I need help to find a reference to prove the the following theorem:
Theorem 6.9. Let $A$ be a $C^*$ algebra. If $a,b\in A_+$ and $a\leq b$ then $a^\alpha\leq b^\alpha$ for $0\leq \alpha\leq 1$. On the other hand if $a>1$ is fixed and $A$ is a $C^*$ algrebra in which $0\leq a\leq b$ implies $0\leq a^\alpha\leq b^\alpha$ then $A$ is commutative.
Thanks!
I do not understand why this got 3 votes to be closed. It's a fine question and it deserves to be on MSE. Anyway, I think I have seen this in Karen Strung's book, but I cannot remember correctly. Here is a proof for the first part:
The trick is to add a very small multiple of the identity so that things become invertible and thus more easy to handle. More specifically: embed $A\subset\tilde{A}$ in the unitization and denote by $1$ the unit. Now if $\varepsilon>0$, we have that $$0\leq a\leq b\leq b+\varepsilon1$$ and $b+\varepsilon1\geq\varepsilon1$ so $b+\varepsilon1$ is invertible in $\tilde{A}$. Define $E=\{t\in[0,\infty): a^t\leq (b+\varepsilon1)^t\}\subset\mathbb{R}$ and note that $0,1\in E$. Also, it follows from standard functional calculus that $E$ is closed; (hint-exercise: if $(f_n)\subset C[0,1]$ is a sequence that converges to $f$ uniformly and $0\leq x\leq 1$, then $f_n(x)\to f(x)$).
Suppose that we have now shown that $s,t\in E\implies\frac{s+t}{2}\in E$. Then we have that $\frac{0+1}{2}=\frac{1}{2}\in E$. Going on, $\frac{1+\frac{1}{2}}{2}=\frac{3}{4}\in E$ and likewise $\frac{0+\frac{1}{2}}{2}=\frac{1}{4}\in E$ and so on, so one shows that $E$ contains the set $$\{\frac{k}{2^n}:k=0,\dots,2^n-1, n\geq1\}$$ and this is dense in $[0,1]$. Using the fact that $E$ is closed, we conclude that $[0,1]\subset E$, which is what we wanted.
So all we need to do is show that $s,t\in E\implies\frac{s+t}{2}\in E$. Set $p=\frac{s+t}{2}$. It suffices to show that $$(b+\varepsilon1)^{-p/2}a^p(b+\varepsilon1)^{-p/2}\leq1$$ and since $0\leq(b+\varepsilon1)^{-p/2}a^p(b+\varepsilon1)^{-p/2}$, we have that $$(b+\varepsilon1)^{-p/2}a^p(b+\varepsilon1)^{-p/2}\leq\|(b+\varepsilon1)^{-p/2}a^p(b+\varepsilon1)^{-p/2}\|\cdot1$$ so we have to estimate the norm. Note that the element is self-adjoint, so its norm is equal to its spectral radius. But it is in general true that $r(xy)=r(yx)$, so $$\|(b+\varepsilon1)^{-p/2}a^p(b+\varepsilon1)^{-p/2}\|=r((b+\varepsilon1)^{-p/2}a^p(b+\varepsilon1)^{-p/2})=r((b+\varepsilon1)^{-p}a^p)=\|(b+\varepsilon1)^{-p}a^p\|=$$ $$=\|a^p(b+\varepsilon1)^{-p}\|=\|a^{s/2}a^{t/2}(b+\varepsilon1)^{-t/2}(b+\varepsilon1)^{-s/2}\|\leq\|(b+\varepsilon1)^{-s/2}a^{s/2}\|\cdot\|(b+\varepsilon1)^{-t/2}a^{t/2}\|$$ where we have used the spectral radius trick again implicitly in the inequality. Now we will be done if we show that $q\in E\implies\|(b+\varepsilon1)^{-q/2}a^{q/2}\|\leq1$. Indeed, $$q\in E\implies 0\leq a^q\leq(b+\varepsilon1)^q\implies0\leq(b+\varepsilon1)^{-q/2}a^q(b+\varepsilon1)^{-q/2}\leq1$$ so $\|(b+\varepsilon1)^{-q/2}a^q(b+\varepsilon1)^{-q/2}\|\leq1$ and the $C^*$-identity yields $\|(b+\varepsilon1)^{-q/2}a^{q/2}\|\leq1$, as we wanted.
We conclude that if $t\in[0,1]$, then $0\leq a^t\leq(b+\varepsilon1)^t$ for all $\varepsilon>0$. Letting $\varepsilon\to0^+$ yields $0\leq a^t\leq b^t$.