Proposition 41.2.10.label Let $H$ be a complex Hilbert space, then:
- (1)
The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$.
- (2)
For any $T \in B(H)$, $w(T) \le 1$ if and only if $2 + (\lambda T) + (\lambda T)^{*} \ge 0$ for all $\lambda \in \partial B_{\complex}(0, 1)$.
- (3)
For each $T \in B(H)$, $w(T) \le \norm{T}_{B(H)}\le 2w(T)$. Both inequalities are sharp in general.
Proof. (1): For any $S, T \in B(H)$ and $\xi \in H$ with $\norm{\xi}_{H} = 1$,
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 polarisation, for any $\xi, \eta \in H$,
so $T = 0$ and $w$ is a norm on $B(H)$.
(2): For any $\xi \in H$ with $\norm{\xi}_{H} = 1$ and $\lambda \in \partial B_{\complex}(0, 1)$,
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 $\lambda = -\ol{\sgn(\dpn{T\xi, \xi}{H})}$,
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 Young’s inequality,
so the above matrix has numerical radius $1$. However, its operator norm is $2$. Thus the second inequality is also sharp.$\square$
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