First draft of two additional exercises.
This commit is contained in:
Bokuan Li
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\input{./completely.tex}
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\input{./completely.tex}
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\input{./complete-upgrade.tex}
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\input{./complete-upgrade.tex}
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\input{./module.tex}
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\input{./module.tex}
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\input{./numerical-radius.tex}
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\input{./schur.tex}
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\section{The Numerical Radius}
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\label{section:numerical-radius}
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\begin{definition}[Numerical Radius]
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\label{definition:bh-numerical-radius}
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Let $H$ be a complex Hilbert space and $T \in B(H)$, then
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\[
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w(T) = \sup_{\substack{\xi \in H \\ \norm{\xi}_H = 1}}|\dpn{T\xi, \xi}{H}|
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\]
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is the \textbf{numerical radius} of $T$.
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\end{definition}
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\begin{proposition}
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\label{proposition:bh-numerical-radius-gymnastics}
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Let $H$ be a complex Hilbert space, then:
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\begin{enumerate}
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\item The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$.
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\item For any $T \in B(H)$, $w(T) \le 1$ if and only if $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$.
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\item For each $T \in B(H)_{sa}$, $w(T) = \norm{T}_{B(H)}$.
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\item For each $T \in B(H)$, $w(T) \le \norm{T}_{B(H)} \le 2w(T)$. Both inequalities are sharp in general.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): For any $S, T \in B(H)$ and $\xi \in H$ with $\norm{\xi}_H = 1$,
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\[
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|\dpn{(S + T)\xi, \xi}{H}| \le |\dpn{S\xi, \xi}{H}| + |\dpn{T\xi, \xi}{H}| \le w(S) + w(T)
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\]
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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 $\xi, \eta \in H$,
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\[
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\dpn{T\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} = 0
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\]
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so $T = 0$ and $w$ is a norm on $B(H)$.
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(2): For any $\xi \in H$ with $\norm{\xi}_H = 1$ and $\lambda \in \partial B_\complex(0, 1)$,
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\begin{align*}
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\dpn{[2 + (\lambda T) + (\lambda T)^*]\xi, \xi}{H} &= 2 + \dpn{\lambda T \xi, \xi}{H} + \dpn{\xi, \lambda T \xi}{H} \\
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&= 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H}
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\end{align*}
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If $w(T) \le 1$, then $2\text{Re}(\dpn{\lambda T\xi, \xi}{H}) \ge -2$, and $2 + (\lambda T) + (\lambda T)^* \ge 0$.
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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 $\lambda = -\ol{\sgn(\dpn{T\xi, \xi}{H})}$,
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\begin{align*}
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\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}| \\
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0 &\le 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H} = 2 - 2|\dpn{T\xi, \xi}{H}|
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\end{align*}
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so $|\dpn{T\xi, \xi}{H}| \le 1$, and $w(T) \le 1$.
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(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 $\xi, \eta \in H$,
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\begin{align*}
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\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} \\
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&= \frac{1}{4}(\dpn{T(\xi + \eta), \xi + \eta}{H} - \dpn{T(\xi - \eta), \xi - \eta}{H}) \\
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&\le \frac{w(T)}{4}(\norm{\xi + \eta}_H^2 + \norm{\xi - \eta}_H^2) \\
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&= \frac{1}{2}(\norm{\xi}_H^2 + \norm{\eta}_H^2)
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\end{align*}
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by \hyperref[polarisation]{proposition:polarisation-complex} and the parallelogram law. In particular,
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\begin{align*}
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\norm{T\xi}_{H}^2 &\le \frac{1}{2}(\norm{\xi}_H^2 + \norm{T\xi}_H^2) \\
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\norm{T\xi}_H^2 &\le \norm{\xi}_H^2
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\end{align*}
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(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 $I_H$.
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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 $\norm{T}_{B(H)} \le 2w(T)$.
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For any $\xi, \eta \in \complex$ with $\norm{(\xi, \eta)}_{\complex^2} = 1$, by \hyperref[Young's inequality]{lemma:young-inequality},
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\[
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\abs{\angles{\begin{bmatrix}
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0 &2 \\
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0 & 0
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\end{bmatrix}\begin{bmatrix}
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\xi \\
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\eta
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\end{bmatrix}, \begin{bmatrix}
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\xi \\
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\eta
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\end{bmatrix}}_{\complex^2}} = 2|\eta \ol{\xi}| \le |\xi|^2 + |\eta|^2 \le 1
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\]
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so the above matrix has numerical radius $1$. However, its operator norm is $2$. Thus the second inequality is also sharp.
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\end{proof}
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\begin{proposition}
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\label{proposition:bh-numerical-radius-positive}
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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 $a \in S$, $w(\phi(a)) \le \norm{a}_A$.
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\end{proposition}
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\begin{proof}
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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}.
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\end{proof}
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\begin{proposition}
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\label{proposition:cyclic-forward-shift-radius}
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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
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\[
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S_n: \complex^n \to \complex^n \quad S_ne_j = e_{j+1 \mod n}
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\]
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then:
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\begin{enumerate}
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\item Let $\omega = e^{2\pi i/n}$ be the primitive $n$-th root of unity, then $S$ is unitarily equivalent to $\text{diag}(\omega^0, \cdots, \omega^{n-1})$.
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\item $w(T) = w(T \otimes S_n)$.
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\item $w(R) \le w(T \otimes S_n)$.
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\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 $w(R) \le 1$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): For each $0 \le k \le n - 1$, let $f_k = \sum_{j = 0}^{n-1}\omega^{-jk}e_j$, then
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\begin{align*}
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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} \\
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&= \omega^{k}\sum_{j = 0}^{n-1}\omega^{-jk}e_{j+1} = \omega^kf_k
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\end{align*}
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For each $0 \le k < l \le n-1$,
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\[
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\dpn{f_k, f_l}{\complex^n} = \sum_{j = 0}^{n-1} \omega^{-jk} \cdot \ol{\omega^{-jl}} = \sum_{j = 0}^{n-1} \omega^{-j(k-l)} = 0
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\]
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so $\bracsn{f_k}_0^{n-1} \subset \complex^n$ is an orthogonal eigenbasis for $S_n$, with eigenvalues $\bracsn{\omega^k}_0^{n-1}$.
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(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
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\begin{align*}
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\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} \\
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&= \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} \\
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|\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
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\end{align*}
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so $w(T \otimes S) \le w(T)$. On the other hand, for any $\xi \in H$,
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\[
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\dpn{T\xi, \xi}{H} = \dpn{(T \otimes S_n)(\xi \otimes f_0), \xi \otimes f_0}{H} \le w(T \otimes S_n)\norm{\xi}_{H}^2
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\]
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so $w(T) \le w(T \otimes S)$.
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(3): Let $\xi = (\xi_1, \cdots, \xi_{n}) \in H^n$, then
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\begin{align*}
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\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} \\
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&= \dpn{T0, \xi_1}{H} + \dpn{T\xi_n, 0}{H} + \sum_{j = 1}^{n-1} \dpn{T\xi_j, \xi_{j+1}}{H} \\
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&= \dpn{(T \otimes S_{n+1})(\xi_1, \cdots, \xi_n, 0), (\xi_1, \cdots, \xi_n, 0)}{H^{n+1}}
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\end{align*}
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so $w(R) \le w(T \otimes S_{n+1}) = w(T) = w(T \otimes S_n)$ by (2).
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(4): Suppose that $\text{Re}\dpn{R\xi, \xi}{H^n} \le 1$ for all $\xi \in H^n$ with $\norm{\xi}_{H^n} \le 1$.
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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
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\begin{align*}
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\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} \\
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&= \ol{\sgn(\dpn{R\xi, \xi}{H^n})} \dpn{R\xi, \xi}{H^n} = |\dpn{R\xi, \xi}{H^n}|
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\end{align*}
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As $\norm{\xi_\omega}_{H^n} = \norm{\xi}_{H^n} = 1$,
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\[
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|\dpn{R\xi, \xi}{H^n}| = \dpn{R\xi_\omega, R\xi_\omega}{H^n} = \text{Re}\dpn{R\xi_\omega, \xi_\omega}{H^n} \le 1
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\]
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Therefore $w(T) \le 1$ as well.
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\end{proof}
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+18
-96
@@ -21,6 +21,24 @@
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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 $S$.
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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 $S$.
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\end{definition}
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\end{definition}
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\begin{lemma}
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\label{lemma:operator-system-positive-approximate}
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Let $A$ be a $C^*$-algebra, $S \subset A$ be an operator system, then $(\ol{S})_+ = \ol{S_+}$.
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\end{lemma}
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\begin{proof}
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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,
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\[
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x = \frac{x + x^*}{2} = \limv{n}\frac{x_n + x_n^*}{2} \in \ol{S_{sa}}
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\]
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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
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\[
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x = \limv{n}x_n = \limv{n}[x_n + \inf(\sigma_A(x_n)) \wedge 0] \in \ol{S_+}
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\]
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by \autoref{corollary:spectrum-characterisation-iff}.
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\end{proof}
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\begin{proposition}
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\begin{proposition}
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\label{proposition:operator-system-positive-span}
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\label{proposition:operator-system-positive-span}
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@@ -131,99 +149,3 @@
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(3): By \autoref{theorem:positive-commutative-norm}, any such extension would have norm $1$.
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(3): By \autoref{theorem:positive-commutative-norm}, any such extension would have norm $1$.
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\end{proof}
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\end{proof}
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\begin{definition}[Numerical Radius]
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\label{definition:bh-numerical-radius}
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Let $H$ be a complex Hilbert space and $T \in B(H)$, then
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\[
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w(T) = \sup_{\substack{\xi \in H \\ \norm{\xi}_H = 1}}|\dpn{T\xi, \xi}{H}|
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\]
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is the \textbf{numerical radius} of $T$.
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\end{definition}
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\begin{proposition}
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\label{proposition:bh-numerical-radius-gymnastics}
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Let $H$ be a complex Hilbert space, then:
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\begin{enumerate}
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\item The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$.
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\item For any $T \in B(H)$, $w(T) \le 1$ if and only if $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$.
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\item For each $T \in B(H)_{sa}$, $w(T) = \norm{T}_{B(H)}$.
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\item For each $T \in B(H)$, $w(T) \le \norm{T}_{B(H)} \le 2w(T)$. Both inequalities are sharp in general.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): For any $S, T \in B(H)$ and $\xi \in H$ with $\norm{\xi}_H = 1$,
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\[
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|\dpn{(S + T)\xi, \xi}{H}| \le |\dpn{S\xi, \xi}{H}| + |\dpn{T\xi, \xi}{H}| \le w(S) + w(T)
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\]
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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 $\xi, \eta \in H$,
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\[
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\dpn{T\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} = 0
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\]
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so $T = 0$ and $w$ is a norm on $B(H)$.
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(2): For any $\xi \in H$ with $\norm{\xi}_H = 1$ and $\lambda \in \partial B_\complex(0, 1)$,
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\begin{align*}
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\dpn{[2 + (\lambda T) + (\lambda T)^*]\xi, \xi}{H} &= 2 + \dpn{\lambda T \xi, \xi}{H} + \dpn{\xi, \lambda T \xi}{H} \\
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&= 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H}
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\end{align*}
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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}
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|||||||
@@ -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}
|
||||||
Reference in New Issue
Block a user