Theorem 36.12.5.label Let $A$ be a unital $C^{*}$-algebra, $I \subset A$ be a left ideal, and $\cf \subset 2^{I}$ be the collection of all finite subsets of $I$, directed under inclusion. For each $F \in \cf$, let
\[p_{F} = \sum_{x \in F}x^{*}x \quad e_{F} = \paren{\frac{1}{|F|} + p_F}^{-1}p_{F}\]
then $\angles{e_F}_{F \in \cf}\subset I \cap \ol{B_A(0, 1)}$ is an increasing net of positive elements such that $xe_{F} \to x$ for all $x \in I$.
Proof, [Theorem 15.2, Zhu93]. Since $p_{F}$ is positive, $1/|F| + p_{F}$ is invertible by Proposition 36.8.2, and the continuous functional calculus implies that $0 \le e_{F} \le 1_{A}$. As $I \subset A$ is a left ideal, $e_{F} \in I \cap \ol{B_A(0, 1)}$. Now,
\begin{align*}&\sum_{x \in F}[x(e_{F} - 1_{A})]^{*}[x(e_{F} - 1_{A})] = \sum_{x \in F}(e_{F} - 1_{A})x^{*}x(e_{F} - 1_{A}) \\&= (e_{F} - 1_{A})p_{F}(e_{F} - 1_{A}) = e_{F}^{2}p_{F} - 2e_{F}p_{F} + p_{F} \\&= (1/|F| + p_{F})^{-2}p_{F} \cdot \braks{p_F^2 - 2\paren{\frac{1}{|F|} + p_F}p_F + \paren{\frac{1}{|F|} + p_F}^2}\end{align*}
where
\begin{align*}&p_{F}^{2} - 2\paren{\frac{1}{|F|} + p_F}p_{F} + \paren{\frac{1}{|F|} + p_F}^{2} \\&= - p_{F}^{2} -\frac{2p_{F}}{|F|}+ \frac{1}{|F|^{2}}+ \frac{2p_{F}}{|F|}+ p_{F}^{2} = \frac{1}{|F|^{2}}\end{align*}
so
\[\sum_{x \in F}[x(e_{F} - 1_{A})]^{*}[x(e_{F} - 1_{A})] = \frac{1}{|F|^{2}}(1/|F| + p_{F})^{-2}p_{F}\]
By (1) of Lemma 36.12.4, $\sup_{t \in [0, \infty)}(1/|F| + t)^{-2}t \le |F|/4$, the continuous functional calculus shows that
\[0 \le \sum_{x \in F}[x(e_{F} - 1_{A})]^{*}[x(e_{F} - 1_{A})] \le \frac{1}{|F|^{2}}\cdot \frac{|F|}{4}= \frac{1}{4|F|}\]
Thus for each $x \in F$, $0 \le [x(e_{F} - 1_{A})]^{*}[x(e_{F} - 1_{A})] \le 1/(4|F|)$. In particular, $\norm{xe_F - x}_{A} \le \sqrt{1/4|F|}$.
To see that $\angles{e_F}_{F \in \cf}$ is increasing, let $F, G \in \cf$ with $F \subset G$, then $p_{F} \le p_{G}$, and $(1/|F| + p_{F})^{-1}\ge (1/|F| + p_{G})^{-1}$ by Lemma 36.8.8. By (2) of Lemma 36.12.4,
\[\frac{1}{|F|}\paren{\frac{1}{|F|} + t}^{-1}\ge \frac{1}{|G|}\paren{\frac{1}{|G|} + t}^{-1}\]
for all $t \in [0, \infty)$. For each $t \in [0, \infty)$, rewrite
\begin{align*}\paren{\frac{1}{|F|} + t}^{-1}t&= 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + t}^{-1}\\ \paren{\frac{1}{|G|} + t}^{-1}t&= 1 - \frac{1}{|G|}\paren{\frac{1}{|G|} + t}^{-1}\end{align*}
Thus by the continuous functional calculus,
\begin{align*}e_{F}&= 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + p_F}^{-1}\le 1 - \frac{1}{|F|}\paren{\frac{1}{|F|} + p_G}^{-1}\\&\le 1 - \frac{1}{|G|}\paren{\frac{1}{|G|} + p_G}^{-1}= e_{G}\end{align*}
$\square$
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