37.2 Von Neumann Algebras

Proof, [Section 17, Zhu93]. (1): After rescaling, assume without loss of generality that $-I \le T \le I$ for all $T \in \cf$. Since $\cf \subset A_{sa}$, $\norm{T}_{B(H)}= [T]_{sp}\le 1$ by Theorem 36.4.3, where the spectral radius is taken with respect to $B(H)$.

Thus $\cf \subset \ol{B_{A}(0, 1)}$, and is relatively compact in the weak operator topology by the Banach-Alaoglu Theorem. As such, $\bigcap_{T \in \cf}\ol{\bracs{S \in \cf|S \ge T}}^{\text{\small WOT}}\ne \emptyset$. Let $R \in \bigcap_{T \in \cf}\ol{\bracs{S \in \cf|S \ge T}}^{\text{\small WOT}}$. Since $A$ is strong-operator closed, so is $A_{sa}$ by Proposition 37.1.3. Thus for each $T \in A_{sa}$, $\bracs{S \in A_{sa}|S \ge T}$ is closed in the weak operator topology, and $R \in A_{sa}$ with $R \ge T$ for all $T \in \cf$.

Let $S \in B(H)$ be self-adjoint such that $S \ge T$ for all $T \in \cf$, then $S \ge T$ for all $T \in \ol{\cf}^{\text{\small WOT}}$. In particular, $S \ge R$, thus $R$ is indeed the supremum of $\cf$.

Finally, let $x \in H$ and $\eps > 0$, then there exists $T \in \cf$ such that $\dpn{(R - T)x, x}{H}\le \eps$. For any $S \in \cf$ with $S \ge T$,

By the continuous functional calculus and Theorem 36.4.3,

so

for all $S \in \cf$ with $S \ge T$. As such a $T$ exists for all $\eps > 0$, $R = \sotlim_{T \in \cf}T$.

(2): Assume without loss of generality that $A$ is the smallest strong-operator closed $C^{*}$-subalgebra of $B(H)$ containing $\cf$. In which case, by separate continuity of composition in the strong operator topology, $A$ is also commutative. Thus $A_{sa}$ is a lattice by the Gelfand-Naimark Theorem, and $\cf$ may be extended into a directed family. By (1), $\sup(\cf)$ exists in $A$.

(3): Assume without loss of generality that $\mathcal{P}$ is directed. By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case,

Thus $x \in \sup(\mathcal{P})(H)$, and $\sup(\mathcal{P})(H) \supset{\ol{\bigcup_{P \in \mathcal{P}}P(H)}}$.

On the other hand, let $Q \in B(H)$ be the orthogonal projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$, then $Q$ is also an upper bound of $\mathcal{P}$. Therefore

(4): As $0 \le T \le I$, $\sigma_{B(H)}(T) \subset [0, 1]$ by the Gelfand-Naimark Theorem. For each $n \in \natp$ and $t \in [0, 1]$, let $f_{n}(t) = t^{1/n}$, then $f_{n}$ is an increasing sequence of continuous functions on $[0, 1]$. By the continuous functional calculus, $\bracsn{f_n(T)}_{1}^{\infty} = \bracsn{T^{1/n}}_{1}^{\infty}$ is an increasing sequence that lies between $0$ and $I$.

Let $n \in \natp$. By the Stone-Weierstrass Theorem, there exist polynomials $\seq{p_{n, k}}\subset \real[x]$ such that:

1. (i)

   For each $k \in \natp$, $p_{n, k}(0) = 0$.

2. (ii)

   $p_{n, k}\to f_{n}$ uniformly on $\sigma_{B(H)}(T)$.

As $A$ is a uniformly closed subalgebra of $B(H)$, $f_{n}(T) \in A$ for all $n \in \natp$.

By (1), the supremum $Q = \sup_{n \in \natp}T^{1/n}= \sotlim_{n \to \infty}T^{1/n}$ exists in $A$. Since

and $Q$ is self-adjoint, $Q$ is a projection.

For each $x \in T(H) \setminus \bracs{0}$, $\dpn{Qx, x}{H}\ge \dpn{Tx, x}{H}> 0$ because $T$ is positive. As such, $\ker(Q) \subset \ker(T)$. For any $x \in \ker(T)$, $\dpn{Qx, x}{H}= \limv{n}\dpn{T^{1/n}x, x}{H}= 0$, so $\ker(Q) \supset \ker(T)$. Since both operators are self-adjoint, $\ol{Q(H)}= \ol{T(H)}= P(H)$, and $P = Q$.

(5): Assume without loss of generality that $T \ne 0$. Let $x \in H$, then

Thus $\ker(T^{*}) = \ker(TT^{*})$, and

By (4) applied to $TT^{*}/\norm{TT^*}_{B(H)}$, the orthogonal projection onto $\ol{T(H)}$ is in $A$.

(6): Let $\mathcal{P}$ be the set of all projections in $A$, then $\mathcal{P}\subset A_{sa}$ is bounded and directed. By (3), $P = \sup_{Q \in \mathcal{P}}Q \in A$, which is the maximum projection in $A$.

Let $T \in A$, then by (5), $P$ is greater than the projection onto $\ol{T(H)}$, so $PT = T$. On the other hand, since $PT^{*} = T^{*}$, $TP = T$ as well. Therefore $P$ is the multiplicative identity in $A$.$\square$