Review of previous parts.

src/op/system/completely.tex

@@ -74,14 +74,12 @@
 
     The sharp case is given by operators of the form $E_{ij} \otimes b$ for $b \in A$. 
     
-    Finally, let $\seqf{e_j} \subset \complex^n$ be the standard orthonormal basis for $\complex^n$. Fix $\alpha \in (0, 1)$. For each $1 \le i, j \le n$, let $\xi_{ij} \in H$ such that $\norm{\xi_{ij}}_H = 1$ and $\norm{a_{ij}\xi_{ij}}_H \ge \alpha \norm{a_{ij}}_A$, then
+    Finally, 
-    \begin{align*}
+    \[
-        \alpha\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2 &\le \sum_{i, j = 1}^n \norm{a_{ij}\xi_{ij}}_H^2 = \sum_{i = 1}^n \norm{a(\xi_{i1}, \cdots, \xi_{in})}_H^2 \\
+    \sum_{i, j = 1}^n \norm{a_{ij}}_A^2 = \sum_{i, j = 1}^n \norm{E_{ii} \cdot a \cdot E_{jj}}_{M_n(A)}^2 \le n^2\norm{a}_{M_n(A)}^2
-        &\le \sum_{i = 1}^n \norm{a}_{M_n(A)}^2 \cdot \norm{(\xi_{i1}, \cdots, \xi_{in})}_{H^n}^2 = n^2\norm{a}_{M_n(A)}^2 \\
+    \]
-        \alpha^{1/2}\braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} &\le n\norm{a}_{M_n(A)}
-    \end{align*}
 
-    As the above holds for all $\alpha \in (0, 1)$, 
+    and
     \[
     \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)}
     \]
@@ -106,7 +104,7 @@
 \begin{proof}
     (1) $\Rightarrow$ (2): For any $\seqf{a_j} \subset A$, let $a = \sum_{i = 1}^n E_{1i} \otimes a_i$, then
     \begin{align*}
-        0 \le aba^* &= \braks{\sum_{i = 1}^n E_{1i} \otimes a_i}\braks{\sum_{j, k = 1}^n E_{jk} \otimes b_{jk}}\braks{\sum_{i = 1}^n E_{i1} \otimes a_i} \\
+        0 \le aba^* &= \braks{\sum_{k = 1}^n E_{k1} \otimes a_k^*}\braks{\sum_{j, k = 1}^n E_{jk} \otimes b_{jk}}\braks{\sum_{i = 1}^n E_{i1} \otimes a_i} \\
         &= \sum_{j, k = 1}^n (E_{1j} \cdot E_{jk} \cdot E_{k1}) \otimes (a_jb_{jk}a_k^*) = E_{11} \otimes \sum_{i, j = 1}^n a_ib_{ij}a_j^*
     \end{align*}
     
@@ -130,7 +128,7 @@
     0 \le \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} = \dpn{b\xi, \xi}{H^n}
     \]
 
-    As $(H_\phi, \pi_\phi)$ is a cyclic representation, the above implies that $\dpn{b\eta, \eta}{H^n} \ge 0$ for all $\eta \in H_\phi$. Since this holds for all $\phi \in S(A)$, $b \ge 0$ in $M_n(A)$. 
+    As $(H_\phi, \pi_\phi)$ is a cyclic representation, the above implies that $\dpn{b\eta, \eta}{H^n} \ge 0$ for all $\eta \in H_\phi^n$. Since this holds for all $\phi \in S(A)$, $b \ge 0$ in $M_n(A)$. 
 \end{proof}
 
 
@@ -390,11 +388,6 @@
             1_A & U & UV \\
             U^* & 1_A & U^*UV \\
             V^*U^* & V^*U^*U & 1_A
-        \end{bmatrix}\\
-        &= \begin{bmatrix}
-            1_A & U & UV \\
-            U^* & 1_A & V \\
-            V^*U^* & V^* & 1_A
         \end{bmatrix}
     \end{align*}
     
@@ -477,16 +470,16 @@
     If the identity is completely positive, then
     \[
     \begin{bmatrix}
-        1_A & U & VU \\
+        1_A & U & V \circ U \\
         U^* & 1_A & V \\
-        U^*V^* & V^* & 1_A
+        U^* \circ V^* & V^* & 1_A
     \end{bmatrix} = \begin{bmatrix}
-        1_A & U & U \circ V \\
+        1_A & U & UV \\
         U^* & 1_A & V \\
-        V^* \circ U^* & V^* & 1_A
+        V^*U^* & V^* & 1_A
     \end{bmatrix} \ge 0
     \]
 
-    in $M_3(A^{\text{op}})$. By (4), $UV = U \circ V = VU$. Therefore $A$ is commutative.
+    in $M_3(A^{\text{op}})$. By (4), $VU = U \circ V = V \circ U = UV$. Therefore $A$ is commutative.
 \end{proof}
 

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}

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}
+
+
+
+

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}
-
-
-
-
-

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}