Let $A$ be a $C^\ast$-algebra with approximate identity $\{e_\lambda\}_{\lambda\in\Lambda}$. The author states the following comment in the trenches of a proof:
Since $t^2\leq t$ on $[0,1]$, it follows $e^2_\lambda\leq e_\lambda$.
Intuitively it makes sense, but I am not sure why this is true. I have worked out the following:
We need the following:
(Spectral Mapping Theorem). Suppose that $a$ is a normal element in a $C^\ast$-algebra. If $f\in C(\sigma(a))$, then let $f(a)$ be the image of $a$ under the functional calculus map. Then $\sigma(f(a))=f(\sigma(a))$.
Now as $a:=e_\lambda$ is self-adjoint, it is normal. We calculate $$ \sigma(a)=\sigma(e_\lambda)=[0,1]. $$ Let $f:=t-t^2$. Note that $f\in C(\sigma(a))=C([0,1])$ and $f(a)=f(e_\lambda)=e_\lambda-e_\lambda^2$. By the Spectral Mapping Theorem: $$ \sigma(e_\lambda-e_\lambda^2)=\sigma(f(a))=f(\sigma(a))=f([0,1])\subseteq[0,\infty). $$ Hence $e_\lambda-e_\lambda^2\geq0$, which gives the result.
What I do not understand is why $\sigma(a)=\sigma(e_\lambda)=[0,1]$.
It is wrong that $\sigma(e_\lambda) = [0,1]$. However, we know that $\|e_\lambda\| \le 1$, which implies that $\sigma(e_\lambda) \subseteq [0,1]$ and this is good enough.
The idea is to use the following lemma:
Lemma: If $A$ is a $C^*$-algebra and if $a\in A$ is a positive element with $\|a\|\le 1$, then $a^2\le a$.
Proof lemma: By replacing $A$ by a unitisation of $A$ (for instance $\widetilde{A}$ or the multiplier algebra $M(A)$), we may assume that $A$ is unital.
In this case, we have a unital $*$-isomorphism $$\Phi: C(\sigma(a)) \to C^*(1,a): f \mapsto f(a)$$ with $\Phi(g) = a$ where $g: \sigma(a)\to \mathbb{C}$ is the identify function $x \mapsto x$ (this isomorphism is called the continuous functional calculus).
Consider the function $f: \sigma(a) \to \mathbb{C}: x \mapsto x^2$ and note that $g^2 \le g$ because this holds in every point of $\sigma(a) \subseteq [0,1]$. Hence, $a^2=\Phi(g)^2 \le \Phi(g)=a$ as desired.