From a80013651a77c0a725ed2b12cfc7df6ba522b295 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 6 Oct 2026 12:08:54 -0400 Subject: [PATCH] Problems 1.3 and 1.4. --- src/op/system/os.tex | 76 ++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 76 insertions(+) diff --git a/src/op/system/os.tex b/src/op/system/os.tex index 8a960a1..e1620cf 100644 --- a/src/op/system/os.tex +++ b/src/op/system/os.tex @@ -131,4 +131,80 @@ (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 $T \in B(H)$, $w(T) \le \norm{T}_{B(H)} \le 2w(T)$. 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 $w(T) \le \norm{T}_H$. The sharp case is given by $I_H$. + + On the other hand, if $w(T) \le 1$, then $2 + T + T^* = 2 + 2\text{Re}(T) \ge 0$ implies that $\norm{\text{Re}(T)}_{B(H)} \le 1$. Similarly, $2 + iT - iT^* = 2 + 2\text{Im}(T) \ge 0$ implies that $\norm{\text{Im}(T)}_{B(H)} \le 1$, 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}