Vrftsjtryk

Let $mathcal{A}$ be a unital $C^*$-algebra with $a,binmathcal{A}_+$ such that $|a|<|b|$. Does it follow that $a-bnotinmathcal{A}_+$?

I can find very little about the spectrum of the sum or difference of two elements. But I believe there should at least be some useful observations to be made for positive elements. For example, I did prove the following.

Let $mathcal{A}$ be a unital $C^*$-algebra with $ainmathcal{A}$ and $binmathcal{A}_+$ such that $|a|<|b|$. If $C^*(a,b)$ is abelian, then $a-bnotinmathcal{A}_+$?

Proof: Since $binmathcal{A}_+$ we have $|b|=r(b)$, so we find $xinsigma(b)$ with $x>|a|$. Then there exists a homomorphism $h$ on $C^*(a,b)$ such that $h(b)=x$. We find $h(a-b)=h(a)-h(b)leq|a|-h(b)<0$. Since $h(a-b)insigma(a-b)$, we get $a-bnotinmathcal{A}_+$.

The claim is false, as is the proof you give for the commutative case (the fallacy lies in $|b|-h(a)<0$). Consider $mathcal A=mathbb C$, $a=4$, and $b=7$.

EDIT

Note that now the claim is equivalent to "If $0leq aleq b$, then $|a|leq|b|$." This can be proven as follows: By functional calculus, we have $bleq|b|$, and thus $aleq|b|$. Applying functional calculus again gives $|a|leq|b|$.