex)$ 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 $\sum_{i, j = 1}^NE_{ij}\lambda_{i - j}$ is positive for all $N \in \natp$. + + (T1) $\Rightarrow$ (T2): Using \hyperref[Stinespring's Theorem]{theorem:stinespring-commutative-completely-positive}, it is sufficient to show that $\phi$ is positive. Using \autoref{lemma:operator-system-positive-approximate} and \autoref{proposition:positive-automatic-continuity}, it is sufficient to show that $\phi(P) \ge 0$ for all $P \in \complex[z, \ol z]$ with $P \ge 0$. + + For each $P = \sum_{-n \le j \le n}\lambda_jz^j \in \complex[z, \ol z]$ 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*} + + +\end{proof} \ No newline at end of file