Approximate identity in a $C^*$-algebra

A positive element $a$ in a $C^*$-algebra $A$ has the property that for every $f \in A^{*}_{+}$ (note that $A^{*}_{+}$ denotes the set of all positive linear functionals) and $f \neq 0$, we have $f(a) \geq 0$. Please show that

$\displaystyle \lim_{\epsilon \to 0^{+}} \| {g_{\epsilon}}(a) x - x \| = 0$ for all $x \in A$, where ${g_{\epsilon}}(t) := \dfrac{t}{t + \epsilon}$ for $\epsilon > 0$.

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What is $\epsilon$, is it a real number or an element of $A$? – Raskolnikov Dec 25 '11 at 16:40
$\epsilon$ is a positive real number. – Hezudao Dec 27 '11 at 13:52
Is this homework? Or if it is an exercise from a book, could you please provide a reference? – user16299 Jan 14 '12 at 19:28

Let $\mathcal{A}$ be a C*-algebra, $\mathcal{A}_{+}$ the set of all positive elements, and $\mathbb{B}(\mathcal{A}_{+})$ the set of all positive elements with norm $\leq 1$. Let $a \in \mathcal{A}_{+}$. Next, define a partial ordering $\preceq$ on $(0,\infty)$ as follows: $$\forall r,s \in (0,\infty): \quad r \preceq s \stackrel{\text{def}}{\iff} r \geq_{\mathbb{R}} s.$$ Clearly, $((0,\infty),\preceq)$ is a directed set.

Question: Is the net $\nu_{a}: ((0,\infty),\preceq) \to \mathbb{B}(\mathcal{A}_{+})$ defined by $\nu_{a} \stackrel{\text{def}}{=} ({g_{\epsilon}}(a))_{\epsilon \in (0,\infty)}$ an approximate identity of $\mathcal{A}$?

We shall show the following:

• If $\mathcal{A}$ is unital, then $\nu_{a}$ is an approximate identity if and only if $0 \notin \sigma(a)$.

• If $\mathcal{A}$ is non-unital, then $\nu_{a}$ may not be an approximate identity for any $a \in \mathcal{A}_{+}$.

Let $\mathcal{A}$ be a unital C*-algebra.

• Suppose that $0 \notin \sigma(a)$. Then $\sigma(a) \subseteq [\delta,\infty)$ for some $\delta > 0$, so we have \begin{align} \forall t \geq \delta, ~ \forall \epsilon > 0: \quad \left| 1 - \frac{t}{t + \epsilon} \right| &= \left| \frac{\epsilon}{t + \epsilon} \right| \\ &\leq \left| \frac{\epsilon}{\delta} \right| \\ &= \frac{\epsilon}{\delta}. \end{align} Hence, $\displaystyle \lim_{\epsilon \to 0^{+}} \| 1_{\sigma(a)} - g_{\epsilon} \|_{\sigma(a),\infty} = 0$. By the Continuous Functional Calculus, $\displaystyle \lim_{\epsilon \to 0^{+}} {g_{\epsilon}}(a) = \mathbf{1}_{\mathcal{A}}$. Therefore, $\nu_{a}$ is an approximate identity.

• Suppose that $0 \in \sigma(a)$. Observe that \begin{align} \forall \epsilon \in (0,\infty): \quad {g_{\epsilon}}(a) \cdot \mathbf{1}_{\mathcal{A}} - \mathbf{1}_{\mathcal{A}} &= {g_{\epsilon}}(a) - \mathbf{1}_{\mathcal{A}} \\ &= (g_{\epsilon} - 1_{\sigma(a)})(a). \end{align} As $\| g_{\epsilon} - 1_{\sigma(a)} \|_{\sigma(a),\infty} = 1$ for all $\epsilon > 0$, by the Continuous Functional Calculus, we do not have $\displaystyle \lim_{\epsilon \to 0^{+}} \| {g_{\epsilon}}(a) \cdot \mathbf{1}_{\mathcal{A}} - \mathbf{1}_{\mathcal{A}} \| = 0$. Therefore, $\nu_{a}$ is not an approximate identity.

Consider the net $\nu_{a}': (\mathbb{N},\leq_{\mathbb{N}}) \to \mathbb{B}(\mathcal{A}_{+})$ defined by $\nu_{a}' \stackrel{\text{def}}{=} ({g_{1/n}}(a))_{n \in \mathbb{N}}$. It is easy to verify that $\nu_{a}'$ is a subnet of $\nu_{a}$. Hence, if $\nu_{a}$ is an approximate identity, then $\nu_{a}'$ must be a countable approximate identity. Consequently, if $\mathcal{A}$ does not possess a countable approximate identity (and is thus non-unital), then $\nu_{a}$ cannot be an approximate identity for any $a \in \mathcal{A}_{+}$.

Construction of a C*-algebra that fails to possess a countable approximate identity

Consider the non-unital C*-algebra $\mathcal{A} := {C_{0}}(X)$, where $X$ is an uncountable discrete space (recall that any discrete space is locally compact and Hausdorff). For the sake of contradiction, assume that $\mathcal{A}$ possesses a countable approximate identity $(f_{\lambda})_{\lambda \in \Lambda}$. Then $$\forall \lambda \in \Lambda: \quad |\text{supp}(f_{\lambda})| = |\{ x \in X ~|~ {f_{\lambda}}(x) > 0 \}| < \aleph_{0}.$$ Hence, $\displaystyle X \setminus \bigcup_{\lambda \in \Lambda} \text{supp}(f_{\lambda}) \neq \varnothing$, so we can choose an $f \in \mathcal{A}$ that satisfies $$\varnothing \neq \text{supp}(f) \subseteq X \setminus \bigcup_{\lambda \in \Lambda} \text{supp}(f_{\lambda}).$$ We therefore obtain $f_{\lambda} f = 0_{X}$ for all $\lambda \in \Lambda$, which is impossible if $(f_{\lambda})_{\lambda \in \Lambda}$ is to be an approximate identity.

We are left with the following conjecture, which we intend to return to at a later time.

Conjecture Let $\mathcal{A}$ be any non-unital C*-algebra. Then $\nu_{a}$ is not an approximate identity for any $a \in \mathcal{A}_{+}$.

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