Lemma 43.4.2 (Radon-Nikodym).label Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, $\omega: M \to \complex$ be a positive linear functional, and $\xi \in H$ such that $\omega(x) \le \dpn{x\xi, \xi}{H}$ for all $x \in M_{+}$, then there exists $x' \in M'_{+}$ such that $\omega(x) = \dpn{x(x'\xi), x'\xi}{H}$ for all $x \in M$.
Proof. By the Cauchy-Schwarz inequality, for each $x, y \in M$,
so the mapping $(x, y) \mapsto \omega(y^{*}x)$ is a bounded sesquilinear form on $M\xi$. By the Riesz Representation Theorem, there exists $z \in B(\ol{M\xi})$ such that $\omega(y^{*}x) = \dpn{zx\xi, y\xi}{H}$ for all $x, y \in M$. In particular, $0 \le \omega(x^{*}x) = \dpn{zx\xi, x\xi}{H}$ for all $x \in M$, so $z \ge 0$. Moreover, for any $x, y, t \in M$,
so $z \in M'_{+}$. Finally, let $p: H \to \ol{M\xi}$ be the orthogonal projection onto $\ol{M\xi}$, and $x' = (zp)^{1/2}\in M'$, then for each $x \in M$,
$\square$
Post a Comment