Theorem 37.3.9 (Kaplansky Density Theorem).label Let $H$ be a Hilbert space, $A \subset B(H)$ be a $C^{*}$-subalgebra, and $B$ be the strong-operator closure of $A$, then:
- (1)
$\ol{B_{A_{sa}}(0, 1)}$ is strong-operator dense in $\ol{B_{B_{sa}}(0, 1)}$.
- (2)
$\bracsn{T \in \ol{B_{A}(0, 1)}|T \ge 0}$ is strong-operator dense in $\bracsn{T \in \ol{B_{B}(0, 1)}|T \ge 0}$.
- (3)
$\ol{B_{A}(0, 1)}$ is strong-operator dense in $\ol{B_{B}(0, 1)}$.
Proof, [Theorem 19.5, Zhu93]. (1): Let $T \in \ol{B_{B_{sa}}(0, 1)}$ and $\angles{T_\gamma}_{\gamma \in C}\subset A$ be a net such that $T_{\gamma} \to T$ in the weak operator topology. For each $\gamma \in C$, let $T_{\gamma}' = (T_{\gamma} + T_{\gamma}^{*})/2$, then $T_{\gamma}' \to T$ in the weak operator topology by continuity of the adjoint map in the weak operator topology.
As $A_{sa}$ is a subspace of $B(H)$, its strong and weak-operator closures coincide. Thus there exists a net $\angles{S_\gamma}_{\gamma \in C}\subset A_{sa}$ such that $S_{\gamma} \to T$ in the strong operator topology. In which case, let
then $f \in C_{0}(\real; \real)$. By Corollary 37.2.3, the mapping $S \mapsto f(S)$ is strong-operator continuous. As $\norm{T}_{B(H)}\le 1$, $\sigma_{B}(T) \subset [-1, 1]$. Thus $f(T) = T$, and $f(S_{\gamma}) \to T$ in the strong operator topology. By the continuous functional calculus, $f(S_{\gamma})$ is in the closed unit ball of $A_{sa}$ for all $\gamma \in C$. Therefore the closed unit ball of $A_{sa}$ is strong-operator dense in the closed unit ball of $B_{sa}$.
(2): Let $T \in \ol{B_{B}(0, 1)}$ with $T \ge 0$ and $\angles{T_\gamma}_{\gamma \in C}\subset A_{sa}$ be a net such that $T_{\gamma} \to T$ in the strong operator topology. Define
then $f \in C_{0}(\real; [0, \infty))$. Since $f \ge 0$, the continuous functional calculus then implies that $\norm{f(T_\gamma)}_{B(H)}\le 1$ and $f(T_{\gamma}) \ge 0$ for all $\gamma \in C$. By Corollary 37.2.3, the mapping $S \mapsto f(S)$ is strong-operator continuous. As $\norm{T}_{B(H)}\le 1$ and $T \ge 0$, $\sigma_{B}(T) \subset [0, 1]$ by Proposition 36.8.2. Thus $f(T) = T$, and $f(T_{\gamma}) \to T$ in the strong operator topology.
(3): For each $\mathcal{T}\subset B(H)$, let
then $M_{2}(B)$ is the strong-operator closure of $M_{2}(A)$ in $B(H^{2})$. For each $T \in \ol{B_{B}(0, 1)}$, let
then $T' \in \ol{B_{M_2(B)_{sa}}(0, 1)}$. By (1), there exists a net $\angles{(R_\gamma, S_\gamma, T_\gamma)}_{\gamma \in C}\subset A^{3}$ such that:
- (i)
For each $\gamma \in C$,
\[\norm{\begin{bmatrix}R_{\gamma}&T_{\gamma} \\ T^{*}_{\gamma}&S_{\gamma}\end{bmatrix}}_{B(H^2)}\le 1\]In particular, $\norm{T_\gamma}_{B(H)}\le 1$.
- (ii)
With respect to the strong operator topology on $B(H^{2})$,
\[\begin{bmatrix}R_{\gamma}&T_{\gamma} \\ T^{*}_{\gamma}&S_{\gamma}\end{bmatrix} \to T'\]
Therefore $\angles{T_\gamma}\subset \ol{B_A(0, 1)}$ is a net that converges to $T$ in the strong-operator topology.$\square$
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