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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 $\sigma(a)=\sigma(e_\lambda)=[0,1]$.

Rough_Manifolds
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  • Have you studied "functional calculus" yet? I think it should follow from that, although there's probably a proof using less powerful tools. – JonathanZ Mar 22 '22 at 07:19
  • @JonathanZsupportsMonicaC: I figured it followed from functional calculus, but I cannot really see it through. Was hoping for a detailed explanation using it... – Rough_Manifolds Mar 22 '22 at 07:20
  • I think one of the facts of functional calculus is that if $f$ and $ g$ are two functions, and $f \le g$ on $\sigma(T)$, then$f(T) \le g(T)$. But I'm really weak with functional calculus - hopefully someone better at it will come along. – JonathanZ Mar 22 '22 at 07:27
  • @JonathanZsupportsMonicaC Your last comment is exactly what is needed. I fleshed down the details in an answer below. – J. De Ro Mar 22 '22 at 15:18
  • Hey, glad it helped. – JonathanZ Mar 22 '22 at 15:48

1 Answers1

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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.

J. De Ro
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  • I don't actually know the property that if $g\leq f$ on $\sigma(a)$ then $\Phi(g)\leq\phi(f)$. Does the answer I wrote up in the question work, given that I replace $\sigma(e_\lambda)=[0,1]$ by $\sigma(e_\lambda)\subseteq [0,1]$? In any case, I am not sure why $\sigma(e_\lambda)\subseteq [0,1]$ is true. I know that $\sigma(e_\lambda)\subseteq [0,\infty)$. And if the $C^\ast$-algebra were unital then we have $\rho(e_\lambda)\leq|e_\lambda|\leq1$, and hence $$ \sigma(e_\lambda)\subseteq[0,\rho(a)]\subseteq[0,1]. $$ But we don't have that the $C^\ast$-algebra is unital. – Rough_Manifolds Mar 23 '22 at 21:18
  • I suppose we pass to the unitization? – Rough_Manifolds Mar 23 '22 at 21:18
  • @D.Math A $*$-morphism preserves positive answers, and yes we pass to the unitisitation (as the first sentence in my proof tells). Also, if $a\in A$, then $\sigma(a)\subseteq {z:|z|\le |a|}$. This shows $\sigma(e_\lambda)\subseteq [0,1]$. – J. De Ro Mar 23 '22 at 21:35