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/Part 6: Operator Algebras/Chapter 36: $C^{*}$-Algebras

36.12 Non-Unital $C^{*}$-Algebras

Proposition 36.12.1.label Let $A$ be a non-unital $C^{*}$-algebra and $\td A$ be its unitisation, then there exists a unique norm $\norm{\cdot}_{\td A}: \td A \to [0, \infty)$ such that:

  1. (1)

    For each $x \in A$, $\norm{x}_{\td A}= \norm{x}_{A}$.

  2. (2)

    $(\td A, \norm{\cdot}_{\td A})$ is a unital $C^{*}$-algebra.

Proof, [Theorem 15.1, Zhu93]. For each $x \in A$ and $\lambda \in \complex$, let

\[\norm{x + \lambda}_{\td A}= \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{xy + \lambda y}_{A}\]

be the operator seminorm corresponding to $\td A$ acting on $A$. If $\norm{xy + \lambda y}_{A} = 0$ for all $y \in A$, then $xy = -\lambda y$ for all $y \in A$. Given that $A$ is non-unital, $\lambda = 0$. Since $A$ is a $C^{*}$-algebra, $\norm{x}_{A}^{2} = \norm{xx^*}_{A} = 0$, and $x = 0$ as well. Thus $\norm{\cdot}_{\td A}$ is indeed a norm on $\td A$.

(1): Let $x \in A \setminus \bracs{0}$, then since $A$ is a Banach algebra, $\norm{x}_{\td A}\le \norm{x}_{A}$. On the other hand, as $A$ is a $C^{*}$-algebra,

\[\norm{x}_{\td A}\ge \norm{x \cdot \frac{x^{*}}{\norm{x}_{A}}}_{A} = \frac{\norm{x}_{A}^{2}}{\norm{x}_{A}}= \norm{x}_{A}\]

(2): As $\norm{\cdot}_{\td A}$ is the operator norm corresponding to $\td A$ acting on $A$, $(\td A, \norm{\cdot}_{\td A})$ is a Banach algebra. Moreover, for any $x \in A$ and $\lambda \in \complex$,

\begin{align*}\norm{x + \lambda}_{\td A}^{2}&= \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{(x + \lambda)y}_{A}^{2} = \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{y^*(x + \lambda)^*(x + \lambda)y}_{A} \\&\le \sup_{\substack{y \in A \\ \norm{y}_A \le 1}}\norm{(x + \lambda)^*(x + \lambda)y}_{A} = \norm{(x + \lambda)^*(x+\lambda)}_{\td A}\end{align*}

so $(\td A, \norm{\cdot}_{\td A})$ is a $C^{*}$-algebra.

Finally, Corollary 36.4.5 implies that there can be at most one norm on $\td A$ making it a unital $C^{*}$-algebra, so the constructed norm is unique.$\square$

Definition 36.12.2 (Approximate Identity).label Let $A$ be a Banach algebra and $\angles{e_\beta}_{\beta \in B}\subset A$ be a net, then $\angles{e_\beta}_{\beta \in B}$ is an approximate identity of $A$ if:

  1. (1)

    For each $\beta \in B$, $\norm{e_\beta}_{A}\le 1$.

  2. (2)

    For every $x \in A$, $e_{\beta} x \to x$ and $x e_{\beta} \to x$.

Definition 36.12.3 (Increasing Approximate Identity).label Let $A$ be a $C^{*}$-algebra and $\angles{e_\beta}_{\beta \in B}\subset A$ be an approximate identity, then $\angles{e_\beta}_{\beta \in B}$ is increasing if:

  1. (1)

    For each $\beta \in B$, $e_{\beta} \ge 0$.

  2. (2)

    For each $\beta, \gamma \in B$ with $\beta \le \gamma$, $e_{\beta} \le e_{\gamma}$.

Lemma 36.12.4.label For each $m, n \in \natp$ with $m \le n$,

  1. (1)

    $\sup_{t \in [0, \infty)}\paren{1/n + t}^{-2}t \le n/4$.

  2. (2)

    For every $t \in [0, \infty)$, $m^{-1}(m^{-1}+ t)^{-1}\ge n^{-1}(n^{-1}+ t)^{-1}$.

  3. (3)

    For every $t \in [0, \infty)$, $(1/n + t)^{-1}t = 1 - (1/n + t)^{-1}/n$.

Proof. (1): For each $t \in [0, \infty)$,

\begin{align*}0 \le \paren{t - \frac{1}{n}}^{2}&= t^{2} - \frac{2t}{n}+ \frac{1}{n^{2}}= t^{2} + \frac{2t}{n}+ \frac{1}{n^{2}}- \frac{4t}{n}\\ 0&\le \paren{t + \frac{1}{n}}^{2} - \frac{4t}{n}\end{align*}

so $\frac{4t}{n}\le \paren{t + \frac{1}{n}}^{2}$ and $\frac{n}{4}\ge \paren{\frac{1}{n} + t}^{-2}t$.

(2): For each $t \in [0, \infty)$,

\[\frac{1}{m}\paren{\frac{1}{m} + t}^{-1}= \frac{1}{1 + mt}\ge \frac{1}{1 + nt}= \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}\]

(3): For every $t \in [0, \infty)$,

\begin{align*}\paren{\frac{1}{n} + t}^{-1}t&= \paren{\frac{1}{n} + t}^{-1}\braks{\paren{\frac{1}{n} + t} - \frac{1}{n}}\\&= \paren{\frac{1}{n} + t}^{-1}\paren{\frac{1}{n} + t}- \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}\\&= 1 - \frac{1}{n}\paren{\frac{1}{n} + t}^{-1}\end{align*}

$\square$

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*}

$\square$

Corollary 36.12.6.label Let $A$ be a $C^{*}$-algebra, then $A$ admits an increasing approximate identity.

Proof. Identify $A$ as a self-adjoint two-sided ideal of its unitisation $\td A$, which is a $C^{*}$-algebra by Proposition 36.12.1. Applying Theorem 36.12.5 to $A$ and $\bracs{x^*|x \in A}$ as left and right ideals, respectively, yields that the net constructed by the theorem is an increasing approximate identity for $A$.$\square$

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