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Let A be a unital $C^∗$ algebra. $a\in A_+$ is called strictly positive if for all positive linear functional $\phi \in A_+^*$ and $\phi \neq 0$, we have $\phi(a)>0$. Prove that $a\in A_+$ is strictly positive iff $a$ is invertible.

My attempt: if $a\in A_+$ is invertible, then there exists $x\in A$ such that $a=x^*x$ and $a^{-1}=x^{-1}(x^{-1})^*$, Using Cauchy-Schwarz inequality: $\forall \phi \in A_+^*$ $$\phi(a)\phi(a^{-1})=\phi(x^*x)\phi((x^{-1})^{**}(x^{-1})^{*})\ge |\phi(x^*(x^{-1})^*)|^2=|\phi(1)|^2=\|\phi\|^2 > 0 $$ then we have $\phi(a)\neq 0$, then $\phi(a)>0$.

But for other direction, i have no idea. I knew that $a\in A_+$ is called strictly positive also can be definted that $\overline{aAa}=A$, and from this definition i can prove that if $a$ is strictly positive, then $axa$ is invertible, then $a$ is invertible. but i want to prove $a$ is invertible from defintion which i give firstly.

I guess if $\phi(a)>0$, then $1-\phi(a) <1$, and let $\phi$ be some specific function ,then apply functional calculus, but i can't prove that.

Any idea will be appreciated!

John
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2 Answers2

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EDITED Since I originally just reproved what you already included in your statement.

Suppose $a\in A$ is strictly positive. Since $A$ is unital, the state space of $A$ is weak$^*$-compact, and it follows that $\varepsilon:=\inf\{\phi(a):\phi\text{ is a state on }A\}>0$. It then follows that $a-\varepsilon\cdot1$ is positive, from which it follows that the spectrum of $a$ is contained in $[\varepsilon,\infty)$, i.e., $0$ is not in the spectrum of $a$, i.e., $a$ is invertible.

Here's another proof of the reverse direction:

If $a\in A$ is positive and invertible, its spectrum is a compact subset of $(0,\infty)$, and thus $a-\varepsilon\cdot 1$ is positive for some $\varepsilon>0$. Thus if $\phi$ is a non-zero positive linear functional on $A$, we have $$\phi(a)=\varepsilon\phi(1)+\phi(a-\varepsilon\cdot1)\geq \varepsilon\|\phi\|+0>0.$$

Aweygan
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  • Thank you, your proof is better than mine, Great! – John Nov 28 '20 at 13:19
  • Whoops, it took me a second to realize that I just gave a different proof of what was shown in your problem statement. I have since edited and included a proof of the converse statement, please take a look at your convenience. – Aweygan Nov 28 '20 at 13:31
  • For this you need that $\phi(x)\geq 0$ for all positive funcitonals $\phi$ implies that $x$ is positive. Is there an elementary way to show this, it seems like you will end upo using functional calculus – Jonathan Hole Nov 28 '20 at 13:41
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    @JonathanHole If $\phi(x)\geq0$ for all positive $\phi$, then its image in the universal representation is a positive operator, and therefore $x$ is positive. – Aweygan Nov 28 '20 at 13:48
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For the other direction, let $a$ be a positive non-invertible element. Then $0\in \text{Spec}(a)$ and there is a unital *-homomorphism $\Phi: C^*(a) \rightarrow C(\text{Spec}(a))$ mapping $a$ to the identity function on $\text{Spec}(a)$. Here $C^*(a)$ is the $C^*$-subalgebra generated by $a$ and $C(\text{Spec}(a))$ the continuous functions on $\text{Spec}(a)$. Define a state $\phi$ on $C^*(a)$ by $\phi(x)=\Phi(x)(0)$.

$\phi$ vanishes on $a$ and we can extend it to all of $A$.

Jonathan Hole
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  • Thank you, but i don't underatand why $\phi$ is a state, even why $\phi$ is positive, as you definite? – John Nov 28 '20 at 13:18
  • If $x\in C^(a)$ is positive $\Phi(x)$ we positive because $\Phi$ is a -homomorphism. This just means that $\Phi(x)$, as a function on Spec$(a)$ only attains positive values on Spec$(a)$, in particular $\Phi(x)(0)\geq 0$. – Jonathan Hole Nov 28 '20 at 13:27
  • $\Phi(a)$ is the identity function so $\Phi(a)(0)=0$. $\phi$ is a state because $\Phi(1_A)$ is the function which is constantly $1$ on Spec$(a)$, and thus $\Phi(1_A)(0)=1$ – Jonathan Hole Nov 28 '20 at 13:30
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    oh yes, I'm unfamiliar with functional calculus, thanks for your patience! – John Nov 28 '20 at 13:51