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Second draft.

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Bokuan Li
2026-10-18 19:57:22 -04:00
parent 51530d0616
commit d09736a42e
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src/op/system/numerical-radius.tex
+17 -17
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@@ -43,7 +43,7 @@
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 $\lambda = -\ol{\sgn(\dpn{T\xi, \xi}{H})}$,
Suppose that $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$. For any $\xi \in H$ with $\norm{\xi}_H = 1$ and $\dpn{T\xi, \xi}{H} \ne 0$, let $\lambda = -\ol{\sgn(\dpn{T\xi, \xi}{H})}$, then
\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}|
@@ -51,12 +51,12 @@
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 $\xi, \eta \in H$,
(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 $\xi, \eta \in H$ with $\dpn{T\xi, \eta}{H} \in \real$,
\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)
&\le \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,
@@ -65,7 +65,7 @@
\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 $I_H$.
(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 $\norm{T}_{B(H)} \le 2w(T)$.
@@ -88,23 +88,23 @@
\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 $a \in S$, $w(\phi(a)) \le \norm{a}_A$.
Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, $H$ 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}.
Let $a \in S$ with $\norm{a}_A \le 1$, then for each $\lambda \in \ol{B_\complex(0, 1)}$, $2 + (\lambda a) + (\lambda a)^* \ge 0$. 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
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 Let $\omega = e^{2\pi i/n}$ be the primitive $n$-th root of unity, then $S_n$ 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 $w(R) \le 1$.
@@ -114,7 +114,7 @@
(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
&= \omega^{k}\sum_{j = 0}^{n-1}\omega^{-jk}e_{j} = \omega^kf_k
\end{align*}
For each $0 \le k < l \le n-1$,
@@ -127,20 +127,20 @@
(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} \\
&= \sum_{j, k = 0}^{n-1}\dpn{T\xi_j, \xi_k}{H} \dpn{S_nf_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 $\xi \in H$,
so $w(T \otimes S_n) \le w(T)$. On the other hand, for any $\xi \in H$,
\[
\dpn{T\xi, \xi}{H} = \dpn{(T \otimes S_n)(\xi \otimes f_0), \xi \otimes f_0}{H} \le w(T \otimes S_n)\norm{\xi}_{H}^2
|\dpn{T\xi, \xi}{H}| = |\dpn{(T \otimes S_n)(\xi \otimes f_0), \xi \otimes f_0}{H^n}| \le w(T \otimes S_n)\norm{\xi}_{H}^2
\]
so $w(T) \le w(T \otimes S)$.
so $w(T) \le w(T \otimes S_n)$.
(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{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*}
@@ -149,7 +149,7 @@
(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
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}|
@@ -157,10 +157,10 @@
As $\norm{\xi_\omega}_{H^n} = \norm{\xi}_{H^n} = 1$,
\[
|\dpn{R\xi, \xi}{H^n}| = \dpn{R\xi_\omega, R\xi_\omega}{H^n} = \text{Re}\dpn{R\xi_\omega, \xi_\omega}{H^n} \le 1
|\dpn{R\xi, \xi}{H^n}| = \dpn{R\xi_\omega, \xi_\omega}{H^n} = \text{Re}\dpn{R\xi_\omega, \xi_\omega}{H^n} \le 1
\]
Therefore $w(T) \le 1$ as well.
Therefore $w(R) \le 1$ as well.
\end{proof}
src/op/system/schur.tex
+16 -11
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@@ -4,15 +4,22 @@
\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.
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 $\Lambda = \sum_{i, j = 1}^n \lambda_{ij}E_{ij} \in M_n(\complex)$ 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,
For any $\xi \in \complex^n$, the projection map $P_\xi$ onto its span is given by $\sum_{i, j = 1}^n \xi_i E_{ij} \ol{\xi_j}$. In which case,
\begin{align*}
S_A(P_\xi) &= \sum_{i, j = 1}^n \xi_i \ol{\xi_j} E_{ij} \otimes a_{ij} \\
&= \braks{\sum_{i = 1}^n \xi_{i}E_{ii}}\braks{\sum_{i, j = 1}^n \xi_i \ol{\xi_j} E_{ij} \otimes a_{ij}}\braks{\sum_{j = 1}^n {\xi_j}E_{jj}}^* \ge 0
\end{align*}
Using the \hyperref[Spectral Theorem]{definition:linfty-functional-calculus}, let $\bracsn{\xi^{(j)}}_1^n \subset \complex^n$ be an orthonormal eigenbasis for $\Lambda$ with eigenvectors $\seqf{\lambda_j} \subset \complex$. For each $1 \le j \le n$, let $P_j \in M_n(\complex)$ be the rank one projection onto the span of $\xi^{(j)}$, then
\[
S_A(\lambda) = \sum_{i = 1}^n \lambda_i E_{ii} \otimes a_{ii}
S_A(\Lambda) = \sum_{j = 1}^n \lambda_j S_A(P_{j}) \ge 0
\]
is a sum of positive operators, and hence positive.
\end{proof}
@@ -25,7 +32,7 @@
\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:
For any complex Hilbert space $H$ and $\bracsn{A_n|n \in \integer} \subset B(H)$, 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.
@@ -34,13 +41,13 @@
\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 $L^2(\mathcal{T}; \complex)$.
(1) $\Leftrightarrow$ (2): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $\complex[z, \ol z]$ is dense in $C(\mathbb{T}; \complex)$. By \autoref{proposition:radon-cc-dense}, $C(\mathbb{T}; \complex)$ is dense in $L^2(\mathbb{T}; \complex)$.
(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}} \\
&= \sum_{-N \le k, l \le N} \lambda_{l - k}\mu_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*}
@@ -48,12 +55,12 @@
(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
For each $P = \sum_{-n \le j \le n}\lambda_jz^j \in \complex[z, \ol z]$ with $P \ge 0$ and $m \ge n+1$, 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
is a Schur product of positive matrices, and hence positive by \autoref{lemma:schur-product-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}
@@ -66,6 +73,4 @@
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}