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 $.

 A: 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}_{+} $.

