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/Part 7: Operator Algebras/Chapter 41: Operator Systems

41.7 The Numerical Radius

Definition 41.7.1 (Numerical Radius).label 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 numerical radius of $T$.

Proposition 41.7.2.label Let $H$ be a complex Hilbert space, then:

  1. (1)

    The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$.

  2. (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. (3)

    For each $T \in B(H)_{sa}$, $w(T) = \norm{T}_{B(H)}$.

  4. (4)

    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 $\norm{\xi}_{H} = 1$,

\[|\dpn{(S + T)\xi, \xi}{H}| \le |\dpn{S\xi, \xi}{H}| + |\dpn{T\xi, \xi}{H}| \le w(S) + w(T)\]

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 polarisation, 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 $w(T) \le \norm{T}_{H}$. 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 polarisation 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 Young’s 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.$\square$

Proposition 41.7.3.label 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 $a \in S$, $w(\phi(a)) \le \norm{a}_{A}$.

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 Proposition 41.7.2.$\square$

Proposition 41.7.4.label Let $H$ be a complex Hilbert space, $T \in B(H)$, $R = \sum_{j = 2}^{n}E_{(j-1)j}\otimes T \in M_{n}(B(H))$, $\bracsn{e_j}_{0}^{n-1}\subset \complex^{n}$ be the canonical basis for $\complex^{n}$, and

\[S_{n}: \complex^{n} \to \complex^{n} \quad S_{n}e_{j} = e_{j+1 \mod n}\]

then:

  1. (1)

    Let $\omega = e^{2\pi i/n}$ be the primitive $n$-th root of unity, then $S$ is unitarily equivalent to $\text{diag}(\omega^{0}, \cdots, \omega^{n-1})$.

  2. (2)

    $w(T) = w(T \otimes S_{n})$.

  3. (3)

    $w(R) \le w(T \otimes S_{n})$.

  4. (4)

    $\text{Re}\dpn{R\xi, \xi}{H^n}\le 1$ for all $\xi \in H^{n}$ with $\norm{\xi}_{H^n}= 1$ if and only if $w(R) \le 1$.

Proof. (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_{n}f_{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^{k}f_{k}\end{align*}

For each $0 \le k < l \le n-1$,

\[\dpn{f_k, f_l}{\complex^n}= \sum_{j = 0}^{n-1}\omega^{-jk}\cdot \ol{\omega^{-jl}}= \sum_{j = 0}^{n-1}\omega^{-j(k-l)}= 0\]

so $\bracsn{f_k}_{0}^{n-1}\subset \complex^{n}$ is an orthogonal eigenbasis for $S_{n}$, with eigenvalues $\bracsn{\omega^k}_{0}^{n-1}$.

(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}\\ |\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 $\xi \in H$,

\[\dpn{T\xi, \xi}{H}= \dpn{(T \otimes S_n)(\xi \otimes f_0), \xi \otimes f_0}{H}\le w(T \otimes S_{n})\norm{\xi}_{H}^{2}\]

so $w(T) \le w(T \otimes S)$.

(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{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*}

so $w(R) \le w(T \otimes S_{n+1}) = w(T) = w(T \otimes S_{n})$ by (2).

(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

\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}|\end{align*}

As $\norm{\xi_\omega}_{H^n}= \norm{\xi}_{H^n}= 1$,

\[|\dpn{R\xi, \xi}{H^n}| = \dpn{R\xi_\omega, R\xi_\omega}{H^n}= \text{Re}\dpn{R\xi_\omega, \xi_\omega}{H^n}\le 1\]

Therefore $w(T) \le 1$ as well.$\square$

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