Proposition 41.3.3 (Schur Complement Condition for Positivity).label Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$, $P^{-1/2}: P^{1/2}(H) \to \ker(P)^{\perp}$ and $Q^{-1/2}: Q^{1/2}(H) \to \ker(Q)^{\perp}$ be partial inverses of $P^{1/2}$ and $Q^{1/2}$, respectively, then the following are equivalent:
- (1)
$\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$.
- (2)
For each $\xi, \eta \in H$, $|\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}$.
- (3)
$P \ge 0$, $A(H) \subset P^{1/2}(H)$, and $Q - (P^{-1/2}A)^{*}(P^{-1/2}A) \ge 0$.
- (4)
$Q \ge 0$, $A^{*}(H) \subset Q^{1/2}(H)$, and $P - (Q^{-1/2}A)^{*}(Q^{-1/2}A) \ge 0$.
Proof, [Exercise 3.2, Po02]. (1) $\Rightarrow$ (2): If $\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$, then $\left[\begin{smallmatrix}P + \eps&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$ for all $\eps > 0$. By (P2) of Proposition 41.3.2, $P + \eps \ge 0$ and $Q - A^{*}(P + \eps)^{-1}A \ge 0$. Thus for any $\xi, \eta \in H$, by the Cauchy-Schwarz inequality,
As the above holds for all $\eps > 0$,
(2) $\Rightarrow$ (3): Let $\xi \in H$, then for all $\eta \in P^{1/2}(H)$,
By the Riesz Representation Theorem, there exists $\theta \in \ol{P^{1/2}(H)}$ such that
for all $\eta \in P^{1/2}(H)$, so
for all $\eta \in H$, and $A\xi \in P^{1/2}(H)$.
Now, for each Cauchy sequence $\seq{\xi_n}\subset H$ such that $\bracsn{P^{-1/2}A\xi_n}_{1}^{\infty} \subset \ker(P)^{\perp}$ is Cauchy, then there exists $\eta \in \ker(P)^{\perp}$ such that $P^{-1/2}A\xi_{n} \to \eta$ as $n \to \infty$. By continuity of $P^{1/2}$, $P^{1/2}P^{-1/2}A\xi_{n} \to P^{1/2}\eta$ as $n \to \infty$. Since $\eta$ is the unique element of $\ker(P)^{\perp}$ with $P^{1/2}\eta = A\xi = A\limv{n}\xi_{n}$, $\eta = P^{-1/2}A\limv{n}\xi_{n}$. By the Closed Graph Theorem, $P^{-1/2}A \in B(H)$, and the second part of the condition is well-defined.
Now, for any $\xi \in H$,
As $P^{-1/2}A\xi \in \ker(P)^{\perp} = \ol{P(H)}$,
so
(3) $\Rightarrow$ (1): For each $\xi \in H$,
and by the continuous functional calculus,
Thus $\dpn{P^{-1/2}\xi, P^{-1/2}\xi}{H}\ge \dpn{(P+\eps)^{-1/2}\xi, (P+\eps)^{-1/2}\xi}{H}$ for all $\xi \in P^{1/2}(H)$. In particular, since $A(H) \subset P^{1/2}(H)$,
so
and $\left[\begin{smallmatrix}P + \eps&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$ by (P2) of Proposition 41.3.2. As this holds for all $\eps > 0$, $\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$.$\square$
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