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. (1)

    $\ol{B_{A_{sa}}(0, 1)}$ is strong-operator dense in $\ol{B_{B_{sa}}(0, 1)}$.

  2. (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. (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

\[f: \real \to \real \quad t \mapsto \begin{cases}t &t \in [-1, 1] \\ 1/t &t \in \real \setminus [-1, 1]\end{cases}\]

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

\[f: \real \to \real \quad t \mapsto \begin{cases}0 &t \le 0 \\ t &t \in [0, 1] \\ 1/t &t \ge 1\end{cases}\]

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

\[M_{2}(\mathcal{T}) = \bracs{\begin{bmatrix}Q & R \\ S & T\end{bmatrix} \bigg | Q, R, S, T \in \mathcal{T}}\]

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

\[T' = \begin{bmatrix}0&T \\ T^{*}&0\end{bmatrix}\]

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:

  1. (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, $\norm{T_\gamma}_{B(H)}\le 1$.

  2. (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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