Why is the canonical approximate identity positive? Let $\{u_n\}$ be an approximative identity for C*-algebra A. Why the element $1-u_n$ is also positive? 
I'm asking about it because in some theorem we use something like this:
$\| a-u_n a \| ^2=\| (1-u_n)^{\frac{1}{2}}(1-u_n)^{\frac{1}{2}}a \| ^2$, but we can do it only if $1-u_n$ is positive.
Thanks for any explanations
 A: In general, $1-u_n$ is not necessarily positive, as you probably have guessed. Consider the $\Bbb C$ $*$-algebra of continuous functions on $[0,1]$ with the $\sup$-norm, and let $u_n = 1 + \dfrac 1 n$. Then
$$\| u_n f - f \| = \| f + \frac 1 n f - f \| = \frac 1 n \| f \| \to 0$$
which shows that $(u_n)_{n \ge 1}$ is an approximate identity, yet $1 - u_n = - \dfrac 1 n$, clearly not positive.
However, in theorem 3.1.1 from Gerald Murphy's "$\Bbb C$ $*$-Algebras and Operator Theory" the net given by $u_\lambda = \lambda \in \Lambda$ is not just an arbitrary approximate identity, but a very special one - so special that it is called "the canonical approximate identity" (here $\Lambda$ is the upwards-directed set of the positive elements with norm $< 1$). Remember that in a $\Bbb C$ $*$-algebra, the spectrum $\sigma(x)$ of a self-adjoint element $x$ is contained in $[- \| x \|, \| x \|]$. Since our $u_\lambda$ are taken to be positive, then (by the definition of "positivity") $\sigma (u_\lambda) \subseteq [0, \| u_\lambda \|]$.
Let us show that $1 - u_\lambda \ge 0$; this means that if $\alpha \in \sigma(1 - u_\lambda)$, then we must prove $\alpha \ge 0$. But $\alpha \in \sigma(1 - u_\lambda)$ means that $1 - u_\lambda - \alpha \cdot 1$ is not invertible (by the definition of "spectrum"), which is equivalent to saying that $u_\lambda - (1 - \alpha) \cdot 1$ is not invertible (I have just changed the sign), which implies that $1 - \alpha \in \sigma (u_\lambda) \subseteq [0, \| u_\lambda \|] \subseteq [0,1)$ (because, by choice, $\| u_\lambda \| < 1$), so that $\alpha \in (0, 1]$, which shows that $\sigma(1 - u_\lambda) \subseteq (0, 1]$, which clearly shows that $1 - u_\lambda$ is positive.
