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Let $A$ be a $C^\ast$-algebra and $f\colon A\to\mathbb{C}$ a positive$^[2]$ linear functional. Suppose $a\leq b^{[1]}$, is it true that $f(a)\leq f(b)$? This is used in a proof, but for approximate identities, maybe they are needed.

$^{[1]}$ Note that $a\leq b$ means $b-a$ is a positive element, i.e. $\sigma(b-a)\subset[0,\infty)$.

$^{[2]}$ Note that positive linear functional means $f(a^\ast a)\geq0$ for every $a\in A$.

Rough_Manifolds
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  • Well, since $f$ is positive, we get $f(b-a)\ge 0$, right? – Berci Mar 21 '22 at 21:29
  • @Berci: Well, I thought you get $f\left((b-a)^\ast(b-a)\right)\geq0$. If I could get what you wrote, I agree I would be done. Maybe I am missing something simple... – Rough_Manifolds Mar 21 '22 at 21:34
  • According to this definition I know: https://en.m.wikipedia.org/wiki/Positive_linear_functional , $f$ maps positive elements to nonnegative elements. As you write in the note [1], the hypothesis is that $b-a$ is a positive element. – Berci Mar 21 '22 at 21:38
  • @Berci: Hm. The definition I have at my disposal is the one in $[2]$ above. – Rough_Manifolds Mar 21 '22 at 21:41
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    Are you aware of the fact that every positive element in a $C^*$-algebra has a unique positive square root? – Aweygan Mar 22 '22 at 03:51
  • @Aweygan: Yes I am aware of this. – Rough_Manifolds Mar 22 '22 at 06:01
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    @Aweygan: I think I see where this is going: we have $b-a\geq0$ and so $b-a=x^\ast x$ for some $x\in A$. Using the definition I have: $$ f(b-a)=f(x^\ast x)\geq0. $$ Then the result follows by the linearity of $f$, correct? – Rough_Manifolds Mar 22 '22 at 07:19
  • That's correct. – Aweygan Mar 22 '22 at 15:09

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