Proposition 41.3.4.label Let $H$ be a complex Hilbert space and $P, Q, A \in B(H)$ such that
\[\begin{bmatrix}P&A \\ A^{*}&Q\end{bmatrix} \ge 0\]
then
- (1)
For each $\xi \in H$,
\[0 \le \dpn{(P + A + A^* + Q)\xi, \xi}{H}\le (\dpn{P\xi, \xi}{H}^{1/2}+ \dpn{Q\xi, \xi}{H}^{1/2})^{2}\] - (2)
$\normn{P + A + A^* + Q}_{B(H)}\le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^{2}$.
- (3)
If $P = Q$, then $A^{*}A \le \norm{P}_{B(H)}P$ and $\norm{A}_{B(H)}\le \norm{P}_{B(H)}$.
Proof, [Exercise 3.2, Po02]. (1): For each $\xi \in H$,
\[0 \le \angles{\begin{bmatrix}P&A \\ A^{*}&Q\end{bmatrix}\begin{bmatrix}\xi \\ \xi\end{bmatrix}, \begin{bmatrix}\xi \\ \xi\end{bmatrix}}_{H^2}= \dpn{(P + A + A^* + Q)\xi, \xi}{H}\]
and
\begin{align*}\dpn{A + A^*\xi, \xi}{H}&= 2\text{Re}\dpn{A + A^*\xi, \xi}{H}\\ |\dpn{A + A^*\xi, \xi}{H}|&\le 2 \dpn{P\xi, \xi}{H}^{1/2}\dpn{Q\xi, \xi}{H}^{1/2}\\ \dpn{(P + A + A^* + Q)\xi, \xi}{H}&\le \dpn{P\xi, \xi}{H}+ 2 \dpn{P\xi, \xi}{H}^{1/2}\dpn{Q\xi, \xi}{H}^{1/2}+ \dpn{Q\xi, \xi}{H}\\&= (\dpn{P\xi, \xi}{H}^{1/2}+ \dpn{Q\xi, \xi}{H}^{1/2})^{2} \\\end{align*}
(2): By (3) of Proposition 41.2.10 and (1),
\begin{align*}\norm{P + A + A^* + Q}_{B(H)}&= \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{(P + A + A^* + Q)\xi, \xi}{H}\\&\le \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}(\dpn{P\xi, \xi}{H}^{1/2}+ \dpn{Q\xi, \xi}{H}^{1/2})^{2} \\&\le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^{2}\end{align*}
(3): Using (2) of Proposition 41.3.2, assume without loss of generality that $P \ne 0$. For each $\xi, \eta \in H$,
\begin{align*}0&\le \angles{\begin{bmatrix}P&A \\ A^{*}&P\end{bmatrix}\begin{bmatrix}\xi \\ \eta\end{bmatrix}, \begin{bmatrix}\xi \\ \eta\end{bmatrix}}_{H^2}\\&= \dpn{P\xi, \xi}{H}+ \dpn{P\eta, \eta}{H}+ \dpn{A\eta, \xi}{H}+ \dpn{A^*\xi, \eta}{H}\end{align*}
In particular, for $\xi = -A\eta/\norm{P}_{B(H)}$,
\begin{align*}0&\le \frac{\dpn{PA\eta, A\eta}{H}}{\norm{P}_{B(H)}^{2}}+ \dpn{P\eta, \eta}{H}- \frac{\dpn{A\eta, A\eta}{H}}{\norm{P}_{B(H)}}- \frac{\dpn{A^*A\eta, \eta}{H}}{\norm{P}_{B(H)}}\\&= \frac{\dpn{PA\eta, A\eta}{H}}{\norm{P}_{B(H)}}+ \norm{P}_{B(H)}\dpn{P\eta, \eta}{H}- 2\norm{A\eta}_{H}^{2} \\&\le\norm{A\eta}_{H}^{2} + \norm{P}_{B(H)}\dpn{P\eta, \eta}{H}- 2\norm{A\eta}_{H}^{2} \\&= \norm{P}_{B(H)}\dpn{P\eta, \eta}{H}- \dpn{A^*A\eta, \eta}{H}\end{align*}
so $A^{*}A \le \norm{P}_{B(H)}P$, and $\norm{A}_{B(H)}\le \norm{P}_{B(H)}$.$\square$
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