Problems 1.3 and 1.4.

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}